Chapter 6 from Textbook Rs Aggarwal 2021 2022 for Class 10 MATHS

Question 1:

Answer:

(i) A(9, 3) and B(15, 11)
The given points are A(9, 3) and B(15, 11).
Then (x₁ = 9, y₁ = 3) and (x₂ = 15, y₂ = 11)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(15 - 9)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(15 - 9)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 units$

(ii) A(7, −4) and B(−5, 1)
The given points are A(7, −4) and B(−5, 1).
Then (x₁= 7, y₁ = −4) and (x₂= −5, y₂ = 1)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(- 5 - 7)}^{2} + {\{1 - (- 4)\}}^{2}} = \sqrt{{(- 5 - 7)}^{2} + {(1 + 4)}^{2}} = \sqrt{{(- 12)}^{2} + {(5)}^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13 units$

(iii) A(−6, −4) and B(9, −12)
The given points are A(−6, −4) and B(9, −12).
Then (x₁ = −6, y₁ = −4) and (x₂ = 9, y₂ = −12)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(9 - (- 6))}^{2} + {\{- 12 - (- 4)\}}^{2}} = \sqrt{{(9 + 6)}^{2} + {(- 12 + 4)}^{2}} = \sqrt{{(15)}^{2} + {(- 8)}^{2}} = \sqrt{225 + 64} = \sqrt{289} = 17 units$

(iv) A(1, −3) and B(4, −6)
The given points are A(1, −3) and B(4, −6).
Then (x₁= 1, y₁ = −3) and (x₂ = 4, y₂ = −6)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(4 - 1)}^{2} + {\{- 6 - (- 3)\}}^{2}} = \sqrt{{(4 - 1)}^{2} + {(- 6 + 3)}^{2}} = \sqrt{{(3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2} units$

(v) P(a + b, a − b) and Q(a −b, a + b)
The given points are P(a + b, a − b) and Q(a − b, a + b).
Then (x₁ = a + b, y₁ = a − b) and (x₂ = a − b, y₂ = a + b)
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{\{(a - b) - (a + b)\}}^{2} + {\{(a + b) - (a - b)\}}^{2}} = \sqrt{{(a - b - a - b)}^{2} + {(a + b - a + b)}^{2}} = \sqrt{{(- 2 b)}^{2} + {(2 b)}^{2}} = \sqrt{4 b^{2} + 4 b^{2}} = \sqrt{8 b^{2}} = \sqrt{4 \times 2 b^{2}} = 2 \sqrt{2} b units$

(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, −a sin α).
Then (x₁ = a sin α, y₁ = a cos α) and (x₂ = a cos α, y₂ = −a sin α)
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(a \cos α - a \sin α)}^{2} + {(- a \sin α - a \cos α)}^{2}} = \sqrt{(a^{2} \cos^{2} α + a^{2} \sin^{2} α - 2 a^{2} \cos α \times \sin α) + (a^{2} \sin^{2} α + a^{2} \cos^{2} α + 2 a^{2} \cos α \times \sin α)} = \sqrt{2 a^{2} \cos^{2} α + 2 a^{2} \sin^{2} α} = \sqrt{2 a^{2} (\cos^{2} α + \sin^{2} α)} = \sqrt{2 a^{2} (1)} (From the identity \cos^{2} α + \sin^{2} α = 1) = \sqrt{2 a^{2}} = \sqrt{2} a units$

Question 2:

(i) A(9, 3) and B(15, 11)
The given points are A(9, 3) and B(15, 11).
Then (x₁ = 9, y₁ = 3) and (x₂ = 15, y₂ = 11)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(15 - 9)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(15 - 9)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 units$

(ii) A(7, −4) and B(−5, 1)
The given points are A(7, −4) and B(−5, 1).
Then (x₁= 7, y₁ = −4) and (x₂= −5, y₂ = 1)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(- 5 - 7)}^{2} + {\{1 - (- 4)\}}^{2}} = \sqrt{{(- 5 - 7)}^{2} + {(1 + 4)}^{2}} = \sqrt{{(- 12)}^{2} + {(5)}^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13 units$

(iii) A(−6, −4) and B(9, −12)
The given points are A(−6, −4) and B(9, −12).
Then (x₁ = −6, y₁ = −4) and (x₂ = 9, y₂ = −12)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(9 - (- 6))}^{2} + {\{- 12 - (- 4)\}}^{2}} = \sqrt{{(9 + 6)}^{2} + {(- 12 + 4)}^{2}} = \sqrt{{(15)}^{2} + {(- 8)}^{2}} = \sqrt{225 + 64} = \sqrt{289} = 17 units$

(iv) A(1, −3) and B(4, −6)
The given points are A(1, −3) and B(4, −6).
Then (x₁= 1, y₁ = −3) and (x₂ = 4, y₂ = −6)
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(4 - 1)}^{2} + {\{- 6 - (- 3)\}}^{2}} = \sqrt{{(4 - 1)}^{2} + {(- 6 + 3)}^{2}} = \sqrt{{(3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2} units$

(v) P(a + b, a − b) and Q(a −b, a + b)
The given points are P(a + b, a − b) and Q(a − b, a + b).
Then (x₁ = a + b, y₁ = a − b) and (x₂ = a − b, y₂ = a + b)
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{\{(a - b) - (a + b)\}}^{2} + {\{(a + b) - (a - b)\}}^{2}} = \sqrt{{(a - b - a - b)}^{2} + {(a + b - a + b)}^{2}} = \sqrt{{(- 2 b)}^{2} + {(2 b)}^{2}} = \sqrt{4 b^{2} + 4 b^{2}} = \sqrt{8 b^{2}} = \sqrt{4 \times 2 b^{2}} = 2 \sqrt{2} b units$

(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)
The given points are P(a sin α, a cos α) and Q(a cos α, −a sin α).
Then (x₁ = a sin α, y₁ = a cos α) and (x₂ = a cos α, y₂ = −a sin α)
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(a \cos α - a \sin α)}^{2} + {(- a \sin α - a \cos α)}^{2}} = \sqrt{(a^{2} \cos^{2} α + a^{2} \sin^{2} α - 2 a^{2} \cos α \times \sin α) + (a^{2} \sin^{2} α + a^{2} \cos^{2} α + 2 a^{2} \cos α \times \sin α)} = \sqrt{2 a^{2} \cos^{2} α + 2 a^{2} \sin^{2} α} = \sqrt{2 a^{2} (\cos^{2} α + \sin^{2} α)} = \sqrt{2 a^{2} (1)} (From the identity \cos^{2} α + \sin^{2} α = 1) = \sqrt{2 a^{2}} = \sqrt{2} a units$

Answer:

(i) A(5, −12)
Let O(0, 0) be the origin.
$O A = \sqrt{{(5 - 0)}^{2} + {(- 12 - 0)}^{2}} = \sqrt{{(5)}^{2} + {(- 12)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 units$

(ii) B(−5, 5)
Let O(0, 0) be the origin.
$O B = \sqrt{{(- 5 - 0)}^{2} + {(5 - 0)}^{2}} = \sqrt{{(- 5)}^{2} + {(5)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = \sqrt{25 \times 2} = 5 \sqrt{2} units$

(iii) C(−4, −6)
Let O(0,0) be the origin.
$O C = \sqrt{{(- 4 - 0)}^{2} + {(- 6 - 0)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 6)}^{2}} = \sqrt{16 + 36} = \sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13} units$

Question 3:

(i) A(5, −12)
Let O(0, 0) be the origin.
$O A = \sqrt{{(5 - 0)}^{2} + {(- 12 - 0)}^{2}} = \sqrt{{(5)}^{2} + {(- 12)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 units$

(ii) B(−5, 5)
Let O(0, 0) be the origin.
$O B = \sqrt{{(- 5 - 0)}^{2} + {(5 - 0)}^{2}} = \sqrt{{(- 5)}^{2} + {(5)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = \sqrt{25 \times 2} = 5 \sqrt{2} units$

(iii) C(−4, −6)
Let O(0,0) be the origin.
$O C = \sqrt{{(- 4 - 0)}^{2} + {(- 6 - 0)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 6)}^{2}} = \sqrt{16 + 36} = \sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13} units$

Answer:

Given AB = 5 units
Therefore, (AB)² = 25 units
$\Rightarrow {(5 - x)}^{2} + {\{3 - (- 1)\}}^{2} = 25 \Rightarrow {(5 - x)}^{2} + {(3 + 1)}^{2} = 25 \Rightarrow {(5 - x)}^{2} + {(4)}^{2} = 25 \Rightarrow {(5 - x)}^{2} + 16 = 25 \Rightarrow {(5 - x)}^{2} = 25 - 16 \Rightarrow {(5 - x)}^{2} = 9 \Rightarrow (5 - x) = \pm \sqrt{9} \Rightarrow 5 - x = \pm 3 \Rightarrow 5 - x = 3 or 5 - x = - 3 \Rightarrow x = 2 or 8$
Therefore, x = 2 or 8.

Question 4:

Given AB = 5 units
Therefore, (AB)² = 25 units
$\Rightarrow {(5 - x)}^{2} + {\{3 - (- 1)\}}^{2} = 25 \Rightarrow {(5 - x)}^{2} + {(3 + 1)}^{2} = 25 \Rightarrow {(5 - x)}^{2} + {(4)}^{2} = 25 \Rightarrow {(5 - x)}^{2} + 16 = 25 \Rightarrow {(5 - x)}^{2} = 25 - 16 \Rightarrow {(5 - x)}^{2} = 9 \Rightarrow (5 - x) = \pm \sqrt{9} \Rightarrow 5 - x = \pm 3 \Rightarrow 5 - x = 3 or 5 - x = - 3 \Rightarrow x = 2 or 8$
Therefore, x = 2 or 8.

Answer:

The given points are $A (2, - 3) and B (10, y)$ .
$∴ A B = \sqrt{{(2 - 10)}^{2} + {(- 3 - y)}^{2}} = \sqrt{{(- 8)}^{2} + {(- 3 - y)}^{2}} = \sqrt{64 + 9 + y^{2} + 6 y}$
$∵ A B = 10 ∴ \sqrt{64 + 9 + y^{2} + 6 y} = 10 \Rightarrow 73 + y^{2} + 6 y = 100 (Squaring both sides) \Rightarrow y^{2} + 6 y - 27 = 0$
$\Rightarrow y^{2} + 9 y - 3 y - 27 = 0 \Rightarrow y (y + 9) - 3 (y + 9) = 0 \Rightarrow (y + 9) (y - 3) = 0 \Rightarrow y + 9 = 0 or y - 3 = 0$
$\Rightarrow y = - 9 or y = 3$
Hence, the possible values of y are $- 9 and 3$ .

Question 5:

The given points are $A (2, - 3) and B (10, y)$ .
$∴ A B = \sqrt{{(2 - 10)}^{2} + {(- 3 - y)}^{2}} = \sqrt{{(- 8)}^{2} + {(- 3 - y)}^{2}} = \sqrt{64 + 9 + y^{2} + 6 y}$
$∵ A B = 10 ∴ \sqrt{64 + 9 + y^{2} + 6 y} = 10 \Rightarrow 73 + y^{2} + 6 y = 100 (Squaring both sides) \Rightarrow y^{2} + 6 y - 27 = 0$
$\Rightarrow y^{2} + 9 y - 3 y - 27 = 0 \Rightarrow y (y + 9) - 3 (y + 9) = 0 \Rightarrow (y + 9) (y - 3) = 0 \Rightarrow y + 9 = 0 or y - 3 = 0$
$\Rightarrow y = - 9 or y = 3$
Hence, the possible values of y are $- 9 and 3$ .

Answer:

The given points are P(x, 4) and Q(9, 10).
$∴ P Q = \sqrt{{(x - 9)}^{2} + {(4 - 10)}^{2}} = \sqrt{{(x - 9)}^{2} + {(- 6)}^{2}} = \sqrt{x^{2} - 18 x + 81 + 36} = \sqrt{x^{2} - 18 x + 117}$
$∵ P Q = 10 ∴ \sqrt{x^{2} - 18 x + 117} = 10 \Rightarrow x^{2} - 18 x + 117 = 100 (Squaring both sides) \Rightarrow x^{2} - 18 x + 17 =$
$\Rightarrow x^{2} - 17 x - x + 17 = 0 \Rightarrow x (x - 17) - 1 (x - 17) = 0 \Rightarrow (x - 17) (x - 1) = 0 \Rightarrow x - 17 = 0 or x - 1 = 0$
$\Rightarrow x = 17 or x = 1$
Hence, the values of x are 1 and 17.

Question 6:

The given points are P(x, 4) and Q(9, 10).
$∴ P Q = \sqrt{{(x - 9)}^{2} + {(4 - 10)}^{2}} = \sqrt{{(x - 9)}^{2} + {(- 6)}^{2}} = \sqrt{x^{2} - 18 x + 81 + 36} = \sqrt{x^{2} - 18 x + 117}$
$∵ P Q = 10 ∴ \sqrt{x^{2} - 18 x + 117} = 10 \Rightarrow x^{2} - 18 x + 117 = 100 (Squaring both sides) \Rightarrow x^{2} - 18 x + 17 =$
$\Rightarrow x^{2} - 17 x - x + 17 = 0 \Rightarrow x (x - 17) - 1 (x - 17) = 0 \Rightarrow (x - 17) (x - 1) = 0 \Rightarrow x - 17 = 0 or x - 1 = 0$
$\Rightarrow x = 17 or x = 1$
Hence, the values of x are 1 and 17.

Answer:

As per the question
$A B = A C \Rightarrow \sqrt{{(x - 8)}^{2} + {(2 + 2)}^{2}} = \sqrt{{(x - 2)}^{2} + {(2 + 2)}^{2}}$
Squaring both sides, we get
${(x - 8)}^{2} + 4^{2} = {(x - 2)}^{2} + 4^{2} \Rightarrow x^{2} - 16 x + 64 + 16 = x^{2} + 4 - 4 x + 16 \Rightarrow 16 x - 4 x = 64 - 4 \Rightarrow x = \frac{60}{12} = 5$
Now,
$A B = \sqrt{{(x - 8)}^{2} + {(2 + 2)}^{2}} = \sqrt{{(5 - 8)}^{2} + {(2 + 2)}^{2}} (∵ x = 2) = \sqrt{{(- 3)}^{2} + {(4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
Hence, x = 5 and AB = 5 units.

Question 7:

As per the question
$A B = A C \Rightarrow \sqrt{{(x - 8)}^{2} + {(2 + 2)}^{2}} = \sqrt{{(x - 2)}^{2} + {(2 + 2)}^{2}}$
Squaring both sides, we get
${(x - 8)}^{2} + 4^{2} = {(x - 2)}^{2} + 4^{2} \Rightarrow x^{2} - 16 x + 64 + 16 = x^{2} + 4 - 4 x + 16 \Rightarrow 16 x - 4 x = 64 - 4 \Rightarrow x = \frac{60}{12} = 5$
Now,
$A B = \sqrt{{(x - 8)}^{2} + {(2 + 2)}^{2}} = \sqrt{{(5 - 8)}^{2} + {(2 + 2)}^{2}} (∵ x = 2) = \sqrt{{(- 3)}^{2} + {(4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
Hence, x = 5 and AB = 5 units.

Answer:

As per the question
$A B = A C \Rightarrow \sqrt{{(0 - 3)}^{2} + {(2 - p)}^{2}} = \sqrt{{(0 - p)}^{2} + {(2 - 5)}^{2}} \Rightarrow \sqrt{{(- 3)}^{2} + {(2 - p)}^{2}} = \sqrt{{(- p)}^{2} + {(- 3)}^{2}}$
Squaring both sides, we get
${(- 3)}^{2} + {(2 - p)}^{2} = {(- p)}^{2} + {(- 3)}^{2} \Rightarrow 9 + 4 + p^{2} - 4 p = p^{2} + 9 \Rightarrow 4 p = 4 \Rightarrow p = 1$
Now,
$A B = \sqrt{{(0 - 3)}^{2} + {(2 - p)}^{2}} = \sqrt{{(- 3)}^{2} + {(2 - 1)}^{2}} (∵ p = 1) = \sqrt{9 + 1} = \sqrt{10} units$
Hence, p = 1 and AB = $\sqrt{10}$ units.

Question 8:

As per the question
$A B = A C \Rightarrow \sqrt{{(0 - 3)}^{2} + {(2 - p)}^{2}} = \sqrt{{(0 - p)}^{2} + {(2 - 5)}^{2}} \Rightarrow \sqrt{{(- 3)}^{2} + {(2 - p)}^{2}} = \sqrt{{(- p)}^{2} + {(- 3)}^{2}}$
Squaring both sides, we get
${(- 3)}^{2} + {(2 - p)}^{2} = {(- p)}^{2} + {(- 3)}^{2} \Rightarrow 9 + 4 + p^{2} - 4 p = p^{2} + 9 \Rightarrow 4 p = 4 \Rightarrow p = 1$
Now,
$A B = \sqrt{{(0 - 3)}^{2} + {(2 - p)}^{2}} = \sqrt{{(- 3)}^{2} + {(2 - 1)}^{2}} (∵ p = 1) = \sqrt{9 + 1} = \sqrt{10} units$
Hence, p = 1 and AB = $\sqrt{10}$ units.

Answer:

Let the point on the x - axis be (x, 0).
$We have A (- 2, 5) and B (2, - 3) AX = BX {AX}^{2} = {BX}^{2} . . . . . (1) Using distance formula : d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} For AX AX = \sqrt{{(x - (- 2))}^{2} + {(0 - 5)}^{2}} {AX}^{2} = {(x + 2)}^{2} + {(- 5)}^{2} {AX}^{2} = x^{2} + 4 x + 29 . . . . . (2)$
$And BX = \sqrt{{(x - 2)}^{2} + {(0 - (- 3))}^{2}} {BX}^{2} = {(x - 2)}^{2} + {(3)}^{2} {BX}^{2} = x^{2} - 4 x + 13 . . . . . (3) From (1), (2) and (3) x^{2} + 4 x + 29 = x^{2} - 4 x + 13 x = - 2$
Point on the x - axis is ( $-$ 2, 0).

Question 9:

Let the point on the x - axis be (x, 0).
$We have A (- 2, 5) and B (2, - 3) AX = BX {AX}^{2} = {BX}^{2} . . . . . (1) Using distance formula : d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} For AX AX = \sqrt{{(x - (- 2))}^{2} + {(0 - 5)}^{2}} {AX}^{2} = {(x + 2)}^{2} + {(- 5)}^{2} {AX}^{2} = x^{2} + 4 x + 29 . . . . . (2)$
$And BX = \sqrt{{(x - 2)}^{2} + {(0 - (- 3))}^{2}} {BX}^{2} = {(x - 2)}^{2} + {(3)}^{2} {BX}^{2} = x^{2} - 4 x + 13 . . . . . (3) From (1), (2) and (3) x^{2} + 4 x + 29 = x^{2} - 4 x + 13 x = - 2$
Point on the x - axis is ( $-$ 2, 0).

Answer:

Let P (x, 0) be the point on the x-axis. Then as per the question, we have
$A P = 10 \Rightarrow \sqrt{{(x - 11)}^{2} + {(0 + 8)}^{2}} = 10 \Rightarrow {(x - 11)}^{2} + 8^{2} = 100 (Squaring both sides) \Rightarrow {(x - 11)}^{2} = 100 - 64 = 36$
$\Rightarrow x - 11 = \pm 6 \Rightarrow x = 11 \pm 6 \Rightarrow x = 11 - 6, 11 + 6 \Rightarrow x = 5, 17$
Hence, the points on the x-axis are (5, 0) and (17, 0).

Question 10:

Let P (x, 0) be the point on the x-axis. Then as per the question, we have
$A P = 10 \Rightarrow \sqrt{{(x - 11)}^{2} + {(0 + 8)}^{2}} = 10 \Rightarrow {(x - 11)}^{2} + 8^{2} = 100 (Squaring both sides) \Rightarrow {(x - 11)}^{2} = 100 - 64 = 36$
$\Rightarrow x - 11 = \pm 6 \Rightarrow x = 11 \pm 6 \Rightarrow x = 11 - 6, 11 + 6 \Rightarrow x = 5, 17$
Hence, the points on the x-axis are (5, 0) and (17, 0).

Answer:

Let P (0, y) be a point on the y-axis. Then as per the question, we have
$A P = B P \Rightarrow \sqrt{{(0 - 6)}^{2} + {(y - 5)}^{2}} = \sqrt{{(0 + 4)}^{2} + {(y - 3)}^{2}} \Rightarrow \sqrt{{(6)}^{2} + {(y - 5)}^{2}} = \sqrt{{(4)}^{2} + {(y - 3)}^{2}} \Rightarrow {(6)}^{2} + {(y - 5)}^{2} = {(4)}^{2} + {(y - 3)}^{2} (Squaring both sides)$
$\Rightarrow 36 + y^{2} - 10 y + 25 = 16 + y^{2} - 6 y + 9 \Rightarrow 4 y = 36 \Rightarrow y = 9$
Hence, the point on the y-axis is (0, 9).

Question 11:

Let P (0, y) be a point on the y-axis. Then as per the question, we have
$A P = B P \Rightarrow \sqrt{{(0 - 6)}^{2} + {(y - 5)}^{2}} = \sqrt{{(0 + 4)}^{2} + {(y - 3)}^{2}} \Rightarrow \sqrt{{(6)}^{2} + {(y - 5)}^{2}} = \sqrt{{(4)}^{2} + {(y - 3)}^{2}} \Rightarrow {(6)}^{2} + {(y - 5)}^{2} = {(4)}^{2} + {(y - 3)}^{2} (Squaring both sides)$
$\Rightarrow 36 + y^{2} - 10 y + 25 = 16 + y^{2} - 6 y + 9 \Rightarrow 4 y = 36 \Rightarrow y = 9$
Hence, the point on the y-axis is (0, 9).

Answer:

As per the question, we have
$A P = B P \Rightarrow \sqrt{{(x - 5)}^{2} + {(y - 1)}^{2}} = \sqrt{{(x + 1)}^{2} + {(y - 5)}^{2}} \Rightarrow {(x - 5)}^{2} + {(y - 1)}^{2} = {(x + 1)}^{2} + {(y - 5)}^{2} (Squaring both sides) \Rightarrow x^{2} - 10 x + 25 + y^{2} - 2 y + 1 = x^{2} + 2 x + 1 + y^{2} - 10 y + 25$
$\Rightarrow - 10 x - 2 y = 2 x - 10 y \Rightarrow 8 y = 12 x \Rightarrow 3 x = 2 y$
Hence, 3x = 2y.

Question 12:

As per the question, we have
$A P = B P \Rightarrow \sqrt{{(x - 5)}^{2} + {(y - 1)}^{2}} = \sqrt{{(x + 1)}^{2} + {(y - 5)}^{2}} \Rightarrow {(x - 5)}^{2} + {(y - 1)}^{2} = {(x + 1)}^{2} + {(y - 5)}^{2} (Squaring both sides) \Rightarrow x^{2} - 10 x + 25 + y^{2} - 2 y + 1 = x^{2} + 2 x + 1 + y^{2} - 10 y + 25$
$\Rightarrow - 10 x - 2 y = 2 x - 10 y \Rightarrow 8 y = 12 x \Rightarrow 3 x = 2 y$
Hence, 3x = 2y.

Answer:

The given points are A(6, −1) and B(2, 3). The point P(x, y) is equidistant from the points A and B. So, PA = PB.
Also, (PA)² = (PB)^{2
$\Rightarrow {(6 - x)}^{2} + {(- 1 - y)}^{2} = {(2 - x)}^{2} + {(3 - y)}^{2} \Rightarrow x^{2} - 12 x + 36 + y^{2} + 2 y + 1 = x^{2} - 4 x + 4 + y^{2} - 6 y + 9 \Rightarrow x^{2} + y^{2} - 12 x + 2 y + 37 = x^{2} + y^{2} - 4 x - 6 y + 13 \Rightarrow x^{2} + y^{2} - 12 x + 2 y - x^{2} - y^{2} + 4 x + 6 y = 13 - 37 \Rightarrow - 8 x + 8 y = - 24 \Rightarrow - 8 (x - y) = - 24 \Rightarrow x - y = \frac{- 24}{- 8} \Rightarrow x - y = 3$}
Hence proved.

Question 13:

The given points are A(6, −1) and B(2, 3). The point P(x, y) is equidistant from the points A and B. So, PA = PB.
Also, (PA)² = (PB)^{2
$\Rightarrow {(6 - x)}^{2} + {(- 1 - y)}^{2} = {(2 - x)}^{2} + {(3 - y)}^{2} \Rightarrow x^{2} - 12 x + 36 + y^{2} + 2 y + 1 = x^{2} - 4 x + 4 + y^{2} - 6 y + 9 \Rightarrow x^{2} + y^{2} - 12 x + 2 y + 37 = x^{2} + y^{2} - 4 x - 6 y + 13 \Rightarrow x^{2} + y^{2} - 12 x + 2 y - x^{2} - y^{2} + 4 x + 6 y = 13 - 37 \Rightarrow - 8 x + 8 y = - 24 \Rightarrow - 8 (x - y) = - 24 \Rightarrow x - y = \frac{- 24}{- 8} \Rightarrow x - y = 3$}
Hence proved.

Answer:

Let the required point be P(x, y). Then AP = BP = CP
That is, (AP)²=(BP)²=(CP)²
This means (AP)²=(BP)²
$\Rightarrow {(x - 5)}^{2} + {(y - 3)}^{2} = {(x - 5)}^{2} + {(y + 5)}^{2} \Rightarrow x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = x^{2} - 10 x + 25 + y^{2} + 10 y + 25 \Rightarrow x^{2} - 10 x + y^{2} - 6 y + 34 = x^{2} - 10 x + y^{2} + 10 y + 50 \Rightarrow x^{2} - 10 x + y^{2} - 6 y - x^{2} + 10 x - y^{2} - 10 y = 50 - 34 \Rightarrow - 16 y = 16 \Rightarrow y = - \frac{16}{16} = - 1 And {(B P)}^{2} = {(C P)}^{2} \Rightarrow {(x - 5)}^{2} + {(y + 5)}^{2} = {(x - 1)}^{2} + {(y + 5)}^{2} \Rightarrow x^{2} - 10 x + 25 + y^{2} + 10 y + 25 = x^{2} - 2 x + 1 + y^{2} + 10 y + 25 \Rightarrow x^{2} - 10 x + y^{2} + 10 y + 50 = x^{2} - 2 x + y^{2} + 10 y + 26 \Rightarrow x^{2} - 10 x + y^{2} + 10 y - x^{2} + 2 x - y^{2} - 10 y = 26 - 50 \Rightarrow - 8 x = - 24 \Rightarrow x = \frac{- 24}{- 8} = 3$
Hence, the required point is (3, −1).

Question 14:

Let the required point be P(x, y). Then AP = BP = CP
That is, (AP)²=(BP)²=(CP)²
This means (AP)²=(BP)²
$\Rightarrow {(x - 5)}^{2} + {(y - 3)}^{2} = {(x - 5)}^{2} + {(y + 5)}^{2} \Rightarrow x^{2} - 10 x + 25 + y^{2} - 6 y + 9 = x^{2} - 10 x + 25 + y^{2} + 10 y + 25 \Rightarrow x^{2} - 10 x + y^{2} - 6 y + 34 = x^{2} - 10 x + y^{2} + 10 y + 50 \Rightarrow x^{2} - 10 x + y^{2} - 6 y - x^{2} + 10 x - y^{2} - 10 y = 50 - 34 \Rightarrow - 16 y = 16 \Rightarrow y = - \frac{16}{16} = - 1 And {(B P)}^{2} = {(C P)}^{2} \Rightarrow {(x - 5)}^{2} + {(y + 5)}^{2} = {(x - 1)}^{2} + {(y + 5)}^{2} \Rightarrow x^{2} - 10 x + 25 + y^{2} + 10 y + 25 = x^{2} - 2 x + 1 + y^{2} + 10 y + 25 \Rightarrow x^{2} - 10 x + y^{2} + 10 y + 50 = x^{2} - 2 x + y^{2} + 10 y + 26 \Rightarrow x^{2} - 10 x + y^{2} + 10 y - x^{2} + 2 x - y^{2} - 10 y = 26 - 50 \Rightarrow - 8 x = - 24 \Rightarrow x = \frac{- 24}{- 8} = 3$
Hence, the required point is (3, −1).

Answer:

Given, the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3).
Then OA = OB
Also (OA)² = (OB)²
$\Rightarrow {(4 - 2)}^{2} + {(3 - 3)}^{2} = {(x - 2)}^{2} + {(5 - 3)}^{2} \Rightarrow {(2)}^{2} + {(0)}^{2} = {(x - 2)}^{2} + {(2)}^{2} \Rightarrow 4 = {(x - 2)}^{2} + 4 \Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$
Therefore, x = 2.

Question 15:

Given, the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3).
Then OA = OB
Also (OA)² = (OB)²
$\Rightarrow {(4 - 2)}^{2} + {(3 - 3)}^{2} = {(x - 2)}^{2} + {(5 - 3)}^{2} \Rightarrow {(2)}^{2} + {(0)}^{2} = {(x - 2)}^{2} + {(2)}^{2} \Rightarrow 4 = {(x - 2)}^{2} + 4 \Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$
Therefore, x = 2.

Answer:

As per the question, we have
$A C = B C \Rightarrow \sqrt{{(- 2 - 3)}^{2} + {(3 + 1)}^{2}} = \sqrt{{(- 2 - x)}^{2} + {(3 - 8)}^{2}} \Rightarrow \sqrt{{(5)}^{2} + {(4)}^{2}} = \sqrt{{(x + 2)}^{2} + {(- 5)}^{2}} \Rightarrow 25 + 16 = {(x + 2)}^{2} + 25 (Squaring both sides)$
$\Rightarrow 25 + 16 = {(x + 2)}^{2} + 25 \Rightarrow {(x + 2)}^{2} = 16 \Rightarrow x + 2 = \pm 4 \Rightarrow x = - 2 \pm 4 = - 2 - 4, - 2 + 4 = - 6, 2$
Now
$B C = \sqrt{{(- 2 - x)}^{2} + {(3 - 8)}^{2}} = \sqrt{{(- 2 - 2)}^{2} + (- 5)} = \sqrt{16 + 25} = \sqrt{41} units$
Hence, x = 2 or −6 and $B C = \sqrt{41} units$ .

Question 16:

As per the question, we have
$A C = B C \Rightarrow \sqrt{{(- 2 - 3)}^{2} + {(3 + 1)}^{2}} = \sqrt{{(- 2 - x)}^{2} + {(3 - 8)}^{2}} \Rightarrow \sqrt{{(5)}^{2} + {(4)}^{2}} = \sqrt{{(x + 2)}^{2} + {(- 5)}^{2}} \Rightarrow 25 + 16 = {(x + 2)}^{2} + 25 (Squaring both sides)$
$\Rightarrow 25 + 16 = {(x + 2)}^{2} + 25 \Rightarrow {(x + 2)}^{2} = 16 \Rightarrow x + 2 = \pm 4 \Rightarrow x = - 2 \pm 4 = - 2 - 4, - 2 + 4 = - 6, 2$
Now
$B C = \sqrt{{(- 2 - x)}^{2} + {(3 - 8)}^{2}} = \sqrt{{(- 2 - 2)}^{2} + (- 5)} = \sqrt{16 + 25} = \sqrt{41} units$
Hence, x = 2 or −6 and $B C = \sqrt{41} units$ .

Answer:

As per the question, we have
$A P = B P \Rightarrow \sqrt{{(2 + 2)}^{2} + {(2 - k)}^{2}} = \sqrt{{(2 + 2 k)}^{2} + {(2 + 3)}^{2}} \Rightarrow \sqrt{{(4)}^{2} + {(2 - k)}^{2}} = \sqrt{{(2 + 2 k)}^{2} + {(5)}^{2}} \Rightarrow 16 + 4 + k^{2} - 4 k = 4 + 4 k^{2} + 8 k + 25 (Squaring both sides)$
$\Rightarrow k^{2} + 4 k + 3 = 0 \Rightarrow (k + 1) (k + 3) = 0 \Rightarrow k = - 3, - 1$
Now for $k = - 1$
$A P = \sqrt{{(2 + 2)}^{2} + {(2 - k)}^{2}} = \sqrt{{(4)}^{2} + {(2 + 1)}^{2}} = \sqrt{16 + 9} = 5 units$
For $k = - 3$
$A P = \sqrt{{(2 + 2)}^{2} + {(2 - k)}^{2}} = \sqrt{{(4)}^{2} + {(2 + 3)}^{2}} = \sqrt{16 + 25} = \sqrt{41} units$
Hence, $k = - 1, - 3; A P = 5 units for k = - 1 and A P = \sqrt{41} units for k = - 3$ .

Question 17:

As per the question, we have
$A P = B P \Rightarrow \sqrt{{(2 + 2)}^{2} + {(2 - k)}^{2}} = \sqrt{{(2 + 2 k)}^{2} + {(2 + 3)}^{2}} \Rightarrow \sqrt{{(4)}^{2} + {(2 - k)}^{2}} = \sqrt{{(2 + 2 k)}^{2} + {(5)}^{2}} \Rightarrow 16 + 4 + k^{2} - 4 k = 4 + 4 k^{2} + 8 k + 25 (Squaring both sides)$
$\Rightarrow k^{2} + 4 k + 3 = 0 \Rightarrow (k + 1) (k + 3) = 0 \Rightarrow k = - 3, - 1$
Now for $k = - 1$
$A P = \sqrt{{(2 + 2)}^{2} + {(2 - k)}^{2}} = \sqrt{{(4)}^{2} + {(2 + 1)}^{2}} = \sqrt{16 + 9} = 5 units$
For $k = - 3$
$A P = \sqrt{{(2 + 2)}^{2} + {(2 - k)}^{2}} = \sqrt{{(4)}^{2} + {(2 + 3)}^{2}} = \sqrt{16 + 25} = \sqrt{41} units$
Hence, $k = - 1, - 3; A P = 5 units for k = - 1 and A P = \sqrt{41} units for k = - 3$ .

Answer:

(i) As per the question, we have
$\sqrt{{(x - a - b)}^{2} + {(y - b + a)}^{2}} = \sqrt{{(x - a + b)}^{2} + {(y - a - b)}^{2}} \Rightarrow {(x - a - b)}^{2} + {(y - b + a)}^{2} = {(x - a + b)}^{2} + {(y - a - b)}^{2} (Squaring both sides) \Rightarrow x^{2} + {(a + b)}^{2} - 2 x (a + b) + y^{2} + {(a - b)}^{2} - 2 y (a - b) = x^{2} + {(a - b)}^{2} - 2 x (a - b) + y^{2} + {(a + b)}^{2} - 2 y (a + b) \Rightarrow - x (a + b) - y (a - b) = - x (a - b) - y (a + b)$
$\Rightarrow - x a - x b - a y + b y = - x a + b x - y a - b y \Rightarrow b y = b x$
Hence, bx = ay.

(ii)
As per the question, we have
$AP = BP \Rightarrow \sqrt{{(x - 5)}^{2} + {(y - 1)}^{2}} = \sqrt{{(x + 1)}^{2} + {(y - 5)}^{2}} \Rightarrow {(x - 5)}^{2} + {(y - 1)}^{2} = {(x + 1)}^{2} + {(y - 5)}^{2} (Squaring both sides) \Rightarrow x^{2} - 10 x + 25 + y^{2} - 2 y + 1 = x^{2} + 2 x + 1 + y^{2} - 10 y + 25$
$\Rightarrow - 10 x - 2 y = 2 x - 10 y \Rightarrow 8 y = 12 x \Rightarrow 3 x = 2 y$
Hence, 3x = 2y.

Question 18:

(i) As per the question, we have
$\sqrt{{(x - a - b)}^{2} + {(y - b + a)}^{2}} = \sqrt{{(x - a + b)}^{2} + {(y - a - b)}^{2}} \Rightarrow {(x - a - b)}^{2} + {(y - b + a)}^{2} = {(x - a + b)}^{2} + {(y - a - b)}^{2} (Squaring both sides) \Rightarrow x^{2} + {(a + b)}^{2} - 2 x (a + b) + y^{2} + {(a - b)}^{2} - 2 y (a - b) = x^{2} + {(a - b)}^{2} - 2 x (a - b) + y^{2} + {(a + b)}^{2} - 2 y (a + b) \Rightarrow - x (a + b) - y (a - b) = - x (a - b) - y (a + b)$
$\Rightarrow - x a - x b - a y + b y = - x a + b x - y a - b y \Rightarrow b y = b x$
Hence, bx = ay.

(ii)
As per the question, we have
$AP = BP \Rightarrow \sqrt{{(x - 5)}^{2} + {(y - 1)}^{2}} = \sqrt{{(x + 1)}^{2} + {(y - 5)}^{2}} \Rightarrow {(x - 5)}^{2} + {(y - 1)}^{2} = {(x + 1)}^{2} + {(y - 5)}^{2} (Squaring both sides) \Rightarrow x^{2} - 10 x + 25 + y^{2} - 2 y + 1 = x^{2} + 2 x + 1 + y^{2} - 10 y + 25$
$\Rightarrow - 10 x - 2 y = 2 x - 10 y \Rightarrow 8 y = 12 x \Rightarrow 3 x = 2 y$
Hence, 3x = 2y.

Answer:

(i)
Let A(1, −1), B(5, 2) and C(9, 5) be the given points. Then
$A B = \sqrt{{(5 - 1)}^{2} + {(2 + 1)}^{2}} = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5 units B C = \sqrt{{(9 - 5)}^{2} + {(5 - 2)}^{2}} = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5 units A C = \sqrt{{(9 - 1)}^{2} + {(5 + 1)}^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{100} = 10 units$
$∴ A B + B C = (5 + 5) units = 10 units = A C$
Hence, the given points are collinear.

(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then
$A B = \sqrt{{(0 - 6)}^{2} + {(1 - 9)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} = \sqrt{100} = 10 units B C = \sqrt{{(- 6 - 0)}^{2} + {(- 7 - 1)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} = \sqrt{100} = 10 units A C = \sqrt{{(- 6 - 6)}^{2} + {(- 7 - 9)}^{2}} = \sqrt{{(- 12)}^{2} + {(- 16)}^{2}} = \sqrt{400} = 20 units$
$∴ A B + B C = (10 + 10) units = 20 units = A C$
Hence, the given points are collinear.

(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then
$A B = \sqrt{{(2 + 1)}^{2} + {(3 + 1)}^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = \sqrt{25} = 5 units B C = \sqrt{{(8 - 2)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = \sqrt{100} = 10 units A C = \sqrt{{(8 + 1)}^{2} + {(11 + 1)}^{2}} = \sqrt{{(9)}^{2} + {(12)}^{2}} = \sqrt{225} = 15 units$
$∴ A B + B C = (5 + 10) units = 15 units = A C$
Hence, the given points are collinear.

(iv)
Let A(−2, 5), B(0, 1) and C(2, −3) be the given points. Then
$A B = \sqrt{{(0 + 2)}^{2} + {(1 - 5)}^{2}} = \sqrt{{(2)}^{2} + {(- 4)}^{2}} = \sqrt{20} = 2 \sqrt{5} units B C = \sqrt{{(2 - 0)}^{2} + {(- 3 - 1)}^{2}} = \sqrt{{(2)}^{2} + {(- 4)}^{2}} = \sqrt{20} = 2 \sqrt{5} units A C = \sqrt{{(2 + 2)}^{2} + {(- 3 - 5)}^{2}} = \sqrt{{(4)}^{2} + {(- 8)}^{2}} = \sqrt{80} = 4 \sqrt{5} units$
$∴ A B + B C = (2 \sqrt{5} + 2 \sqrt{5}) units = 4 \sqrt{5} units = A C$
Hence, the given points are collinear.

Question 19:

(i)
Let A(1, −1), B(5, 2) and C(9, 5) be the given points. Then
$A B = \sqrt{{(5 - 1)}^{2} + {(2 + 1)}^{2}} = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5 units B C = \sqrt{{(9 - 5)}^{2} + {(5 - 2)}^{2}} = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5 units A C = \sqrt{{(9 - 1)}^{2} + {(5 + 1)}^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{100} = 10 units$
$∴ A B + B C = (5 + 5) units = 10 units = A C$
Hence, the given points are collinear.

(ii)
Let A(6, 9), B(0, 1) and C(−6, −7) be the given points. Then
$A B = \sqrt{{(0 - 6)}^{2} + {(1 - 9)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} = \sqrt{100} = 10 units B C = \sqrt{{(- 6 - 0)}^{2} + {(- 7 - 1)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} = \sqrt{100} = 10 units A C = \sqrt{{(- 6 - 6)}^{2} + {(- 7 - 9)}^{2}} = \sqrt{{(- 12)}^{2} + {(- 16)}^{2}} = \sqrt{400} = 20 units$
$∴ A B + B C = (10 + 10) units = 20 units = A C$
Hence, the given points are collinear.

(iii)
Let A(−1, −1), B(2, 3) and C(8, 11) be the given points. Then
$A B = \sqrt{{(2 + 1)}^{2} + {(3 + 1)}^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = \sqrt{25} = 5 units B C = \sqrt{{(8 - 2)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = \sqrt{100} = 10 units A C = \sqrt{{(8 + 1)}^{2} + {(11 + 1)}^{2}} = \sqrt{{(9)}^{2} + {(12)}^{2}} = \sqrt{225} = 15 units$
$∴ A B + B C = (5 + 10) units = 15 units = A C$
Hence, the given points are collinear.

(iv)
Let A(−2, 5), B(0, 1) and C(2, −3) be the given points. Then
$A B = \sqrt{{(0 + 2)}^{2} + {(1 - 5)}^{2}} = \sqrt{{(2)}^{2} + {(- 4)}^{2}} = \sqrt{20} = 2 \sqrt{5} units B C = \sqrt{{(2 - 0)}^{2} + {(- 3 - 1)}^{2}} = \sqrt{{(2)}^{2} + {(- 4)}^{2}} = \sqrt{20} = 2 \sqrt{5} units A C = \sqrt{{(2 + 2)}^{2} + {(- 3 - 5)}^{2}} = \sqrt{{(4)}^{2} + {(- 8)}^{2}} = \sqrt{80} = 4 \sqrt{5} units$
$∴ A B + B C = (2 \sqrt{5} + 2 \sqrt{5}) units = 4 \sqrt{5} units = A C$
Hence, the given points are collinear.

Answer:

The given points are A(7, 10), B(−2, 5) and C(3, −4).
$A B = \sqrt{{(- 2 - 7)}^{2} + {(5 - 10)}^{2}} = \sqrt{{(- 9)}^{2} + {(- 5)}^{2}} = \sqrt{81 + 25} = \sqrt{106} B C = \sqrt{{(3 - (- 2))}^{2} + {(- 4 - 5)}^{2}} = \sqrt{{(5)}^{2} + {(- 9)}^{2}} = \sqrt{25 + 81} = \sqrt{106} A C = \sqrt{{(3 - 7)}^{2} + {(- 4 - 10)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 14)}^{2}} = \sqrt{16 + 196} = \sqrt{212}$
Since, AB and BC are equal, they form the vertices of an isosceles triangle.
Also, (AB)²+(BC)² = ${(\sqrt{106})}^{2} + {(\sqrt{106})}^{2} = 212$
and (AC)² = ${(\sqrt{212})}^{2}$ = 212
Thus, (AB)²+(BC)² = (AC)²
This show that $∆ A B C$ is right- angled at B.
Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.

Question 20:

The given points are A(7, 10), B(−2, 5) and C(3, −4).
$A B = \sqrt{{(- 2 - 7)}^{2} + {(5 - 10)}^{2}} = \sqrt{{(- 9)}^{2} + {(- 5)}^{2}} = \sqrt{81 + 25} = \sqrt{106} B C = \sqrt{{(3 - (- 2))}^{2} + {(- 4 - 5)}^{2}} = \sqrt{{(5)}^{2} + {(- 9)}^{2}} = \sqrt{25 + 81} = \sqrt{106} A C = \sqrt{{(3 - 7)}^{2} + {(- 4 - 10)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 14)}^{2}} = \sqrt{16 + 196} = \sqrt{212}$
Since, AB and BC are equal, they form the vertices of an isosceles triangle.
Also, (AB)²+(BC)² = ${(\sqrt{106})}^{2} + {(\sqrt{106})}^{2} = 212$
and (AC)² = ${(\sqrt{212})}^{2}$ = 212
Thus, (AB)²+(BC)² = (AC)²
This show that $∆ A B C$ is right- angled at B.
Therefore, the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right-angled triangle.

Answer:

The given points are A(3, 0), B(6, 4) and C(− 1, 3). Now
$A B = \sqrt{{(3 - 6)}^{2} + {(0 - 4)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 B C = \sqrt{{(6 + 1)}^{2} + {(4 - 3)}^{2}} = \sqrt{{(7)}^{2} + {(1)}^{2}} = \sqrt{49 + 1} = \sqrt{50} = 5 \sqrt{2}$
$A C = \sqrt{{(3 + 1)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$
$∵ A B = A C and A B^{2} + A C^{2} = B C^{2}$
Therefore, A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.

Question 21:

The given points are A(3, 0), B(6, 4) and C(− 1, 3). Now
$A B = \sqrt{{(3 - 6)}^{2} + {(0 - 4)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 B C = \sqrt{{(6 + 1)}^{2} + {(4 - 3)}^{2}} = \sqrt{{(7)}^{2} + {(1)}^{2}} = \sqrt{49 + 1} = \sqrt{50} = 5 \sqrt{2}$
$A C = \sqrt{{(3 + 1)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$
$∵ A B = A C and A B^{2} + A C^{2} = B C^{2}$
Therefore, A(3, 0), B(6, 4) and C(− 1, 3) are the vertices of an isosceles right triangle.

Answer:

$∵ ∠ B = 90^{\circ} ∴ A C^{2} = A B^{2} + B C^{2} \Rightarrow {(5 + 2)}^{2} + {(2 - t)}^{2} = {(5 - 2)}^{2} + {(2 + 2)}^{2} + {(2 + 2)}^{2} + {(- 2 - t)}^{2} \Rightarrow {(7)}^{2} + {(t - 2)}^{2} = {(3)}^{2} + {(4)}^{2} + {(4)}^{2} + {(t + 2)}^{2}$
$\Rightarrow 49 + t^{2} - 4 t + 4 = 9 + 16 + 16 + t^{2} + 4 t + 4 \Rightarrow 8 - 4 t = 4 t \Rightarrow 8 t = 8 \Rightarrow t = 1$
Hence, t = 1.

Question 22:

$∵ ∠ B = 90^{\circ} ∴ A C^{2} = A B^{2} + B C^{2} \Rightarrow {(5 + 2)}^{2} + {(2 - t)}^{2} = {(5 - 2)}^{2} + {(2 + 2)}^{2} + {(2 + 2)}^{2} + {(- 2 - t)}^{2} \Rightarrow {(7)}^{2} + {(t - 2)}^{2} = {(3)}^{2} + {(4)}^{2} + {(4)}^{2} + {(t + 2)}^{2}$
$\Rightarrow 49 + t^{2} - 4 t + 4 = 9 + 16 + 16 + t^{2} + 4 t + 4 \Rightarrow 8 - 4 t = 4 t \Rightarrow 8 t = 8 \Rightarrow t = 1$
Hence, t = 1.

Answer:

The given points are A(2, 4), B(2, 6) and $C (2 + \sqrt{3}, 5)$ . Now
$A B = \sqrt{{(2 - 2)}^{2} + {(4 - 6)}^{2}} = \sqrt{{(0)}^{2} + {(- 2)}^{2}} = \sqrt{0 + 4} = 2 B C = \sqrt{{(2 - 2 - \sqrt{3})}^{2} + {(6 - 5)}^{2}} = \sqrt{{(- \sqrt{3})}^{2} + {(1)}^{2}} = \sqrt{3 + 1} = 2$
$A C = \sqrt{{(2 - 2 - \sqrt{3})}^{2} + {(4 - 5)}^{2}} = \sqrt{{(- \sqrt{3})}^{2} + {(- 1)}^{2}} = \sqrt{3 + 1} = 2$
Hence, the points A(2, 4), B(2, 6) and $C (2 + \sqrt{3}, 5)$ are the vertices of an equilateral triangle.

Question 23:

The given points are A(2, 4), B(2, 6) and $C (2 + \sqrt{3}, 5)$ . Now
$A B = \sqrt{{(2 - 2)}^{2} + {(4 - 6)}^{2}} = \sqrt{{(0)}^{2} + {(- 2)}^{2}} = \sqrt{0 + 4} = 2 B C = \sqrt{{(2 - 2 - \sqrt{3})}^{2} + {(6 - 5)}^{2}} = \sqrt{{(- \sqrt{3})}^{2} + {(1)}^{2}} = \sqrt{3 + 1} = 2$
$A C = \sqrt{{(2 - 2 - \sqrt{3})}^{2} + {(4 - 5)}^{2}} = \sqrt{{(- \sqrt{3})}^{2} + {(- 1)}^{2}} = \sqrt{3 + 1} = 2$
Hence, the points A(2, 4), B(2, 6) and $C (2 + \sqrt{3}, 5)$ are the vertices of an equilateral triangle.

Answer:

Let the given points be A(− 3, − 3), B(3, 3) and C $(- 3 \sqrt{3}, 3 \sqrt{3})$ . Now
$A B = \sqrt{{(- 3 - 3)}^{2} + {(- 3 - 3)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 6)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 6 \sqrt{2} B C = \sqrt{{(3 + 3 \sqrt{3})}^{2} + {(3 - 3 \sqrt{3})}^{2}} = \sqrt{9 + 27 + 18 \sqrt{3} + 9 + 27 - 18 \sqrt{3}} = \sqrt{72} = 6 \sqrt{2}$
$A C = \sqrt{{(- 3 + 3 \sqrt{3})}^{2} + {(- 3 - 3 \sqrt{3})}^{2}} = \sqrt{{(3 - 3 \sqrt{3})}^{2} + {(3 + 3 \sqrt{3})}^{2}} = \sqrt{9 + 27 - 18 \sqrt{3} + 9 + 27 + 18 \sqrt{3}} = \sqrt{72} = 6 \sqrt{2}$
Hence, the given points are the vertices of an equilateral triangle.

Question 24:

Let the given points be A(− 3, − 3), B(3, 3) and C $(- 3 \sqrt{3}, 3 \sqrt{3})$ . Now
$A B = \sqrt{{(- 3 - 3)}^{2} + {(- 3 - 3)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 6)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 6 \sqrt{2} B C = \sqrt{{(3 + 3 \sqrt{3})}^{2} + {(3 - 3 \sqrt{3})}^{2}} = \sqrt{9 + 27 + 18 \sqrt{3} + 9 + 27 - 18 \sqrt{3}} = \sqrt{72} = 6 \sqrt{2}$
$A C = \sqrt{{(- 3 + 3 \sqrt{3})}^{2} + {(- 3 - 3 \sqrt{3})}^{2}} = \sqrt{{(3 - 3 \sqrt{3})}^{2} + {(3 + 3 \sqrt{3})}^{2}} = \sqrt{9 + 27 - 18 \sqrt{3} + 9 + 27 + 18 \sqrt{3}} = \sqrt{72} = 6 \sqrt{2}$
Hence, the given points are the vertices of an equilateral triangle.

Answer:

Let the given points be A(−5, 6) B(3, 0) and C(9, 8).
$A B = \sqrt{{(3 - (- 5))}^{2} + {(0 - 6)}^{2}} = \sqrt{{(8)}^{2} + {(- 6)}^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 units B C = \sqrt{{(9 - 3)}^{2} + {(8 - 0)}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 units A C = \sqrt{{(9 - (- 5))}^{2} + {(8 - 6)}^{2}} = \sqrt{{(14)}^{2} + {(2)}^{2}} = \sqrt{196 + 4} = \sqrt{200} = 10 \sqrt{2} units Therefore, A B = B C = 10 units$

Also, (AB)²+(BC)² = ${(10)}^{2} + {(10)}^{2} = 200$
and (AC)² = ${(10 \sqrt{2})}^{2} = 200$
Thus, (AB)²+(BC)² = (AC)²
This show that $∆ A B C$ is right- angled at B.
Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also, $a rea of a triangle = \frac{1}{2} \times base \times height$
$If A B is the height and B C is the base, Area = \frac{1}{2} \times 10 \times 10 = 50 square units$

Question 25:

Let the given points be A(−5, 6) B(3, 0) and C(9, 8).
$A B = \sqrt{{(3 - (- 5))}^{2} + {(0 - 6)}^{2}} = \sqrt{{(8)}^{2} + {(- 6)}^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 units B C = \sqrt{{(9 - 3)}^{2} + {(8 - 0)}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 units A C = \sqrt{{(9 - (- 5))}^{2} + {(8 - 6)}^{2}} = \sqrt{{(14)}^{2} + {(2)}^{2}} = \sqrt{196 + 4} = \sqrt{200} = 10 \sqrt{2} units Therefore, A B = B C = 10 units$

Also, (AB)²+(BC)² = ${(10)}^{2} + {(10)}^{2} = 200$
and (AC)² = ${(10 \sqrt{2})}^{2} = 200$
Thus, (AB)²+(BC)² = (AC)²
This show that $∆ A B C$ is right- angled at B.
Therefore, the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle.
Also, $a rea of a triangle = \frac{1}{2} \times base \times height$
$If A B is the height and B C is the base, Area = \frac{1}{2} \times 10 \times 10 = 50 square units$

Answer:

The given points are O(0, 0) A(3, $\sqrt{3}$ ) and B(3, − $\sqrt{3}$ ).
$O A = \sqrt{{(3 - 0)}^{2} + {\{(\sqrt{3}) - 0\}}^{2}} = \sqrt{{(3)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{9 + 3} = \sqrt{12} = 2 \sqrt{3} units A B = \sqrt{{(3 - 3)}^{2} + {(- \sqrt{3} - \sqrt{3})}^{2}} = \sqrt{0 + {(2 \sqrt{3})}^{2}} = \sqrt{4 (3)} = \sqrt{12} = 2 \sqrt{3} units O B = \sqrt{{(3 - 0)}^{2} + {(- \sqrt{3} - 0)}^{2}} = \sqrt{{(3)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{9 + 3} = \sqrt{12} = 2 \sqrt{3} units Therefore, O A = A B = O B = 2 \sqrt{3} units$
Thus, the points O(0, 0) A(3, $\sqrt{3}$ )and B(3, − $\sqrt{3}$ ) are the vertices of an equilateral triangle.
Also, the area of the triangle OAB = $\frac{\sqrt{3}}{4} \times {(side)}^{2}$
$\frac{\sqrt{3}}{4} \times {(2 \sqrt{3})}^{2} = \frac{\sqrt{3}}{4} \times 12 = 3 \sqrt{3} square units$

Question 26:

The given points are O(0, 0) A(3, $\sqrt{3}$ ) and B(3, − $\sqrt{3}$ ).
$O A = \sqrt{{(3 - 0)}^{2} + {\{(\sqrt{3}) - 0\}}^{2}} = \sqrt{{(3)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{9 + 3} = \sqrt{12} = 2 \sqrt{3} units A B = \sqrt{{(3 - 3)}^{2} + {(- \sqrt{3} - \sqrt{3})}^{2}} = \sqrt{0 + {(2 \sqrt{3})}^{2}} = \sqrt{4 (3)} = \sqrt{12} = 2 \sqrt{3} units O B = \sqrt{{(3 - 0)}^{2} + {(- \sqrt{3} - 0)}^{2}} = \sqrt{{(3)}^{2} + {(\sqrt{3})}^{2}} = \sqrt{9 + 3} = \sqrt{12} = 2 \sqrt{3} units Therefore, O A = A B = O B = 2 \sqrt{3} units$
Thus, the points O(0, 0) A(3, $\sqrt{3}$ )and B(3, − $\sqrt{3}$ ) are the vertices of an equilateral triangle.
Also, the area of the triangle OAB = $\frac{\sqrt{3}}{4} \times {(side)}^{2}$
$\frac{\sqrt{3}}{4} \times {(2 \sqrt{3})}^{2} = \frac{\sqrt{3}}{4} \times 12 = 3 \sqrt{3} square units$

Answer:

(i) The given points are A(3, 2), B(0, 5), C(−3, 2) and D(0, −1).
$A B = \sqrt{{(0 - 3)}^{2} + {(5 - 2)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units B C = \sqrt{{(- 3 - 0)}^{2} + {(2 - 5)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units C D = \sqrt{{(0 + 3)}^{2} + {(- 1 - 2)}^{2}} = \sqrt{{(3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units D A = \sqrt{{(0 - 3)}^{2} + {(- 1 - 2)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Therefore, A B = B C = C D = D A = 3 \sqrt{2} units Also, A C = \sqrt{{(- 3 - 3)}^{2} + {(2 - 2)}^{2}} = \sqrt{{(- 6)}^{2} + {(0)}^{2}} = \sqrt{36} = 6 units B D = \sqrt{{(0 - 0)}^{2} + {(- 1 - 5)}^{2}} = \sqrt{{(0)}^{2} + {(- 6)}^{2}} = \sqrt{36} = 6 units Thus, diagonal A C = diagonal B D$
Therefore, the given points form a square.

(ii) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6).
$A B = \sqrt{{(2 - 6)}^{2} + {(1 - 2)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 1)}^{2}} = \sqrt{16 + 1} = \sqrt{17} units B C = \sqrt{{(1 - 2)}^{2} + {(5 - 1)}^{2}} = \sqrt{{(- 1)}^{2} + {(- 4)}^{2}} = \sqrt{1 + 16} = \sqrt{17} units C D = \sqrt{{(5 - 1)}^{2} + {(6 - 5)}^{2}} = \sqrt{{(4)}^{2} + {(1)}^{2}} = \sqrt{16 + 1} = \sqrt{17} units D A = \sqrt{{(5 - 6)}^{2} + {(6 - 2)}^{2}} = \sqrt{{(1)}^{2} + {(4)}^{2}} = \sqrt{1 + 16} = \sqrt{17} units Therefore A B = B C = C D = D A = \sqrt{17} units Also A C = \sqrt{{(1 - 6)}^{2} + {(5 - 2)}^{2}} = \sqrt{{(- 5)}^{2} + {(3)}^{2}} = \sqrt{25 + 9} = \sqrt{34} units B D = \sqrt{{(5 - 2)}^{2} + {(6 - 1)}^{2}} = \sqrt{{(3)}^{2} + {(5)}^{2}} = \sqrt{9 + 25} = \sqrt{34} units Thus, diagonal A C = diagonal B D$
Therefore, the given points form a square.

(iii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1).
$P Q = \sqrt{{(3 - 0)}^{2} + {(1 + 2)}^{2}} = \sqrt{{(3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Q R = \sqrt{{(0 - 3)}^{2} + {(4 - 1)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units R S = \sqrt{{(- 3 - 0)}^{2} + {(1 - 4)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units S P = \sqrt{{(- 3 - 0)}^{2} + {(1 + 2)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Therefore, P Q = Q R = R S = S P = 3 \sqrt{2} units Also, P R = \sqrt{{(0 - 0)}^{2} + {(4 + 2)}^{2}} = \sqrt{{(0)}^{2} + {(6)}^{2}} = \sqrt{36} = 6 units Q S = \sqrt{{(- 3 - 3)}^{2} + {(1 - 1)}^{2}} = \sqrt{{(- 6)}^{2} + {(0)}^{2}} = \sqrt{36} = 6 units Thus, diagonal P R = diagonal Q S$
Therefore, the given points form a square.

Question 27:

(i) The given points are A(3, 2), B(0, 5), C(−3, 2) and D(0, −1).
$A B = \sqrt{{(0 - 3)}^{2} + {(5 - 2)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units B C = \sqrt{{(- 3 - 0)}^{2} + {(2 - 5)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units C D = \sqrt{{(0 + 3)}^{2} + {(- 1 - 2)}^{2}} = \sqrt{{(3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units D A = \sqrt{{(0 - 3)}^{2} + {(- 1 - 2)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Therefore, A B = B C = C D = D A = 3 \sqrt{2} units Also, A C = \sqrt{{(- 3 - 3)}^{2} + {(2 - 2)}^{2}} = \sqrt{{(- 6)}^{2} + {(0)}^{2}} = \sqrt{36} = 6 units B D = \sqrt{{(0 - 0)}^{2} + {(- 1 - 5)}^{2}} = \sqrt{{(0)}^{2} + {(- 6)}^{2}} = \sqrt{36} = 6 units Thus, diagonal A C = diagonal B D$
Therefore, the given points form a square.

(ii) The given points are A(6, 2), B(2, 1), C(1, 5) and D(5, 6).
$A B = \sqrt{{(2 - 6)}^{2} + {(1 - 2)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 1)}^{2}} = \sqrt{16 + 1} = \sqrt{17} units B C = \sqrt{{(1 - 2)}^{2} + {(5 - 1)}^{2}} = \sqrt{{(- 1)}^{2} + {(- 4)}^{2}} = \sqrt{1 + 16} = \sqrt{17} units C D = \sqrt{{(5 - 1)}^{2} + {(6 - 5)}^{2}} = \sqrt{{(4)}^{2} + {(1)}^{2}} = \sqrt{16 + 1} = \sqrt{17} units D A = \sqrt{{(5 - 6)}^{2} + {(6 - 2)}^{2}} = \sqrt{{(1)}^{2} + {(4)}^{2}} = \sqrt{1 + 16} = \sqrt{17} units Therefore A B = B C = C D = D A = \sqrt{17} units Also A C = \sqrt{{(1 - 6)}^{2} + {(5 - 2)}^{2}} = \sqrt{{(- 5)}^{2} + {(3)}^{2}} = \sqrt{25 + 9} = \sqrt{34} units B D = \sqrt{{(5 - 2)}^{2} + {(6 - 1)}^{2}} = \sqrt{{(3)}^{2} + {(5)}^{2}} = \sqrt{9 + 25} = \sqrt{34} units Thus, diagonal A C = diagonal B D$
Therefore, the given points form a square.

(iii) The given points are P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1).
$P Q = \sqrt{{(3 - 0)}^{2} + {(1 + 2)}^{2}} = \sqrt{{(3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Q R = \sqrt{{(0 - 3)}^{2} + {(4 - 1)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units R S = \sqrt{{(- 3 - 0)}^{2} + {(1 - 4)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units S P = \sqrt{{(- 3 - 0)}^{2} + {(1 + 2)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Therefore, P Q = Q R = R S = S P = 3 \sqrt{2} units Also, P R = \sqrt{{(0 - 0)}^{2} + {(4 + 2)}^{2}} = \sqrt{{(0)}^{2} + {(6)}^{2}} = \sqrt{36} = 6 units Q S = \sqrt{{(- 3 - 3)}^{2} + {(1 - 1)}^{2}} = \sqrt{{(- 6)}^{2} + {(0)}^{2}} = \sqrt{36} = 6 units Thus, diagonal P R = diagonal Q S$
Therefore, the given points form a square.

Answer:

The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).
$A B = \sqrt{{(- 5 + 3)}^{2} + {(- 5 - 2)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 7)}^{2}} = \sqrt{4 + 49} = \sqrt{53} units . B C = \sqrt{{(2 + 5)}^{2} + {(- 3 + 5)}^{2}} = \sqrt{{(7)}^{2} + {(2)}^{2}} = \sqrt{49 + 4} = \sqrt{53} units . C D = \sqrt{{(4 - 2)}^{2} + {(4 + 3)}^{2}} = \sqrt{{(2)}^{2} + {(7)}^{2}} = \sqrt{4 + 49} = \sqrt{53} units . D A = \sqrt{{(4 + 3)}^{2} + {(4 - 2)}^{2}} = \sqrt{{(7)}^{2} + {(2)}^{2}} = \sqrt{49 + 4} = \sqrt{53} units . Therefore, A B = B C = C D = D A = \sqrt{53} units A C = \sqrt{{(2 + 3)}^{2} + {(- 3 - 2)}^{2}} = \sqrt{{(5)}^{2} + {(- 5)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = \sqrt{25 \times 2} = 5 \sqrt{2} units B D = \sqrt{{(4 + 5)}^{2} + {(4 + 5)}^{2}} = \sqrt{{(9)}^{2} + {(9)}^{2}} = \sqrt{81 + 81} = \sqrt{162} = \sqrt{81 \times 2} = 9 \sqrt{2} units Thus, diagonal A C is not equal to diagonal B D .$
Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.
$Area of a rhombus = \frac{1}{2} \times (product of diagonals) = \frac{1}{2} \times (5 \sqrt{2}) \times (9 \sqrt{2}) = \frac{45 (2)}{2} = 45 square units$

Question 28:

The given points are A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4).
$A B = \sqrt{{(- 5 + 3)}^{2} + {(- 5 - 2)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 7)}^{2}} = \sqrt{4 + 49} = \sqrt{53} units . B C = \sqrt{{(2 + 5)}^{2} + {(- 3 + 5)}^{2}} = \sqrt{{(7)}^{2} + {(2)}^{2}} = \sqrt{49 + 4} = \sqrt{53} units . C D = \sqrt{{(4 - 2)}^{2} + {(4 + 3)}^{2}} = \sqrt{{(2)}^{2} + {(7)}^{2}} = \sqrt{4 + 49} = \sqrt{53} units . D A = \sqrt{{(4 + 3)}^{2} + {(4 - 2)}^{2}} = \sqrt{{(7)}^{2} + {(2)}^{2}} = \sqrt{49 + 4} = \sqrt{53} units . Therefore, A B = B C = C D = D A = \sqrt{53} units A C = \sqrt{{(2 + 3)}^{2} + {(- 3 - 2)}^{2}} = \sqrt{{(5)}^{2} + {(- 5)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = \sqrt{25 \times 2} = 5 \sqrt{2} units B D = \sqrt{{(4 + 5)}^{2} + {(4 + 5)}^{2}} = \sqrt{{(9)}^{2} + {(9)}^{2}} = \sqrt{81 + 81} = \sqrt{162} = \sqrt{81 \times 2} = 9 \sqrt{2} units Thus, diagonal A C is not equal to diagonal B D .$
Therefore, ABCD is a quadrilateral with equal sides and unequal diagonals.
Hence, ABCD is a rhombus.
$Area of a rhombus = \frac{1}{2} \times (product of diagonals) = \frac{1}{2} \times (5 \sqrt{2}) \times (9 \sqrt{2}) = \frac{45 (2)}{2} = 45 square units$

Answer:

The given points are A(3, 0), B(4, 5), C(− 1, 4) and D(− 2, − 1).
$A B = \sqrt{{(3 - 4)}^{2} + {(0 - 5)}^{2}} = \sqrt{{(- 1)}^{2} + {(- 5)}^{2}} = \sqrt{1 + 25} = \sqrt{26}$
$B C = \sqrt{{(4 + 1)}^{2} + {(5 - 4)}^{2}} = \sqrt{{(5)}^{2} + {(1)}^{2}} = \sqrt{25 + 1} = \sqrt{26}$
$C D = \sqrt{{(- 1 + 2)}^{2} + {(4 + 1)}^{2}} = \sqrt{{(1)}^{2} + {(5)}^{2}} = \sqrt{1 + 25} = \sqrt{26} A D = \sqrt{{(3 + 2)}^{2} + {(0 + 1)}^{2}} = \sqrt{{(5)}^{2} + {(1)}^{2}} = \sqrt{25 + 1} = \sqrt{26}$
$A C = \sqrt{{(3 + 1)}^{2} + {(0 - 4)}^{2}} = \sqrt{{(4)}^{2} + {(- 4)}^{2}} = \sqrt{16 + 16} = 4 \sqrt{2} B D = \sqrt{{(4 + 2)}^{2} + {(5 + 1)}^{2}} = \sqrt{{(6)}^{2} + {(6)}^{2}} = \sqrt{36 + 36} = 6 \sqrt{2}$
$∵ A B = B C = C D = A D = 6 \sqrt{2} and A C \neq B D$
Therefore, the given points are the vertices of a rhombus.
$Area (▱ A B C D) = \frac{1}{2} \times A C \times B D = \frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2} = 24 sq . units$
Hence, the area of the rhombus is 24 sq. units.

Question 29:

The given points are A(3, 0), B(4, 5), C(− 1, 4) and D(− 2, − 1).
$A B = \sqrt{{(3 - 4)}^{2} + {(0 - 5)}^{2}} = \sqrt{{(- 1)}^{2} + {(- 5)}^{2}} = \sqrt{1 + 25} = \sqrt{26}$
$B C = \sqrt{{(4 + 1)}^{2} + {(5 - 4)}^{2}} = \sqrt{{(5)}^{2} + {(1)}^{2}} = \sqrt{25 + 1} = \sqrt{26}$
$C D = \sqrt{{(- 1 + 2)}^{2} + {(4 + 1)}^{2}} = \sqrt{{(1)}^{2} + {(5)}^{2}} = \sqrt{1 + 25} = \sqrt{26} A D = \sqrt{{(3 + 2)}^{2} + {(0 + 1)}^{2}} = \sqrt{{(5)}^{2} + {(1)}^{2}} = \sqrt{25 + 1} = \sqrt{26}$
$A C = \sqrt{{(3 + 1)}^{2} + {(0 - 4)}^{2}} = \sqrt{{(4)}^{2} + {(- 4)}^{2}} = \sqrt{16 + 16} = 4 \sqrt{2} B D = \sqrt{{(4 + 2)}^{2} + {(5 + 1)}^{2}} = \sqrt{{(6)}^{2} + {(6)}^{2}} = \sqrt{36 + 36} = 6 \sqrt{2}$
$∵ A B = B C = C D = A D = 6 \sqrt{2} and A C \neq B D$
Therefore, the given points are the vertices of a rhombus.
$Area (▱ A B C D) = \frac{1}{2} \times A C \times B D = \frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2} = 24 sq . units$
Hence, the area of the rhombus is 24 sq. units.

Answer:

The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3).
$A B = \sqrt{{(6 - 8)}^{2} + {(1 - 2)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 1)}^{2}} = \sqrt{4 + 1} = \sqrt{5}$
$B C = \sqrt{{(8 - 9)}^{2} + {(2 - 4)}^{2}} = \sqrt{{(- 1)}^{2} + {(- 2)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
$C D = \sqrt{{(9 - 7)}^{2} + {(4 - 3)}^{2}} = \sqrt{{(2)}^{2} + {(1)}^{2}} = \sqrt{4 + 1} = \sqrt{5} A D = \sqrt{{(7 - 6)}^{2} + {(3 - 1)}^{2}} = \sqrt{{(1)}^{2} + {(2)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
$A C = \sqrt{{(6 - 9)}^{2} + {(1 - 4)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = 3 \sqrt{2} B D = \sqrt{{(8 - 7)}^{2} + {(2 - 3)}^{2}} = \sqrt{{(1)}^{2} + {(- 1)}^{2}} = \sqrt{1 + 1} = \sqrt{2}$
$∵ A B = B C = C D = A D = \sqrt{5} and A C \neq B D$
Therefore, the given points are the vertices of a rhombus. Now
$Area (▱ A B C D) = \frac{1}{2} \times A C \times B D = \frac{1}{2} \times 3 \sqrt{2} \times \sqrt{2} = 3 sq . units$
Hence, the area of the rhombus is 3 sq. units.

Question 30:

The given points are A(6, 1), B(8, 2), C(9, 4) and D(7, 3).
$A B = \sqrt{{(6 - 8)}^{2} + {(1 - 2)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 1)}^{2}} = \sqrt{4 + 1} = \sqrt{5}$
$B C = \sqrt{{(8 - 9)}^{2} + {(2 - 4)}^{2}} = \sqrt{{(- 1)}^{2} + {(- 2)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
$C D = \sqrt{{(9 - 7)}^{2} + {(4 - 3)}^{2}} = \sqrt{{(2)}^{2} + {(1)}^{2}} = \sqrt{4 + 1} = \sqrt{5} A D = \sqrt{{(7 - 6)}^{2} + {(3 - 1)}^{2}} = \sqrt{{(1)}^{2} + {(2)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
$A C = \sqrt{{(6 - 9)}^{2} + {(1 - 4)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 3)}^{2}} = \sqrt{9 + 9} = 3 \sqrt{2} B D = \sqrt{{(8 - 7)}^{2} + {(2 - 3)}^{2}} = \sqrt{{(1)}^{2} + {(- 1)}^{2}} = \sqrt{1 + 1} = \sqrt{2}$
$∵ A B = B C = C D = A D = \sqrt{5} and A C \neq B D$
Therefore, the given points are the vertices of a rhombus. Now
$Area (▱ A B C D) = \frac{1}{2} \times A C \times B D = \frac{1}{2} \times 3 \sqrt{2} \times \sqrt{2} = 3 sq . units$
Hence, the area of the rhombus is 3 sq. units.

Answer:

The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3).
$A B = \sqrt{{(5 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{{(3)}^{2} + {(1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} units B C = \sqrt{{(6 - 5)}^{2} + {(4 - 2)}^{2}} = \sqrt{{(1)}^{2} + {(2)}^{2}} = \sqrt{1 + 4} = \sqrt{5} units C D = \sqrt{{(3 - 6)}^{2} + {(3 - 4)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} units A D = \sqrt{{(3 - 2)}^{2} + {(3 - 1)}^{2}} = \sqrt{{(1)}^{2} + {(2)}^{2}} = \sqrt{1 + 4} = \sqrt{5} units Thus, A B = C D = \sqrt{10} units and B C = A D = \sqrt{5} units So, quadrilateral A B C D is a parallelogram Also, A C = \sqrt{{(6 - 2)}^{2} + {(4 - 1)}^{2}} = \sqrt{{(4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units B D = \sqrt{{(3 - 5)}^{2} + {(3 - 2)}^{2}} = \sqrt{{(- 2)}^{2} + {(1)}^{2}} = \sqrt{4 + 1} = \sqrt{5} units$
But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.

Question 31:

The given points are A(2, 1), B(5, 2), C(6, 4) and D(3, 3).
$A B = \sqrt{{(5 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{{(3)}^{2} + {(1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} units B C = \sqrt{{(6 - 5)}^{2} + {(4 - 2)}^{2}} = \sqrt{{(1)}^{2} + {(2)}^{2}} = \sqrt{1 + 4} = \sqrt{5} units C D = \sqrt{{(3 - 6)}^{2} + {(3 - 4)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} units A D = \sqrt{{(3 - 2)}^{2} + {(3 - 1)}^{2}} = \sqrt{{(1)}^{2} + {(2)}^{2}} = \sqrt{1 + 4} = \sqrt{5} units Thus, A B = C D = \sqrt{10} units and B C = A D = \sqrt{5} units So, quadrilateral A B C D is a parallelogram Also, A C = \sqrt{{(6 - 2)}^{2} + {(4 - 1)}^{2}} = \sqrt{{(4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units B D = \sqrt{{(3 - 5)}^{2} + {(3 - 2)}^{2}} = \sqrt{{(- 2)}^{2} + {(1)}^{2}} = \sqrt{4 + 1} = \sqrt{5} units$
But diagonal AC is not equal to diagonal BD.
Hence, the given points do not form a rectangle.

Answer:

The given vertices are A(1, 2), B(4, 3), C(6, 6) and D(3, 5).
$A B = \sqrt{{(1 - 4)}^{2} + {(2 - 3)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 1)}^{2}} = \sqrt{9 + 1} = \sqrt{10}$
$B C = \sqrt{{(4 - 6)}^{2} + {(3 - 6)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 3)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$
$C D = \sqrt{{(6 - 3)}^{2} + {(6 - 5)}^{2}} = \sqrt{{(3)}^{2} + {(1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} A D = \sqrt{{(1 - 3)}^{2} + {(2 - 5)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 3)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$
$∵ A B = C D = \sqrt{10} units and B C = A D = \sqrt{13} units$
Therefore, ABCD is a parallelogram. Now
$A C = \sqrt{{(1 - 6)}^{2} + {(2 - 6)}^{2}} = \sqrt{{(- 5)}^{2} + {(- 4)}^{2}} = \sqrt{25 + 16} = \sqrt{41} B D = \sqrt{{(4 - 3)}^{2} + {(3 - 5)}^{2}} = \sqrt{{(1)}^{2} + {(- 2)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.

Question 32:

The given vertices are A(1, 2), B(4, 3), C(6, 6) and D(3, 5).
$A B = \sqrt{{(1 - 4)}^{2} + {(2 - 3)}^{2}} = \sqrt{{(- 3)}^{2} + {(- 1)}^{2}} = \sqrt{9 + 1} = \sqrt{10}$
$B C = \sqrt{{(4 - 6)}^{2} + {(3 - 6)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 3)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$
$C D = \sqrt{{(6 - 3)}^{2} + {(6 - 5)}^{2}} = \sqrt{{(3)}^{2} + {(1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} A D = \sqrt{{(1 - 3)}^{2} + {(2 - 5)}^{2}} = \sqrt{{(- 2)}^{2} + {(- 3)}^{2}} = \sqrt{4 + 9} = \sqrt{13}$
$∵ A B = C D = \sqrt{10} units and B C = A D = \sqrt{13} units$
Therefore, ABCD is a parallelogram. Now
$A C = \sqrt{{(1 - 6)}^{2} + {(2 - 6)}^{2}} = \sqrt{{(- 5)}^{2} + {(- 4)}^{2}} = \sqrt{25 + 16} = \sqrt{41} B D = \sqrt{{(4 - 3)}^{2} + {(3 - 5)}^{2}} = \sqrt{{(1)}^{2} + {(- 2)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.

Answer:

(i)
$Given : A (- 4, - 1), B (- 2, - 4), C (4, 0) and D (2, 3) In rectangle the opposite sides are equal Using d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} AB = \sqrt{{(- 4 + 2)}^{2} + {(- 1 + 4)}^{2}} AB = \sqrt{13} BC = \sqrt{{(4 + 2)}^{2} + {(0 + 4)}^{2}} BC = \sqrt{52} CD = \sqrt{{(2 - 4)}^{2} + {(3 - 0)}^{2}} CD = \sqrt{13} AD = \sqrt{{(2 + 4)}^{2} + {(3 + 1)}^{2}} AD = \sqrt{52} AB = CD and BC = AD$
$Now, we will prove for rectangles because diagnols in a rectangle are also equal AC = \sqrt{{(4 - (- 4))}^{2} + {(0 - (- 1))}^{2}} AC = \sqrt{{(4 + 4)}^{2} + {(1)}^{2}} AC = \sqrt{65} And BD = \sqrt{{(2 - (- 2))}^{2} + {(3 - (- 4))}^{2}} BD = \sqrt{{(4)}^{2} + {(7)}^{2}} BD = \sqrt{65} We can see that diagonal s are also equal Therefore, given coordinates are of rectangle .$

(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).
$A B = \sqrt{{(14 - 2)}^{2} + {\{10 - (- 2)\}}^{2}} = \sqrt{{(12)}^{2} + {(12)}^{2}} = \sqrt{144 + 144} = \sqrt{288} = 12 \sqrt{2} units B C = \sqrt{{(11 - 14)}^{2} + {(13 - 10)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units C D = \sqrt{{(- 1 - 11)}^{2} + {(1 - 13)}^{2}} = \sqrt{{(- 12)}^{2} + {(- 12)}^{2}} = \sqrt{144 + 144} = \sqrt{288} = 12 \sqrt{2} units A D = \sqrt{{(- 1 - 2)}^{2} + {\{1 - (- 2)\}}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Thus, A B = C D = 12 \sqrt{2} units and B C = A D = 3 \sqrt{2} units Also, A C = \sqrt{{(11 - 2)}^{2} + {\{13 - (- 2)\}}^{2}} = \sqrt{{(9)}^{2} + {(15)}^{2}} = \sqrt{81 + 225} = \sqrt{306} = 3 \sqrt{34} units B D = \sqrt{{(- 1 - 14)}^{2} + {(1 - 10)}^{2}} = \sqrt{{(- 15)}^{2} + {(- 9)}^{2}} = \sqrt{81 + 225} = \sqrt{306} = 3 \sqrt{34} units$
Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

(iii) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).
$A B = \sqrt{{(6 - 0)}^{2} + {\{2 - (- 4)\}}^{2}} = \sqrt{{(6)}^{2} + {(6)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 6 \sqrt{2} units B C = \sqrt{{(3 - 6)}^{2} + {(5 - 2)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units C D = \sqrt{{(- 3 - 3)}^{2} + {(- 1 - 5)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 6)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 6 \sqrt{2} units A D = \sqrt{{(- 3 - 0)}^{2} + {\{- 1 - (- 4)\}}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Thus, A B = C D = \sqrt{10} units and B C = A D = \sqrt{5} units Also, A C = \sqrt{{(3 - 0)}^{2} + {\{5 - (- 4)\}}^{2}} = \sqrt{{(3)}^{2} + {(9)}^{2}} = \sqrt{9 + 81} = \sqrt{90} = 3 \sqrt{10} units B D = \sqrt{{(- 3 - 6)}^{2} + {(- 1 - 2)}^{2}} = \sqrt{{(- 9)}^{2} + {(- 3)}^{2}} = \sqrt{81 + 9} = \sqrt{90} = 3 \sqrt{10} units$
Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.

Question 33:

(i)
$Given : A (- 4, - 1), B (- 2, - 4), C (4, 0) and D (2, 3) In rectangle the opposite sides are equal Using d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} AB = \sqrt{{(- 4 + 2)}^{2} + {(- 1 + 4)}^{2}} AB = \sqrt{13} BC = \sqrt{{(4 + 2)}^{2} + {(0 + 4)}^{2}} BC = \sqrt{52} CD = \sqrt{{(2 - 4)}^{2} + {(3 - 0)}^{2}} CD = \sqrt{13} AD = \sqrt{{(2 + 4)}^{2} + {(3 + 1)}^{2}} AD = \sqrt{52} AB = CD and BC = AD$
$Now, we will prove for rectangles because diagnols in a rectangle are also equal AC = \sqrt{{(4 - (- 4))}^{2} + {(0 - (- 1))}^{2}} AC = \sqrt{{(4 + 4)}^{2} + {(1)}^{2}} AC = \sqrt{65} And BD = \sqrt{{(2 - (- 2))}^{2} + {(3 - (- 4))}^{2}} BD = \sqrt{{(4)}^{2} + {(7)}^{2}} BD = \sqrt{65} We can see that diagonal s are also equal Therefore, given coordinates are of rectangle .$

(ii) The given points are A(2, −2), B(14, 10) C(11, 13) and D(−1, 1).
$A B = \sqrt{{(14 - 2)}^{2} + {\{10 - (- 2)\}}^{2}} = \sqrt{{(12)}^{2} + {(12)}^{2}} = \sqrt{144 + 144} = \sqrt{288} = 12 \sqrt{2} units B C = \sqrt{{(11 - 14)}^{2} + {(13 - 10)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units C D = \sqrt{{(- 1 - 11)}^{2} + {(1 - 13)}^{2}} = \sqrt{{(- 12)}^{2} + {(- 12)}^{2}} = \sqrt{144 + 144} = \sqrt{288} = 12 \sqrt{2} units A D = \sqrt{{(- 1 - 2)}^{2} + {\{1 - (- 2)\}}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Thus, A B = C D = 12 \sqrt{2} units and B C = A D = 3 \sqrt{2} units Also, A C = \sqrt{{(11 - 2)}^{2} + {\{13 - (- 2)\}}^{2}} = \sqrt{{(9)}^{2} + {(15)}^{2}} = \sqrt{81 + 225} = \sqrt{306} = 3 \sqrt{34} units B D = \sqrt{{(- 1 - 14)}^{2} + {(1 - 10)}^{2}} = \sqrt{{(- 15)}^{2} + {(- 9)}^{2}} = \sqrt{81 + 225} = \sqrt{306} = 3 \sqrt{34} units$
Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

(iii) The given points are A(0, −4), B(6, 2) C(3, 5) and D(−3, −1).
$A B = \sqrt{{(6 - 0)}^{2} + {\{2 - (- 4)\}}^{2}} = \sqrt{{(6)}^{2} + {(6)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 6 \sqrt{2} units B C = \sqrt{{(3 - 6)}^{2} + {(5 - 2)}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units C D = \sqrt{{(- 3 - 3)}^{2} + {(- 1 - 5)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 6)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 6 \sqrt{2} units A D = \sqrt{{(- 3 - 0)}^{2} + {\{- 1 - (- 4)\}}^{2}} = \sqrt{{(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2} units Thus, A B = C D = \sqrt{10} units and B C = A D = \sqrt{5} units Also, A C = \sqrt{{(3 - 0)}^{2} + {\{5 - (- 4)\}}^{2}} = \sqrt{{(3)}^{2} + {(9)}^{2}} = \sqrt{9 + 81} = \sqrt{90} = 3 \sqrt{10} units B D = \sqrt{{(- 3 - 6)}^{2} + {(- 1 - 2)}^{2}} = \sqrt{{(- 9)}^{2} + {(- 3)}^{2}} = \sqrt{81 + 9} = \sqrt{90} = 3 \sqrt{10} units$
Also, diagonal AC = diagonal BD.
Hence, the given points form a rectangle.

Answer:

In ΔABC, the coordinates of the vertices are A(–2, 0), B(0,2), C(2,0).

$AB = \sqrt{{(0 + 2)}^{2} + {(2 - 0)}^{2}} = \sqrt{8} = 2 \sqrt{2} CB = \sqrt{{(0 - 2)}^{2} + {(2 - 0)}^{2}} = \sqrt{8} = 2 \sqrt{2} AC = \sqrt{{(2 + 2)}^{2} + {(0 - 0)}^{2}} = 4$

In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).

$DF = \sqrt{{(4 + 4)}^{2} + {(0 - 0)}^{2}} = 8 FE = \sqrt{{(0 - 4)}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2} DE = \sqrt{{(0 + 4)}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2}$

Now, for ΔABC and ΔDEF to be similar, the corresponding sides should be proportional.

$So, \frac{AC}{DF} = \frac{BC}{FE} = \frac{AB}{DE} \Rightarrow \frac{4}{8} = \frac{2 \sqrt{2}}{4 \sqrt{2}} = \frac{2 \sqrt{2}}{4 \sqrt{2}} \Rightarrow \frac{1}{2} = \frac{1}{2} = \frac{1}{2} Since, the corresponding sides are proportional Therefore, given two triangles are similar .$

Question 1:

In ΔABC, the coordinates of the vertices are A(–2, 0), B(0,2), C(2,0).

$AB = \sqrt{{(0 + 2)}^{2} + {(2 - 0)}^{2}} = \sqrt{8} = 2 \sqrt{2} CB = \sqrt{{(0 - 2)}^{2} + {(2 - 0)}^{2}} = \sqrt{8} = 2 \sqrt{2} AC = \sqrt{{(2 + 2)}^{2} + {(0 - 0)}^{2}} = 4$

In ΔDEF, the coordinates of the vertices are D(–4, 0), E(4, 0), F(0, 4).

$DF = \sqrt{{(4 + 4)}^{2} + {(0 - 0)}^{2}} = 8 FE = \sqrt{{(0 - 4)}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2} DE = \sqrt{{(0 + 4)}^{2} + {(4 - 0)}^{2}} = 4 \sqrt{2}$

Now, for ΔABC and ΔDEF to be similar, the corresponding sides should be proportional.

$So, \frac{AC}{DF} = \frac{BC}{FE} = \frac{AB}{DE} \Rightarrow \frac{4}{8} = \frac{2 \sqrt{2}}{4 \sqrt{2}} = \frac{2 \sqrt{2}}{4 \sqrt{2}} \Rightarrow \frac{1}{2} = \frac{1}{2} = \frac{1}{2} Since, the corresponding sides are proportional Therefore, given two triangles are similar .$

Answer:

(i) The end points of AB are A(−1, 7) and B(4, −3).
Therefore, (x₁ = −1, y₁ = 7) and (x₂ = 4, y₂ = −3)
Also, m = 2 and n = 3
Let the required point be P(x, y).
By section formula, we get:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{2 \times 4 + 3 \times (- 1)\}}{2 + 3}, y = \frac{\{2 \times (- 3) + 3 \times 7\}}{2 + 3} \Rightarrow x = \frac{8 - 3}{5}, y = \frac{- 6 + 21}{5} \Rightarrow x = \frac{5}{5}, y = \frac{15}{5} Therefore, x = 1 and y = 3$
Hence, the coordinates of the required point are (1, 3).

(ii) The end points of AB are A(−5, 11) and B(4, −7).
Therefore, (x₁ = −5, y₁ = 11) and (x₂ = 4, y₂ = −7).
Also, m = 7 and n = 2
Let the required point be P(x, y).
By section formula, we have:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{7 \times 4 + 2 \times (- 5)\}}{7 + 2}, y = \frac{\{7 \times (- 7) + 2 \times 11\}}{7 + 2} \Rightarrow x = \frac{28 - 10}{9}, y = \frac{- 49 + 22}{9} \Rightarrow x = \frac{18}{9}, y = - \frac{27}{9} Therefore, x = 2 and y = - 3$
Hence, the required point is P(2, −3).

Question 2:

(i) The end points of AB are A(−1, 7) and B(4, −3).
Therefore, (x₁ = −1, y₁ = 7) and (x₂ = 4, y₂ = −3)
Also, m = 2 and n = 3
Let the required point be P(x, y).
By section formula, we get:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{2 \times 4 + 3 \times (- 1)\}}{2 + 3}, y = \frac{\{2 \times (- 3) + 3 \times 7\}}{2 + 3} \Rightarrow x = \frac{8 - 3}{5}, y = \frac{- 6 + 21}{5} \Rightarrow x = \frac{5}{5}, y = \frac{15}{5} Therefore, x = 1 and y = 3$
Hence, the coordinates of the required point are (1, 3).

(ii) The end points of AB are A(−5, 11) and B(4, −7).
Therefore, (x₁ = −5, y₁ = 11) and (x₂ = 4, y₂ = −7).
Also, m = 7 and n = 2
Let the required point be P(x, y).
By section formula, we have:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{7 \times 4 + 2 \times (- 5)\}}{7 + 2}, y = \frac{\{7 \times (- 7) + 2 \times 11\}}{7 + 2} \Rightarrow x = \frac{28 - 10}{9}, y = \frac{- 49 + 22}{9} \Rightarrow x = \frac{18}{9}, y = - \frac{27}{9} Therefore, x = 2 and y = - 3$
Hence, the required point is P(2, −3).

Answer:

Consider the figure.
Here points P and Q trisect AB.
Therefore, P divides AB into 1 : 2 and Q divides AB into 2 : 1.
Using section formula, coordinates of P are;
$P (x, y) = (\frac{1 \times 1 + 2 \times 7}{3}, \frac{1 \times - 5 + 2 \times - 2}{3}) P (x, y) = (\frac{15}{3}, \frac{- 9}{3}) = (5, - 3)$
Similarly, coordinates of Q are;
$Q (a, b) = (\frac{2 \times 1 + 1 \times 7}{3}, \frac{2 \times - 5 + 1 \times - 2}{3}) Q (a, b) = (\frac{9}{3}, \frac{- 12}{3}) = (3, - 4)$
Therefore, coordinates of points P and Q are (5, $-$ 3) and (3, $-$ 4) respectively.

Question 3:

Consider the figure.
Here points P and Q trisect AB.
Therefore, P divides AB into 1 : 2 and Q divides AB into 2 : 1.
Using section formula, coordinates of P are;
$P (x, y) = (\frac{1 \times 1 + 2 \times 7}{3}, \frac{1 \times - 5 + 2 \times - 2}{3}) P (x, y) = (\frac{15}{3}, \frac{- 9}{3}) = (5, - 3)$
Similarly, coordinates of Q are;
$Q (a, b) = (\frac{2 \times 1 + 1 \times 7}{3}, \frac{2 \times - 5 + 1 \times - 2}{3}) Q (a, b) = (\frac{9}{3}, \frac{- 12}{3}) = (3, - 4)$
Therefore, coordinates of points P and Q are (5, $-$ 3) and (3, $-$ 4) respectively.

Answer:

The coordinates of the points A and B are (−2, −2) and (2, −4) respectively, where $A P = \frac{3}{7} A B$ and P lies on the line segment AB. So
$A P + B P = A B \Rightarrow A P + B P = \frac{7 A P}{3} ∵ A P = \frac{3}{7} A B \Rightarrow B P = \frac{7 A P}{3} - A P \Rightarrow \frac{A P}{B P} = \frac{3}{4}$
Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then
$x = \frac{3 \times 2 + 4 \times (- 2)}{3 + 4} = \frac{6 - 8}{7} = - \frac{2}{7} y = \frac{3 \times (- 4) + 4 \times (- 2)}{3 + 4} = \frac{- 12 - 8}{7} = - \frac{20}{7}$
Hence, the coordinates of point P are $(- \frac{2}{7}, - \frac{20}{7})$ .

Question 4:

The coordinates of the points A and B are (−2, −2) and (2, −4) respectively, where $A P = \frac{3}{7} A B$ and P lies on the line segment AB. So
$A P + B P = A B \Rightarrow A P + B P = \frac{7 A P}{3} ∵ A P = \frac{3}{7} A B \Rightarrow B P = \frac{7 A P}{3} - A P \Rightarrow \frac{A P}{B P} = \frac{3}{4}$
Let (x, y) be the coordinates of P which divides AB in the ratio 3 : 4 internally. Then
$x = \frac{3 \times 2 + 4 \times (- 2)}{3 + 4} = \frac{6 - 8}{7} = - \frac{2}{7} y = \frac{3 \times (- 4) + 4 \times (- 2)}{3 + 4} = \frac{- 12 - 8}{7} = - \frac{20}{7}$
Hence, the coordinates of point P are $(- \frac{2}{7}, - \frac{20}{7})$ .

Answer:

Let the coordinates of A be (x, y). Here, $\frac{P A}{P Q} = \frac{2}{5}$ . So,
$P A + A Q = P Q \Rightarrow P A + A Q = \frac{5 P A}{2} [∵ P A = \frac{2}{5} P Q] \Rightarrow A Q = \frac{5 P A}{2} - P A \Rightarrow \frac{A Q}{P A} = \frac{3}{2} \Rightarrow \frac{P A}{A Q} = \frac{2}{3}$
Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get
$x = \frac{2 \times (- 4) + 3 \times (6)}{2 + 3} = \frac{- 8 + 18}{5} = \frac{10}{5} = 2 y = \frac{2 \times (- 1) + 3 \times (- 6)}{2 + 3} = \frac{- 2 - 18}{5} = \frac{- 20}{5} = - 4$
Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore
$3 \times 2 + k (- 4 + 1) = 0 \Rightarrow 3 k = 6 \Rightarrow k = \frac{6}{3} = 2$
Hence, k = 2.

Question 5:

Let the coordinates of A be (x, y). Here, $\frac{P A}{P Q} = \frac{2}{5}$ . So,
$P A + A Q = P Q \Rightarrow P A + A Q = \frac{5 P A}{2} [∵ P A = \frac{2}{5} P Q] \Rightarrow A Q = \frac{5 P A}{2} - P A \Rightarrow \frac{A Q}{P A} = \frac{3}{2} \Rightarrow \frac{P A}{A Q} = \frac{2}{3}$
Let (x, y) be the coordinates of A, which divides PQ in the ratio 2 : 3 internally. Then using section formula, we get
$x = \frac{2 \times (- 4) + 3 \times (6)}{2 + 3} = \frac{- 8 + 18}{5} = \frac{10}{5} = 2 y = \frac{2 \times (- 1) + 3 \times (- 6)}{2 + 3} = \frac{- 2 - 18}{5} = \frac{- 20}{5} = - 4$
Now, the point (2, −4) lies on the line 3x + k (y + 1) = 0, therefore
$3 \times 2 + k (- 4 + 1) = 0 \Rightarrow 3 k = 6 \Rightarrow k = \frac{6}{3} = 2$
Hence, k = 2.

Answer:

Since, the points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts, so
AP = PQ = QR = RS = SB
Here, point P divides AB in the ratio of 1 : 4 internally. So using section formula, we get
$Coordinates of P = (\frac{1 \times (6) + 4 \times (1)}{1 + 4}, \frac{1 \times (7) + 4 \times (2)}{1 + 4}) = (\frac{6 + 4}{5}, \frac{7 + 8}{5}) = (2, 3)$
The point Q divides AB in the ratio of 2 : 3 internally. So using section formula, we get
$Coordinates of Q = (\frac{2 \times (6) + 3 \times (1)}{2 + 3}, \frac{2 \times (7) + 3 \times (2)}{2 + 3}) = (\frac{12 + 3}{5}, \frac{14 + 6}{5}) = (3, 4)$
The point R divides AB in the ratio of 3 : 2 internally. So using section formula, we get
$Coordinates of R = (\frac{3 \times (6) + 2 \times (1)}{3 + 2}, \frac{3 \times (7) + 2 \times (2)}{3 + 2}) = (\frac{18 + 2}{5}, \frac{21 + 4}{5}) = (4, 5)$
Hence, the coordinates of the points P, Q and R are (2, 3), (3, 4) and (4, 5) respectively.

Question 6:

Since, the points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in five equal parts, so
AP = PQ = QR = RS = SB
Here, point P divides AB in the ratio of 1 : 4 internally. So using section formula, we get
$Coordinates of P = (\frac{1 \times (6) + 4 \times (1)}{1 + 4}, \frac{1 \times (7) + 4 \times (2)}{1 + 4}) = (\frac{6 + 4}{5}, \frac{7 + 8}{5}) = (2, 3)$
The point Q divides AB in the ratio of 2 : 3 internally. So using section formula, we get
$Coordinates of Q = (\frac{2 \times (6) + 3 \times (1)}{2 + 3}, \frac{2 \times (7) + 3 \times (2)}{2 + 3}) = (\frac{12 + 3}{5}, \frac{14 + 6}{5}) = (3, 4)$
The point R divides AB in the ratio of 3 : 2 internally. So using section formula, we get
$Coordinates of R = (\frac{3 \times (6) + 2 \times (1)}{3 + 2}, \frac{3 \times (7) + 2 \times (2)}{3 + 2}) = (\frac{18 + 2}{5}, \frac{21 + 4}{5}) = (4, 5)$
Hence, the coordinates of the points P, Q and R are (2, 3), (3, 4) and (4, 5) respectively.

Answer:

The given points are A(1, 6) and B(5, −2).
Then, P(x, y) is a point that divides the line AB in the ratio 1:3.
By the section formula:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{(1 \times 5 + 3 \times 1)}{1 + 3}, y = \frac{(1 \times (- 2) + 3 \times 6)}{1 + 3} \Rightarrow x = \frac{5 + 3}{4}, y = \frac{- 2 + 18}{4} \Rightarrow x = \frac{8}{4}, y = \frac{16}{4} \Rightarrow x = 2 and y = 4$
Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{1 + 5}{2}, y = \frac{6 + (- 2)}{2} \Rightarrow x = \frac{6}{2}, y = \frac{4}{2} \Rightarrow x = 3, y = 2$
Therefore, the coordinates of Q are (3, 2).
Let R (x, y) be a point that divides AB in the ratio 3:1.
Then, by the section formula:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{(3 \times 5 + 1 \times 1)}{3 + 1}, y = \frac{(3 \times (- 2) + 1 \times 6)}{3 + 1} \Rightarrow x = \frac{15 + 1}{4}, y = \frac{- 6 + 6}{4} \Rightarrow x = \frac{16}{4}, y = \frac{0}{4} \Rightarrow x = 4 and y = 0$
Therefore, the coordinates of R are (4, 0).
Hence, the coordinates of point P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.

Question 7:

The given points are A(1, 6) and B(5, −2).
Then, P(x, y) is a point that divides the line AB in the ratio 1:3.
By the section formula:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{(1 \times 5 + 3 \times 1)}{1 + 3}, y = \frac{(1 \times (- 2) + 3 \times 6)}{1 + 3} \Rightarrow x = \frac{5 + 3}{4}, y = \frac{- 2 + 18}{4} \Rightarrow x = \frac{8}{4}, y = \frac{16}{4} \Rightarrow x = 2 and y = 4$
Therefore, the coordinates of point P are (2, 4).
Let Q be the mid point of AB.
Then, Q(x, y):
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{1 + 5}{2}, y = \frac{6 + (- 2)}{2} \Rightarrow x = \frac{6}{2}, y = \frac{4}{2} \Rightarrow x = 3, y = 2$
Therefore, the coordinates of Q are (3, 2).
Let R (x, y) be a point that divides AB in the ratio 3:1.
Then, by the section formula:
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{(3 \times 5 + 1 \times 1)}{3 + 1}, y = \frac{(3 \times (- 2) + 1 \times 6)}{3 + 1} \Rightarrow x = \frac{15 + 1}{4}, y = \frac{- 6 + 6}{4} \Rightarrow x = \frac{16}{4}, y = \frac{0}{4} \Rightarrow x = 4 and y = 0$
Therefore, the coordinates of R are (4, 0).
Hence, the coordinates of point P, Q and R are (2, 4), (3, 2) and (4, 0) respectively.

Answer:

Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{1 \times 1 + 2 \times (3)\}}{1 + 2}, y = \frac{\{1 \times 2 + 2 \times (- 4)\}}{1 + 2} \Rightarrow x = \frac{1 + 6}{3}, y = \frac{2 - 8}{3} \Rightarrow x = \frac{7}{3}, y = - \frac{6}{3} \Rightarrow x = \frac{7}{3}, y = - 2$
Hence, the coordinates of P are ( $\frac{7}{3}$ , −2).
But (p, −2) are the coordinates of P.
So, $p = \frac{7}{3}$
Also, Q divides the line AB in the ratio 2:1.
So, the coordinates of Q are
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{(2 \times 1 + 1 \times 3)}{2 + 1}, y = \frac{\{2 \times 2 + 1 \times (- 4)\}}{2 + 1} \Rightarrow x = \frac{2 + 3}{3}, y = \frac{4 - 4}{3} \Rightarrow x = \frac{5}{3}, y = 0 Hence, coordinates of Q are (\frac{5}{3}, 0) .$
But the given coordinates of $Q are (\frac{5}{3}, q) .$
So, q = 0
Thus, $p = \frac{7}{3}$ and $q = 0$ 53

Question 8:

Let P and Q be the points of trisection of AB.
Then, P divides AB in the ratio 1:2.
So, the coordinates of P are
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{1 \times 1 + 2 \times (3)\}}{1 + 2}, y = \frac{\{1 \times 2 + 2 \times (- 4)\}}{1 + 2} \Rightarrow x = \frac{1 + 6}{3}, y = \frac{2 - 8}{3} \Rightarrow x = \frac{7}{3}, y = - \frac{6}{3} \Rightarrow x = \frac{7}{3}, y = - 2$
Hence, the coordinates of P are ( $\frac{7}{3}$ , −2).
But (p, −2) are the coordinates of P.
So, $p = \frac{7}{3}$
Also, Q divides the line AB in the ratio 2:1.
So, the coordinates of Q are
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{(2 \times 1 + 1 \times 3)}{2 + 1}, y = \frac{\{2 \times 2 + 1 \times (- 4)\}}{2 + 1} \Rightarrow x = \frac{2 + 3}{3}, y = \frac{4 - 4}{3} \Rightarrow x = \frac{5}{3}, y = 0 Hence, coordinates of Q are (\frac{5}{3}, 0) .$
But the given coordinates of $Q are (\frac{5}{3}, q) .$
So, q = 0
Thus, $p = \frac{7}{3}$ and $q = 0$ 53

Answer:

(i) The given points are A(3, 0) and B(−5, 4).
Let (x, y) be the mid point of AB. Then:
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{3 + (- 5)}{2}, y = \frac{0 + 4}{2} \Rightarrow x = \frac{- 2}{2}, y = \frac{4}{2} \Rightarrow x = - 1, y = 2$
Therefore, (−1, 2) are the coordinates of mid point of AB.

(ii) The given points are P(−11, −8) and Q(8, −2).
Let (x, y) be the mid point of PQ. Then:
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{- 11 + 8}{2}, y = \frac{- 8 - 2}{2} \Rightarrow x = - \frac{3}{2}, y = - \frac{10}{2} \Rightarrow x = - \frac{3}{2}, y = - 5$
Therefore, $(- \frac{3}{2}, - 5)$ are the coordinates of midpoint of PQ.

Question 9:

(i) The given points are A(3, 0) and B(−5, 4).
Let (x, y) be the mid point of AB. Then:
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{3 + (- 5)}{2}, y = \frac{0 + 4}{2} \Rightarrow x = \frac{- 2}{2}, y = \frac{4}{2} \Rightarrow x = - 1, y = 2$
Therefore, (−1, 2) are the coordinates of mid point of AB.

(ii) The given points are P(−11, −8) and Q(8, −2).
Let (x, y) be the mid point of PQ. Then:
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{- 11 + 8}{2}, y = \frac{- 8 - 2}{2} \Rightarrow x = - \frac{3}{2}, y = - \frac{10}{2} \Rightarrow x = - \frac{3}{2}, y = - 5$
Therefore, $(- \frac{3}{2}, - 5)$ are the coordinates of midpoint of PQ.

Answer:

The given points are A(6, −5) and B(−2, 11).
Let (x, y) be the mid point of AB. Then:
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{6 + (- 2)}{2}, y = \frac{- 5 + 11}{2} \Rightarrow x = \frac{6 - 2}{2}, y = \frac{- 5 + 11}{2} \Rightarrow x = \frac{4}{2}, y = \frac{6}{2} \Rightarrow x = 2, y = 3$
So, the midpoint of AB is (2, 3).
But it is given that the midpoint of AB is (2, p).
Therefore, the value of p = 3.

Question 10:

The given points are A(6, −5) and B(−2, 11).
Let (x, y) be the mid point of AB. Then:
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{6 + (- 2)}{2}, y = \frac{- 5 + 11}{2} \Rightarrow x = \frac{6 - 2}{2}, y = \frac{- 5 + 11}{2} \Rightarrow x = \frac{4}{2}, y = \frac{6}{2} \Rightarrow x = 2, y = 3$
So, the midpoint of AB is (2, 3).
But it is given that the midpoint of AB is (2, p).
Therefore, the value of p = 3.

Answer:

The points are A(2a, 4) and B(−2, 3b).
Let C(1, 2a + 1) be the mid point of AB. Then:
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow 1 = \frac{2 a + (- 2)}{2}, 2 a + 1 = \frac{4 + 3 b}{2} \Rightarrow 2 = 2 a - 2, 4 a + 2 = 4 + 3 b \Rightarrow 2 a = 2 + 2, 4 a - 3 b = 4 - 2 \Rightarrow a = \frac{4}{2}, 4 a - 3 b = 2 \Rightarrow a = 2, 4 a - 3 b = 2 Putting the value of a in the equation 4 a + 3 b = 2, we get : 4 (2) - 3 b = 2 \Rightarrow - 3 b = 2 - 8 = - 6 \Rightarrow b = \frac{6}{3} = 2 Therefore, a = 2 and b = 2 .$

Question 11:

The points are A(2a, 4) and B(−2, 3b).
Let C(1, 2a + 1) be the mid point of AB. Then:
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow 1 = \frac{2 a + (- 2)}{2}, 2 a + 1 = \frac{4 + 3 b}{2} \Rightarrow 2 = 2 a - 2, 4 a + 2 = 4 + 3 b \Rightarrow 2 a = 2 + 2, 4 a - 3 b = 4 - 2 \Rightarrow a = \frac{4}{2}, 4 a - 3 b = 2 \Rightarrow a = 2, 4 a - 3 b = 2 Putting the value of a in the equation 4 a + 3 b = 2, we get : 4 (2) - 3 b = 2 \Rightarrow - 3 b = 2 - 8 = - 6 \Rightarrow b = \frac{6}{3} = 2 Therefore, a = 2 and b = 2 .$

Answer:

The given points are A(−2, 9) and B(6, 3).
Then, C(x, y) is the midpoint of AB.
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{- 2 + 6}{2}, y = \frac{9 + 3}{2} \Rightarrow x = \frac{4}{2}, y = \frac{12}{2} \Rightarrow x = 2, y = 6$
Therefore, the coordinates of point C are (2, 6).

Question 12:

The given points are A(−2, 9) and B(6, 3).
Then, C(x, y) is the midpoint of AB.
$x = \frac{x_{1} + x_{2}}{2}, y = \frac{y_{1} + y_{2}}{2} \Rightarrow x = \frac{- 2 + 6}{2}, y = \frac{9 + 3}{2} \Rightarrow x = \frac{4}{2}, y = \frac{12}{2} \Rightarrow x = 2, y = 6$
Therefore, the coordinates of point C are (2, 6).

Answer:

C(2, −3) is the centre of the given circle. Let A(a, b) and B(1, 4) be the two end-points of the given diameter AB. Then, the coordinates of C are
$x = \frac{a + 1}{2}, y = \frac{b + 4}{2} It is given that x = 2 and y = - 3 . \Rightarrow 2 = \frac{a + 1}{2}, - 3 = \frac{b + 4}{2} \Rightarrow 4 = a + 1, - 6 = b + 4 \Rightarrow a = 4 - 1, b = - 6 - 4 \Rightarrow a = 3, b = - 10$
Therefore, the coordinates of point A are (3, -10).

Question 13:

C(2, −3) is the centre of the given circle. Let A(a, b) and B(1, 4) be the two end-points of the given diameter AB. Then, the coordinates of C are
$x = \frac{a + 1}{2}, y = \frac{b + 4}{2} It is given that x = 2 and y = - 3 . \Rightarrow 2 = \frac{a + 1}{2}, - 3 = \frac{b + 4}{2} \Rightarrow 4 = a + 1, - 6 = b + 4 \Rightarrow a = 4 - 1, b = - 6 - 4 \Rightarrow a = 3, b = - 10$
Therefore, the coordinates of point A are (3, -10).

Answer:

Let the point P(2, 5) divide AB in the ratio k : 1.
Then, by section formula, the coordinates of P are
$x = \frac{- 6 k + 8}{k + 1}, y = \frac{9 k + 2}{k + 1} It is given that the coordinates of P a r e P (2, 5) . \Rightarrow 2 = \frac{- 6 k + 8}{k + 1}, 5 = \frac{9 k + 2}{k + 1} \Rightarrow 2 k + 2 = - 6 k + 8, 5 k + 5 = 9 k + 2 \Rightarrow 2 k + 6 k = 8 - 2, 5 - 2 = 9 k - 5 k \Rightarrow 8 k = 6, 4 k = 3 \Rightarrow k = \frac{6}{8}, k = \frac{3}{4} \Rightarrow k = \frac{3}{4} in each case .$
Therefore, the point P(2, 5) divides AB in the ratio 3 : 4.

Question 14:

Let the point P(2, 5) divide AB in the ratio k : 1.
Then, by section formula, the coordinates of P are
$x = \frac{- 6 k + 8}{k + 1}, y = \frac{9 k + 2}{k + 1} It is given that the coordinates of P a r e P (2, 5) . \Rightarrow 2 = \frac{- 6 k + 8}{k + 1}, 5 = \frac{9 k + 2}{k + 1} \Rightarrow 2 k + 2 = - 6 k + 8, 5 k + 5 = 9 k + 2 \Rightarrow 2 k + 6 k = 8 - 2, 5 - 2 = 9 k - 5 k \Rightarrow 8 k = 6, 4 k = 3 \Rightarrow k = \frac{6}{8}, k = \frac{3}{4} \Rightarrow k = \frac{3}{4} in each case .$
Therefore, the point P(2, 5) divides AB in the ratio 3 : 4.

Answer:

Let k : 1 be the ratio in which the point $P (\frac{3}{4}, \frac{5}{12})$ divides the line segment joining the points $A (\frac{1}{2}, \frac{3}{2})$ and $(2, - 5)$ . Then
$(\frac{3}{4}, \frac{5}{12}) = (\frac{k (2) + \frac{1}{2}}{k + 1}, \frac{k (- 5) + \frac{3}{2}}{k + 1}) \Rightarrow \frac{k (2) + \frac{1}{2}}{k + 1} = \frac{3}{4} and \frac{k (- 5) + \frac{3}{2}}{k + 1} = \frac{5}{12} \Rightarrow 8 k + 2 = 3 k + 3 and - 60 k + 18 = 5 k + 5 \Rightarrow k = \frac{1}{5} and k = \frac{1}{5}$
Hence, the required ratio is 1 : 5.

Question 15:

Let k : 1 be the ratio in which the point $P (\frac{3}{4}, \frac{5}{12})$ divides the line segment joining the points $A (\frac{1}{2}, \frac{3}{2})$ and $(2, - 5)$ . Then
$(\frac{3}{4}, \frac{5}{12}) = (\frac{k (2) + \frac{1}{2}}{k + 1}, \frac{k (- 5) + \frac{3}{2}}{k + 1}) \Rightarrow \frac{k (2) + \frac{1}{2}}{k + 1} = \frac{3}{4} and \frac{k (- 5) + \frac{3}{2}}{k + 1} = \frac{5}{12} \Rightarrow 8 k + 2 = 3 k + 3 and - 60 k + 18 = 5 k + 5 \Rightarrow k = \frac{1}{5} and k = \frac{1}{5}$
Hence, the required ratio is 1 : 5.

Answer:

Let the point P(m, 6) divide the line AB in the ratio k : 1.
Then, by the section formula:
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n} The coordinates of P are (m, 6) . m = \frac{2 k - 4}{k + 1}, 6 = \frac{8 k + 3}{k + 1} \Rightarrow m (k + 1) = 2 k - 4, 6 k + 6 = 8 k + 3 \Rightarrow m (k + 1) = 2 k - 4, 6 - 3 = 8 k - 6 k \Rightarrow m (k + 1) = 2 k - 4, 2 k = 3 \Rightarrow m (k + 1) = 2 k - 4, k = \frac{3}{2} Therefore, the point P divides the line A B in the ratio 3 : 2 . Now, put ting the value of k in the equation m (k + 1) = 2 k - 4, we get: m (\frac{3}{2} + 1) = 2 (\frac{3}{2}) - 4 \Rightarrow m (\frac{3 + 2}{2}) = 3 - 4 \Rightarrow \frac{5 m}{2} = - 1 \Rightarrow 5 m = - 2 \Rightarrow m = - \frac{2}{5} Therefore, the value of m = - \frac{2}{5} So, the coordinates of P a r e (- \frac{2}{5}, 6) .$

Question 16:

Let the point P(m, 6) divide the line AB in the ratio k : 1.
Then, by the section formula:
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n} The coordinates of P are (m, 6) . m = \frac{2 k - 4}{k + 1}, 6 = \frac{8 k + 3}{k + 1} \Rightarrow m (k + 1) = 2 k - 4, 6 k + 6 = 8 k + 3 \Rightarrow m (k + 1) = 2 k - 4, 6 - 3 = 8 k - 6 k \Rightarrow m (k + 1) = 2 k - 4, 2 k = 3 \Rightarrow m (k + 1) = 2 k - 4, k = \frac{3}{2} Therefore, the point P divides the line A B in the ratio 3 : 2 . Now, put ting the value of k in the equation m (k + 1) = 2 k - 4, we get: m (\frac{3}{2} + 1) = 2 (\frac{3}{2}) - 4 \Rightarrow m (\frac{3 + 2}{2}) = 3 - 4 \Rightarrow \frac{5 m}{2} = - 1 \Rightarrow 5 m = - 2 \Rightarrow m = - \frac{2}{5} Therefore, the value of m = - \frac{2}{5} So, the coordinates of P a r e (- \frac{2}{5}, 6) .$

Answer:

Let the point P(−3, k) divide the line AB in the ratio s : 1.
Then, by the section formula:
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n} The coordinates of P are (- 3, k) . - 3 = \frac{- 2 s - 5}{s + 1}, k = \frac{3 s - 4}{s + 1} \Rightarrow - 3 s - 3 = - 2 s - 5, k (s + 1) = 3 s - 4 \Rightarrow - 3 s + 2 s = - 5 + 3, k (s + 1) = 3 s - 4 \Rightarrow - s = - 2, k (s + 1) = 3 s - 4 \Rightarrow s = 2, k (s + 1) = 3 s - 4 Therefore, the point P divides the line A B in the ratio 2 : 1 . Now, put ting the value of s in the equation k (s + 1) = 3 s - 4, we get: k (2 + 1) = 3 (2) - 4 \Rightarrow 3 k = 6 - 4 \Rightarrow 3 k = 2 \Rightarrow k = \frac{2}{3} Therefore, the value of k = \frac{2}{3} That is, the coordinates of P are (- 3, \frac{2}{3}) .$

Question 17:

Let the point P(−3, k) divide the line AB in the ratio s : 1.
Then, by the section formula:
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n} The coordinates of P are (- 3, k) . - 3 = \frac{- 2 s - 5}{s + 1}, k = \frac{3 s - 4}{s + 1} \Rightarrow - 3 s - 3 = - 2 s - 5, k (s + 1) = 3 s - 4 \Rightarrow - 3 s + 2 s = - 5 + 3, k (s + 1) = 3 s - 4 \Rightarrow - s = - 2, k (s + 1) = 3 s - 4 \Rightarrow s = 2, k (s + 1) = 3 s - 4 Therefore, the point P divides the line A B in the ratio 2 : 1 . Now, put ting the value of s in the equation k (s + 1) = 3 s - 4, we get: k (2 + 1) = 3 (2) - 4 \Rightarrow 3 k = 6 - 4 \Rightarrow 3 k = 2 \Rightarrow k = \frac{2}{3} Therefore, the value of k = \frac{2}{3} That is, the coordinates of P are (- 3, \frac{2}{3}) .$

Answer:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
$P = (\frac{5 k + 2}{k + 1}, \frac{6 k - 3}{k + 1})$
But P lies on the x-axis; so, its ordinate is 0.
$Therefore, \frac{6 k - 3}{k + 1} = 0 \Rightarrow 6 k - 3 = 0 \Rightarrow 6 k = 3 \Rightarrow k = \frac{3}{6} \Rightarrow k = \frac{1}{2}$
Therefore, the required ratio is $\frac{1}{2}$ : 1, which is same as 1 : 2.
Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.
Applying k = $\frac{1}{2}$ , we get the coordinates of point :
$P (\frac{5 k + 2}{k + 1}, 0) = P (\frac{5 \times \frac{1}{2} + 2}{\frac{1}{2} + 1}, 0) = P (\frac{\frac{5 + 4}{2}}{\frac{1 + 2}{2}}, 0) = P (\frac{9}{3}, 0) = P (3, 0)$
Hence, the point of intersection of AB and the x-axis is P(3, 0).

Question 18:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
$P = (\frac{5 k + 2}{k + 1}, \frac{6 k - 3}{k + 1})$
But P lies on the x-axis; so, its ordinate is 0.
$Therefore, \frac{6 k - 3}{k + 1} = 0 \Rightarrow 6 k - 3 = 0 \Rightarrow 6 k = 3 \Rightarrow k = \frac{3}{6} \Rightarrow k = \frac{1}{2}$
Therefore, the required ratio is $\frac{1}{2}$ : 1, which is same as 1 : 2.
Thus, the x-axis divides the line AB in the ratio 1 : 2 at the point P.
Applying k = $\frac{1}{2}$ , we get the coordinates of point :
$P (\frac{5 k + 2}{k + 1}, 0) = P (\frac{5 \times \frac{1}{2} + 2}{\frac{1}{2} + 1}, 0) = P (\frac{\frac{5 + 4}{2}}{\frac{1 + 2}{2}}, 0) = P (\frac{9}{3}, 0) = P (3, 0)$
Hence, the point of intersection of AB and the x-axis is P(3, 0).

Answer:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
$P (\frac{3 k - 2}{k + 1}, \frac{7 k - 3}{k + 1})$
But P lies on the y-axis; so, its abscissa is 0.
$Therefore, \frac{3 k - 2}{k + 1} = 0 \Rightarrow 3 k - 2 = 0 \Rightarrow 3 k = 2 \Rightarrow k = \frac{2}{3} \Rightarrow k = \frac{2}{3}$
Therefore, the required ratio is $\frac{2}{3}$ : 1, which is same as 2 : 3.
Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.
Applying k= $\frac{2}{3}$ , we get the coordinates of point P:
$P (0, \frac{7 k - 3}{k + 1}) = P (0, \frac{7 \times \frac{2}{3} - 3}{\frac{2}{3} + 1}) = P (0, \frac{\frac{14 - 9}{3}}{\frac{2 + 3}{3}}) = P (0, \frac{5}{5}) = P (0, 1)$
Hence, the point of intersection of AB and the x-axis is P(0, 1).

Question 19:

Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordinates of P are
$P (\frac{3 k - 2}{k + 1}, \frac{7 k - 3}{k + 1})$
But P lies on the y-axis; so, its abscissa is 0.
$Therefore, \frac{3 k - 2}{k + 1} = 0 \Rightarrow 3 k - 2 = 0 \Rightarrow 3 k = 2 \Rightarrow k = \frac{2}{3} \Rightarrow k = \frac{2}{3}$
Therefore, the required ratio is $\frac{2}{3}$ : 1, which is same as 2 : 3.
Thus, the x-axis divides the line AB in the ratio 2:3 at the point P.
Applying k= $\frac{2}{3}$ , we get the coordinates of point P:
$P (0, \frac{7 k - 3}{k + 1}) = P (0, \frac{7 \times \frac{2}{3} - 3}{\frac{2}{3} + 1}) = P (0, \frac{\frac{14 - 9}{3}}{\frac{2 + 3}{3}}) = P (0, \frac{5}{5}) = P (0, 1)$
Hence, the point of intersection of AB and the x-axis is P(0, 1).

Answer:

Let the line x − y − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9) in the ratio k : 1 at P.
Then, the coordinates of P are
$P (\frac{8 k + 3}{k + 1}, \frac{9 k - 1}{k + 1})$
Since, P lies on the line x − y − 2 = 0, we have:

$(\frac{8 k + 3}{k + 1}) - (\frac{9 k - 1}{k + 1}) - 2 = 0 \Rightarrow 8 k + 3 - 9 k + 1 - 2 k - 2 = 0 \Rightarrow 8 k - 9 k - 2 k + 3 + 1 - 2 = 0 \Rightarrow - 3 k + 2 = 0 \Rightarrow - 3 k = - 2 \Rightarrow k = \frac{2}{3}$
So, the required ratio is $\frac{2}{3}$ : 1, which is equal to 2 : 3.

Question 20:

Let the line x − y − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9) in the ratio k : 1 at P.
Then, the coordinates of P are
$P (\frac{8 k + 3}{k + 1}, \frac{9 k - 1}{k + 1})$
Since, P lies on the line x − y − 2 = 0, we have:

$(\frac{8 k + 3}{k + 1}) - (\frac{9 k - 1}{k + 1}) - 2 = 0 \Rightarrow 8 k + 3 - 9 k + 1 - 2 k - 2 = 0 \Rightarrow 8 k - 9 k - 2 k + 3 + 1 - 2 = 0 \Rightarrow - 3 k + 2 = 0 \Rightarrow - 3 k = - 2 \Rightarrow k = \frac{2}{3}$
So, the required ratio is $\frac{2}{3}$ : 1, which is equal to 2 : 3.

Answer:

The vertices of âˆ†ABC are A(0, −1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of âˆ†ABC.

Let D be the midpoint of BC. So, the coordinates of D are
$D (\frac{2 + 0}{2}, \frac{1 + 3}{2}) i . e . D (\frac{2}{2}, \frac{4}{2}) i . e . D (1, 2)$
Let E be the midpoint of AC. So, the coordinates of E are
$E (\frac{0 + 0}{2}, \frac{- 1 + 3}{2}) i . e . E (\frac{0}{2}, \frac{2}{2}) i . e . E (0, 1)$
Let F be the midpoint of AB. So, the coordinates of F are
$F (\frac{0 + 2}{2}, \frac{- 1 + 1}{2}) i . e . F (\frac{2}{2}, \frac{0}{2}) i . e . F (1, 0)$
$A D = \sqrt{{(1 - 0)}^{2} + {(2 - (- 1))}^{2}} = \sqrt{{(1)}^{2} + {(3)}^{2}} = \sqrt{1 + 9} = \sqrt{10} units . B E = \sqrt{{(0 - 2)}^{2} + {(1 - 1)}^{2}} = \sqrt{{(- 2)}^{2} + {(0)}^{2}} = \sqrt{4 + 0} = \sqrt{4} = 2 units . C F = \sqrt{{(1 - 0)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(1)}^{2} + {(- 3)}^{2}} = \sqrt{1 + 9} = \sqrt{10} units .$
Therefore, the lengths of the medians: AD = $\sqrt{10}$ units, BE = 2 units and CF = $\sqrt{10}$ units

Question 21:

The vertices of âˆ†ABC are A(0, −1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of âˆ†ABC.

Let D be the midpoint of BC. So, the coordinates of D are
$D (\frac{2 + 0}{2}, \frac{1 + 3}{2}) i . e . D (\frac{2}{2}, \frac{4}{2}) i . e . D (1, 2)$
Let E be the midpoint of AC. So, the coordinates of E are
$E (\frac{0 + 0}{2}, \frac{- 1 + 3}{2}) i . e . E (\frac{0}{2}, \frac{2}{2}) i . e . E (0, 1)$
Let F be the midpoint of AB. So, the coordinates of F are
$F (\frac{0 + 2}{2}, \frac{- 1 + 1}{2}) i . e . F (\frac{2}{2}, \frac{0}{2}) i . e . F (1, 0)$
$A D = \sqrt{{(1 - 0)}^{2} + {(2 - (- 1))}^{2}} = \sqrt{{(1)}^{2} + {(3)}^{2}} = \sqrt{1 + 9} = \sqrt{10} units . B E = \sqrt{{(0 - 2)}^{2} + {(1 - 1)}^{2}} = \sqrt{{(- 2)}^{2} + {(0)}^{2}} = \sqrt{4 + 0} = \sqrt{4} = 2 units . C F = \sqrt{{(1 - 0)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(1)}^{2} + {(- 3)}^{2}} = \sqrt{1 + 9} = \sqrt{10} units .$
Therefore, the lengths of the medians: AD = $\sqrt{10}$ units, BE = 2 units and CF = $\sqrt{10}$ units

Answer:

Here, (x₁ = −1, y₁ = 0), (x₂ = 5, y₂ = −2) and (x₃ = 8, y₃ = 2).
Let G(x, y) be the centroid of the âˆ†ABC. Then,
$x = \frac{1}{3} (x_{1} + x_{2} + x_{3}) = \frac{1}{3} (- 1 + 5 + 8) = \frac{1}{3} (12) = 4 y = \frac{1}{3} (y_{1} + y_{2} + y_{3}) = \frac{1}{3} (0 - 2 + 2) = \frac{1}{3} (0) = 0$
Hence, the centroid of âˆ†ABC is G(4, 0).

Question 22:

Here, (x₁ = −1, y₁ = 0), (x₂ = 5, y₂ = −2) and (x₃ = 8, y₃ = 2).
Let G(x, y) be the centroid of the âˆ†ABC. Then,
$x = \frac{1}{3} (x_{1} + x_{2} + x_{3}) = \frac{1}{3} (- 1 + 5 + 8) = \frac{1}{3} (12) = 4 y = \frac{1}{3} (y_{1} + y_{2} + y_{3}) = \frac{1}{3} (0 - 2 + 2) = \frac{1}{3} (0) = 0$
Hence, the centroid of âˆ†ABC is G(4, 0).

Answer:

Two vertices of âˆ†ABC are A(1, −6) and B(−5, 2). Let the third vertex be C(a, b).
Then the coordinates of its centroid are
$C (\frac{1 - 5 + a}{3}, \frac{- 6 + 2 + b}{3}) C (\frac{- 4 + a}{3}, \frac{- 4 + b}{3})$
But it is given that G(−2, 1) is the centroid. Therefore,
$- 2 = \frac{- 4 + a}{3}, 1 = \frac{- 4 + b}{3} \Rightarrow - 6 = - 4 + a, 3 = - 4 + b \Rightarrow - 6 + 4 = a, 3 + 4 = b \Rightarrow a = - 2, b = 7$
Therefore, the third vertex of âˆ†ABC is C(−2, 7).

Question 23:

Two vertices of âˆ†ABC are A(1, −6) and B(−5, 2). Let the third vertex be C(a, b).
Then the coordinates of its centroid are
$C (\frac{1 - 5 + a}{3}, \frac{- 6 + 2 + b}{3}) C (\frac{- 4 + a}{3}, \frac{- 4 + b}{3})$
But it is given that G(−2, 1) is the centroid. Therefore,
$- 2 = \frac{- 4 + a}{3}, 1 = \frac{- 4 + b}{3} \Rightarrow - 6 = - 4 + a, 3 = - 4 + b \Rightarrow - 6 + 4 = a, 3 + 4 = b \Rightarrow a = - 2, b = 7$
Therefore, the third vertex of âˆ†ABC is C(−2, 7).

Answer:

Two vertices of âˆ†ABC are B(−3,1) and C(0, −2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are
$(\frac{- 3 + 0 + a}{3}, \frac{1 - 2 + b}{3}) i . e . (\frac{- 3 + a}{3}, \frac{- 1 + b}{3})$
But it is given that the centroid is at the origin, that is G(0, 0). Therefore,
$0 = \frac{- 3 + a}{3}, 0 = \frac{- 1 + b}{3} \Rightarrow 0 = - 3 + a, 0 = - 1 + b \Rightarrow 3 = a, 1 = b \Rightarrow a = 3, b = 1$
Therefore, the third vertex of âˆ†ABC is A(3, 1).

Question 24:

Two vertices of âˆ†ABC are B(−3,1) and C(0, −2). Let the third vertex be A(a, b).
Then, the coordinates of its centroid are
$(\frac{- 3 + 0 + a}{3}, \frac{1 - 2 + b}{3}) i . e . (\frac{- 3 + a}{3}, \frac{- 1 + b}{3})$
But it is given that the centroid is at the origin, that is G(0, 0). Therefore,
$0 = \frac{- 3 + a}{3}, 0 = \frac{- 1 + b}{3} \Rightarrow 0 = - 3 + a, 0 = - 1 + b \Rightarrow 3 = a, 1 = b \Rightarrow a = 3, b = 1$
Therefore, the third vertex of âˆ†ABC is A(3, 1).

Answer:

The points are A(3, 1), B(0, −2), C(1, 1) and D(4, 4).
Join AC and BD, intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
$Midpoint of A C = (\frac{3 + 1}{2}, \frac{1 + 1}{2}) = (\frac{4}{2}, \frac{2}{2}) = (2, 1) Midpoint of B D = (\frac{0 + 4}{2}, \frac{- 2 + 4}{2}) = (\frac{4}{2}, \frac{2}{2}) = (2, 1)$
Thus, the diagonals AC and BD have the same midpoint.
Therefore, ABCD is a parallelogram.

Question 25:

The points are A(3, 1), B(0, −2), C(1, 1) and D(4, 4).
Join AC and BD, intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
$Midpoint of A C = (\frac{3 + 1}{2}, \frac{1 + 1}{2}) = (\frac{4}{2}, \frac{2}{2}) = (2, 1) Midpoint of B D = (\frac{0 + 4}{2}, \frac{- 2 + 4}{2}) = (\frac{4}{2}, \frac{2}{2}) = (2, 1)$
Thus, the diagonals AC and BD have the same midpoint.
Therefore, ABCD is a parallelogram.

Answer:

The points are P(a, −11), Q(5, b), R(2, 15) and S(1, 1).

Join PR and QS, intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of PR as well as QS.
$Midpoint of P R = (\frac{a + 2}{2}, \frac{- 11 + 15}{2}) = (\frac{a + 2}{2}, \frac{4}{2}) = (\frac{a + 2}{2}, 2) Midpoint of Q S = (\frac{5 + 1}{2}, \frac{b + 1}{2}) = (\frac{6}{2}, \frac{b + 1}{2}) = (3, \frac{b + 1}{2}) Therefore, \frac{a + 2}{2} = 3, \frac{b + 1}{2} = 2 \Rightarrow a + 2 = 6, b + 1 = 4 \Rightarrow a = 6 - 2, b = 4 - 1 \Rightarrow a = 4 and b = 3$

Question 26:

The points are P(a, −11), Q(5, b), R(2, 15) and S(1, 1).

Join PR and QS, intersecting at O.
We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of PR as well as QS.
$Midpoint of P R = (\frac{a + 2}{2}, \frac{- 11 + 15}{2}) = (\frac{a + 2}{2}, \frac{4}{2}) = (\frac{a + 2}{2}, 2) Midpoint of Q S = (\frac{5 + 1}{2}, \frac{b + 1}{2}) = (\frac{6}{2}, \frac{b + 1}{2}) = (3, \frac{b + 1}{2}) Therefore, \frac{a + 2}{2} = 3, \frac{b + 1}{2} = 2 \Rightarrow a + 2 = 6, b + 1 = 4 \Rightarrow a = 6 - 2, b = 4 - 1 \Rightarrow a = 4 and b = 3$

Answer:

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b).
Join AC and BD intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of AC as well as BD.
$Midpoint of A C = (\frac{1 + 5}{2}, \frac{- 2 + 10}{2}) = (\frac{6}{2}, \frac{8}{2}) = (3, 4) Midpoint of B D = (\frac{3 + a}{2}, \frac{6 + b}{2}) Therefore, \frac{3 + a}{2} = 3 and \frac{6 + b}{2} = 4 \Rightarrow 3 + a = 6 and 6 + b = 8 \Rightarrow a = 6 - 3 and b = 8 - 6 \Rightarrow a = 3 and b = 2$
Therefore, the fourth vertex is D(3, 2).

Question 27:

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b).
Join AC and BD intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of AC as well as BD.
$Midpoint of A C = (\frac{1 + 5}{2}, \frac{- 2 + 10}{2}) = (\frac{6}{2}, \frac{8}{2}) = (3, 4) Midpoint of B D = (\frac{3 + a}{2}, \frac{6 + b}{2}) Therefore, \frac{3 + a}{2} = 3 and \frac{6 + b}{2} = 4 \Rightarrow 3 + a = 6 and 6 + b = 8 \Rightarrow a = 6 - 3 and b = 8 - 6 \Rightarrow a = 3 and b = 2$
Therefore, the fourth vertex is D(3, 2).

Answer:

Let y-axis divides the line segment joining the points (−4, 7) and (3, −7) in the ratio k : 1. Then
$0 = \frac{3 k - 4}{k + 1} \Rightarrow 3 k = 4 \Rightarrow k = \frac{4}{3}$
Hence, the required ratio is 4 : 3.

Question 28:

Let y-axis divides the line segment joining the points (−4, 7) and (3, −7) in the ratio k : 1. Then
$0 = \frac{3 k - 4}{k + 1} \Rightarrow 3 k = 4 \Rightarrow k = \frac{4}{3}$
Hence, the required ratio is 4 : 3.

Answer:

Let the point $P (\frac{1}{2}, y)$ divides the line segment joining the points A(3, −5) and B(−7, 9) in the ratio k : 1. Then
$(\frac{1}{2}, y) = (\frac{k (- 7) + 3}{k + 1}, \frac{k (9) - 5}{k + 1}) \Rightarrow \frac{- 7 k + 3}{k + 1} = \frac{1}{2} and \frac{9 k - 5}{k + 1} = y \Rightarrow k + 1 = - 14 k + 6 \Rightarrow k = \frac{1}{3}$
Now, substituting $k = \frac{1}{3}$ in $\frac{9 k - 5}{k + 1} = y$ , we get
$\frac{\frac{9}{3} - 5}{\frac{1}{3} + 1} = y \Rightarrow y = \frac{9 - 15}{1 + 3} = - \frac{3}{2}$
Hence, required ratio is 1 : 3 and $y = - \frac{3}{2}$ .

Question 29:

Let the point $P (\frac{1}{2}, y)$ divides the line segment joining the points A(3, −5) and B(−7, 9) in the ratio k : 1. Then
$(\frac{1}{2}, y) = (\frac{k (- 7) + 3}{k + 1}, \frac{k (9) - 5}{k + 1}) \Rightarrow \frac{- 7 k + 3}{k + 1} = \frac{1}{2} and \frac{9 k - 5}{k + 1} = y \Rightarrow k + 1 = - 14 k + 6 \Rightarrow k = \frac{1}{3}$
Now, substituting $k = \frac{1}{3}$ in $\frac{9 k - 5}{k + 1} = y$ , we get
$\frac{\frac{9}{3} - 5}{\frac{1}{3} + 1} = y \Rightarrow y = \frac{9 - 15}{1 + 3} = - \frac{3}{2}$
Hence, required ratio is 1 : 3 and $y = - \frac{3}{2}$ .

Answer:

The line segment joining the points A(3, − 3) and B(− 2, 7) is divided by x-axis. Let the required ratio be k : 1. So,
$0 = \frac{k (7) - 3}{k + 1} \Rightarrow k = \frac{3}{7}$
Now
$Point of division = (\frac{k (- 2) + 3}{k + 1}, \frac{k (7) - 3}{k + 1}) = (\frac{\frac{3}{7} \times (- 2) + 3}{\frac{3}{7} + 1}, \frac{\frac{3}{7} \times (7) - 3}{\frac{3}{7} + 1}) (∵ k = \frac{3}{7}) = (\frac{- 6 + 21}{3 + 7}, \frac{21 - 21}{3 + 7}) = (\frac{3}{2}, 0)$
Hence, the required ratio is 3 : 7 and the point of division is $(\frac{3}{2}, 0)$ .

Question 30:

The line segment joining the points A(3, − 3) and B(− 2, 7) is divided by x-axis. Let the required ratio be k : 1. So,
$0 = \frac{k (7) - 3}{k + 1} \Rightarrow k = \frac{3}{7}$
Now
$Point of division = (\frac{k (- 2) + 3}{k + 1}, \frac{k (7) - 3}{k + 1}) = (\frac{\frac{3}{7} \times (- 2) + 3}{\frac{3}{7} + 1}, \frac{\frac{3}{7} \times (7) - 3}{\frac{3}{7} + 1}) (∵ k = \frac{3}{7}) = (\frac{- 6 + 21}{3 + 7}, \frac{21 - 21}{3 + 7}) = (\frac{3}{2}, 0)$
Hence, the required ratio is 3 : 7 and the point of division is $(\frac{3}{2}, 0)$ .

Answer:

Let (x, 0) be the coordinates of R. Then
$0 = \frac{- 4 + x}{2} \Rightarrow x = 4$
Thus, the coordinates of R are (4, 0).
Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then
$P Q = Q R \Rightarrow P Q^{2} = Q R^{2} \Rightarrow {(0 + 4)}^{2} + {(y - 0)}^{2} = 8^{2} \Rightarrow y^{2} = 64 - 16 = 48 \Rightarrow y = \pm 4 \sqrt{3}$
Hence, the required coordinates are $R (4, 0) and P (0, 4 \sqrt{3}) or P (0, - 4 \sqrt{3})$ .

Question 31:

Let (x, 0) be the coordinates of R. Then
$0 = \frac{- 4 + x}{2} \Rightarrow x = 4$
Thus, the coordinates of R are (4, 0).
Here, PQ = QR = PR and the coordinates of P lies on y-axis. Let the coordinates of P be (0, y). Then
$P Q = Q R \Rightarrow P Q^{2} = Q R^{2} \Rightarrow {(0 + 4)}^{2} + {(y - 0)}^{2} = 8^{2} \Rightarrow y^{2} = 64 - 16 = 48 \Rightarrow y = \pm 4 \sqrt{3}$
Hence, the required coordinates are $R (4, 0) and P (0, 4 \sqrt{3}) or P (0, - 4 \sqrt{3})$ .

Answer:

Let (0, y) be the coordinates of B. Then
$0 = \frac{- 3 + y}{2} \Rightarrow y = 3$
Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then
$A B = B C \Rightarrow A B^{2} = B C^{2} \Rightarrow {(x - 0)}^{2} + {(0 - 3)}^{2} = 6^{2} \Rightarrow x^{2} = 36 - 9 = 27 \Rightarrow x = \pm 3 \sqrt{3}$
If the coordinates of point A are $(3 \sqrt{3}, 0)$ , then the coordinates of D are $(- 3 \sqrt{3}, 0)$ .
If the coordinates of point A are $(- 3 \sqrt{3}, 0)$ , then the coordinates of D are $(3 \sqrt{3}, 0)$ .
Hence, the required coordinates are $A (3 \sqrt{3}, 0), B (0, 3) and D (- 3 \sqrt{3}, 0)$ or $A (- 3 \sqrt{3}, 0), B (0, 3) and D (3 \sqrt{3}, 0)$ .

Question 32:

Let (0, y) be the coordinates of B. Then
$0 = \frac{- 3 + y}{2} \Rightarrow y = 3$
Thus, the coordinates of B are (0, 3).
Here, AB = BC = AC and by symmetry the coordinates of A lies on x-axis. Let the coordinates of A be (x, 0). Then
$A B = B C \Rightarrow A B^{2} = B C^{2} \Rightarrow {(x - 0)}^{2} + {(0 - 3)}^{2} = 6^{2} \Rightarrow x^{2} = 36 - 9 = 27 \Rightarrow x = \pm 3 \sqrt{3}$
If the coordinates of point A are $(3 \sqrt{3}, 0)$ , then the coordinates of D are $(- 3 \sqrt{3}, 0)$ .
If the coordinates of point A are $(- 3 \sqrt{3}, 0)$ , then the coordinates of D are $(3 \sqrt{3}, 0)$ .
Hence, the required coordinates are $A (3 \sqrt{3}, 0), B (0, 3) and D (- 3 \sqrt{3}, 0)$ or $A (- 3 \sqrt{3}, 0), B (0, 3) and D (3 \sqrt{3}, 0)$ .

Answer:

Let k be the ratio in which P( −1, y) divides the line segment joining the points A(−3, 10) and B(6, −8). Then
$(- 1, y) = (\frac{k (6) - 3}{k + 1}, \frac{k (- 8) + 10}{k + 1}) \Rightarrow \frac{k (6) - 3}{k + 1} = - 1 and y = \frac{k (- 8) + 10}{k + 1} \Rightarrow k = \frac{2}{7}$
Substituting $k = \frac{2}{7}$ in $y = \frac{k (- 8) + 10}{k + 1}$ , we get
$y = \frac{\frac{- 8 \times 2}{7} + 10}{\frac{2}{7} + 1} = \frac{- 16 + 70}{9} = 6$
Hence, the required ratio is 2 : 7 and y = 6.

Question 33:

Let k be the ratio in which P( −1, y) divides the line segment joining the points A(−3, 10) and B(6, −8). Then
$(- 1, y) = (\frac{k (6) - 3}{k + 1}, \frac{k (- 8) + 10}{k + 1}) \Rightarrow \frac{k (6) - 3}{k + 1} = - 1 and y = \frac{k (- 8) + 10}{k + 1} \Rightarrow k = \frac{2}{7}$
Substituting $k = \frac{2}{7}$ in $y = \frac{k (- 8) + 10}{k + 1}$ , we get
$y = \frac{\frac{- 8 \times 2}{7} + 10}{\frac{2}{7} + 1} = \frac{- 16 + 70}{9} = 6$
Hence, the required ratio is 2 : 7 and y = 6.

Answer:

Here, the points P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Then
$Coordinates of P = (\frac{- 1 - 1}{2}, \frac{- 1 + 4}{2}) = (- 1, \frac{3}{2}) Coordinates of Q = (\frac{- 1 + 5}{2}, \frac{4 + 4}{2}) = (2, 4) Coordinates o f R = (\frac{5 + 5}{2}, \frac{4 - 1}{2}) = (5, \frac{3}{2}) Coordinates of S = (\frac{- 1 + 5}{2}, \frac{- 1 - 1}{2}) = (2, - 1)$
Now
$P Q = \sqrt{{(2 + 1)}^{2} + {(4 - \frac{3}{2})}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2} Q R = \sqrt{{(5 - 2)}^{2} + {(\frac{3}{2} - 4)}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2} R S = \sqrt{{(5 - 2)}^{2} + {(\frac{3}{2} + 1)}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2} S P = \sqrt{{(2 + 1)}^{2} + {(- 1 - \frac{3}{2})}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2} P R = \sqrt{{(5 + 1)}^{2} + {(\frac{3}{2} - \frac{3}{2})}^{2}} = \sqrt{36} = 6 Q S = \sqrt{{(2 - 2)}^{2} + {(- 1 - 4)}^{2}} = \sqrt{25} = 5$
Thus, PQ = QR = RS = SP and $P R \neq Q S$ therefore PQRS is a rhombus.

Question 34:

Here, the points P, Q, R and S are the mid points of AB, BC, CD and DA respectively. Then
$Coordinates of P = (\frac{- 1 - 1}{2}, \frac{- 1 + 4}{2}) = (- 1, \frac{3}{2}) Coordinates of Q = (\frac{- 1 + 5}{2}, \frac{4 + 4}{2}) = (2, 4) Coordinates o f R = (\frac{5 + 5}{2}, \frac{4 - 1}{2}) = (5, \frac{3}{2}) Coordinates of S = (\frac{- 1 + 5}{2}, \frac{- 1 - 1}{2}) = (2, - 1)$
Now
$P Q = \sqrt{{(2 + 1)}^{2} + {(4 - \frac{3}{2})}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2} Q R = \sqrt{{(5 - 2)}^{2} + {(\frac{3}{2} - 4)}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2} R S = \sqrt{{(5 - 2)}^{2} + {(\frac{3}{2} + 1)}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2} S P = \sqrt{{(2 + 1)}^{2} + {(- 1 - \frac{3}{2})}^{2}} = \sqrt{9 + \frac{25}{4}} = \frac{\sqrt{61}}{2} P R = \sqrt{{(5 + 1)}^{2} + {(\frac{3}{2} - \frac{3}{2})}^{2}} = \sqrt{36} = 6 Q S = \sqrt{{(2 - 2)}^{2} + {(- 1 - 4)}^{2}} = \sqrt{25} = 5$
Thus, PQ = QR = RS = SP and $P R \neq Q S$ therefore PQRS is a rhombus.

Answer:

The midpoint of AB is $(\frac{- 10 - 2}{2}, \frac{4 + 0}{2}) = P (- 6, 2)$ .
Let k be the ratio in which P divides CD. So
$(- 6, 2) = (\frac{k (- 4) - 9}{k + 1}, \frac{k (y) - 4}{k + 1}) \Rightarrow \frac{k (- 4) - 9}{k + 1} = - 6 and \frac{k (y) - 4}{k + 1} = 2 \Rightarrow k = \frac{3}{2}$
Now, substituting $k = \frac{3}{2}$ in $\frac{k (y) - 4}{k + 1} = 2$ , we get
$\frac{y \times \frac{3}{2} - 4}{\frac{3}{2} + 1} = 2 \Rightarrow \frac{3 y - 8}{5} = 2 \Rightarrow y = \frac{10 + 8}{3} = 6$
Hence, the required ratio is 3 : 2 and y = 6.

Question 35:

The midpoint of AB is $(\frac{- 10 - 2}{2}, \frac{4 + 0}{2}) = P (- 6, 2)$ .
Let k be the ratio in which P divides CD. So
$(- 6, 2) = (\frac{k (- 4) - 9}{k + 1}, \frac{k (y) - 4}{k + 1}) \Rightarrow \frac{k (- 4) - 9}{k + 1} = - 6 and \frac{k (y) - 4}{k + 1} = 2 \Rightarrow k = \frac{3}{2}$
Now, substituting $k = \frac{3}{2}$ in $\frac{k (y) - 4}{k + 1} = 2$ , we get
$\frac{y \times \frac{3}{2} - 4}{\frac{3}{2} + 1} = 2 \Rightarrow \frac{3 y - 8}{5} = 2 \Rightarrow y = \frac{10 + 8}{3} = 6$
Hence, the required ratio is 3 : 2 and y = 6.

Answer:

Suppose the line intersects the y-axis at P(0, y) and the x-axis at Q(x, 0).

It is given that (2, –5) is the mid-point of PQ.

Using mid-point formula, we have

(\frac{x + 0}{2}, \frac{0 + y}{2}) = (2, - 5) \Rightarrow (\frac{x}{2}, \frac{y}{2}) = (2, - 5) \Rightarrow \frac{x}{2} = 2 and \frac{y}{2} = - 5 \Rightarrow x = 4, y = - 10

Thus, the coordinates of P and Q are (0, −10) and (4, 0), respectively.

Question 36:

Suppose the line intersects the y-axis at P(0, y) and the x-axis at Q(x, 0).

It is given that (2, –5) is the mid-point of PQ.

Using mid-point formula, we have

(\frac{x + 0}{2}, \frac{0 + y}{2}) = (2, - 5) \Rightarrow (\frac{x}{2}, \frac{y}{2}) = (2, - 5) \Rightarrow \frac{x}{2} = 2 and \frac{y}{2} = - 5 \Rightarrow x = 4, y = - 10

Thus, the coordinates of P and Q are (0, −10) and (4, 0), respectively.

Answer:

Let the point P $(\frac{24}{11}, y)$ divides the line PQ in the ratio k : 1.
Then, by the section formula:
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n} The coordinates of R are (\frac{24}{11}, y) . \frac{24}{11} = \frac{3 k + 2}{k + 1}, y = \frac{7 k - 2}{k + 1} \Rightarrow 24 (k + 1) = 33 k + 22, y (k + 1) = 7 k - 2 \Rightarrow 24 k + 24 = 33 k + 22, y k + y = 7 k - 2 \Rightarrow 2 = 9 k \Rightarrow k = \frac{2}{9}$
$Now consider the equation y k + y = 7 k - 2 and put k = \frac{2}{9} . \Rightarrow \frac{2}{9} y + y = \frac{14}{9} - 2 \Rightarrow \frac{11}{9} y = \frac{- 4}{9} \Rightarrow y = \frac{- 4}{11} Therefore, the point R divides the line PQ in the ratio 2 : 9 . And, the coordinates of R are (\frac{24}{11}, \frac{- 4}{11}) .$

Question 37:

Let the point P $(\frac{24}{11}, y)$ divides the line PQ in the ratio k : 1.
Then, by the section formula:
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n} The coordinates of R are (\frac{24}{11}, y) . \frac{24}{11} = \frac{3 k + 2}{k + 1}, y = \frac{7 k - 2}{k + 1} \Rightarrow 24 (k + 1) = 33 k + 22, y (k + 1) = 7 k - 2 \Rightarrow 24 k + 24 = 33 k + 22, y k + y = 7 k - 2 \Rightarrow 2 = 9 k \Rightarrow k = \frac{2}{9}$
$Now consider the equation y k + y = 7 k - 2 and put k = \frac{2}{9} . \Rightarrow \frac{2}{9} y + y = \frac{14}{9} - 2 \Rightarrow \frac{11}{9} y = \frac{- 4}{9} \Rightarrow y = \frac{- 4}{11} Therefore, the point R divides the line PQ in the ratio 2 : 9 . And, the coordinates of R are (\frac{24}{11}, \frac{- 4}{11}) .$

Answer:

Let the coordinates of A, B, C be $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) respectively .$
Because D is the mid-point of BC, using mid-point formula, we have
$\frac{x_{2} + x_{3}}{2} = 3 and \frac{y_{2} + y_{3}}{2} = 4 \Rightarrow x_{2} + x_{3} = 6 and y_{2} + y_{3} = 8 . . . . . (i)$
Similarly, E is the mid point of AC. Using mid-point formula, we have;
$\frac{x_{1} + x_{3}}{2} = 8 and \frac{y_{1} + y_{3}}{2} = 9 \Rightarrow x_{1} + x_{3} = 16 and y_{1} + y_{3} = 18 . . . . . (ii)$
Again, F is the mid point of AB. Using mid point formula, we have
$\frac{x_{1} + x_{2}}{2} = 6 and \frac{y_{1} + y_{2}}{2} = 7 \Rightarrow x_{1} + x_{2} = 12 and y_{1} + y_{2} = 14 . . . . . (iii)$
Adding (i), (ii) and (iii), we get
$2 (x_{1} + x_{2} + x_{3}) = 34 and 2 (y_{1} + y_{2} + y_{3}) = 40 \Rightarrow x_{1} + x_{2} + x_{3} = 17 and y_{1} + y_{2} + y_{3} = 20 . . . . . (iv)$
On solving equation (iv), using equations (i), (ii) and (iii), we get
$x_{1} = 11 x_{2} = 1 x_{3} = 5$
Similarly,
$y_{1} = 12 y_{2} = 2 y_{3} = 6$
Hence, the points are: A(11,12), B(1,2) and C(5,6).

Question 38:

Let the coordinates of A, B, C be $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) respectively .$
Because D is the mid-point of BC, using mid-point formula, we have
$\frac{x_{2} + x_{3}}{2} = 3 and \frac{y_{2} + y_{3}}{2} = 4 \Rightarrow x_{2} + x_{3} = 6 and y_{2} + y_{3} = 8 . . . . . (i)$
Similarly, E is the mid point of AC. Using mid-point formula, we have;
$\frac{x_{1} + x_{3}}{2} = 8 and \frac{y_{1} + y_{3}}{2} = 9 \Rightarrow x_{1} + x_{3} = 16 and y_{1} + y_{3} = 18 . . . . . (ii)$
Again, F is the mid point of AB. Using mid point formula, we have
$\frac{x_{1} + x_{2}}{2} = 6 and \frac{y_{1} + y_{2}}{2} = 7 \Rightarrow x_{1} + x_{2} = 12 and y_{1} + y_{2} = 14 . . . . . (iii)$
Adding (i), (ii) and (iii), we get
$2 (x_{1} + x_{2} + x_{3}) = 34 and 2 (y_{1} + y_{2} + y_{3}) = 40 \Rightarrow x_{1} + x_{2} + x_{3} = 17 and y_{1} + y_{2} + y_{3} = 20 . . . . . (iv)$
On solving equation (iv), using equations (i), (ii) and (iii), we get
$x_{1} = 11 x_{2} = 1 x_{3} = 5$
Similarly,
$y_{1} = 12 y_{2} = 2 y_{3} = 6$
Hence, the points are: A(11,12), B(1,2) and C(5,6).

Answer:

Let ABCD be the parallelogram with two adjacent vertices A(3, 2) and B(−1, 0). Suppose O(2, −5) be the point of intersection of the diagonals AC and BD.
Let C(x₁, y₁) and D(x₂, y₂) be the coordinates of the other vertices of the parallelogram.

We know that the diagonals of the parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.
Using the mid-point formula, we have
$(\frac{x_{1} + 3}{2}, \frac{y_{1} + 2}{2}) = (2, - 5) \Rightarrow \frac{x_{1} + 3}{2} = 2 and \frac{y_{1} + 2}{2} = - 5 \Rightarrow x_{1} + 3 = 4 and y_{1} + 2 = - 10 \Rightarrow x_{1} = 4 - 3 = 1 and y_{1} = - 10 - 2 = - 12$
So, the coordinates of C are (1, −12).
Also,
$(\frac{x_{2} + (- 1)}{2}, \frac{y_{2} + 0}{2}) = (2, - 5) \Rightarrow \frac{x_{2} - 1}{2} = 2 and \frac{y_{2}}{2} = - 5 \Rightarrow x_{2} - 1 = 4 and y_{2} = - 10 \Rightarrow x_{2} = 4 + 1 = 5 and y_{2} = - 10$
So, the coordinates of D are (5, −10).

Question 39:

Let ABCD be the parallelogram with two adjacent vertices A(3, 2) and B(−1, 0). Suppose O(2, −5) be the point of intersection of the diagonals AC and BD.
Let C(x₁, y₁) and D(x₂, y₂) be the coordinates of the other vertices of the parallelogram.

We know that the diagonals of the parallelogram bisect each other. Therefore, O is the mid-point of AC and BD.
Using the mid-point formula, we have
$(\frac{x_{1} + 3}{2}, \frac{y_{1} + 2}{2}) = (2, - 5) \Rightarrow \frac{x_{1} + 3}{2} = 2 and \frac{y_{1} + 2}{2} = - 5 \Rightarrow x_{1} + 3 = 4 and y_{1} + 2 = - 10 \Rightarrow x_{1} = 4 - 3 = 1 and y_{1} = - 10 - 2 = - 12$
So, the coordinates of C are (1, −12).
Also,
$(\frac{x_{2} + (- 1)}{2}, \frac{y_{2} + 0}{2}) = (2, - 5) \Rightarrow \frac{x_{2} - 1}{2} = 2 and \frac{y_{2}}{2} = - 5 \Rightarrow x_{2} - 1 = 4 and y_{2} = - 10 \Rightarrow x_{2} = 4 + 1 = 5 and y_{2} = - 10$
So, the coordinates of D are (5, −10).

Answer:

Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.

Diagonals of a parallelogram bisect each other.

∴ Mid point of AC = Mid point of BD

$Mid point of (x_{1}, y_{1}) and (x_{2}, y_{2}) is (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) . Mid point of A C = (\frac{3 + a}{2}, \frac{1 + b}{2}) Mid point of B D = (\frac{5 + 4}{2}, \frac{1 + 3}{2}) = (\frac{9}{2}, \frac{4}{2}) = (\frac{9}{2}, 2) ∴ (\frac{3 + a}{2}, \frac{1 + b}{2}) = (\frac{9}{2}, 2) \Rightarrow \frac{3 + a}{2} = \frac{9}{2} and \frac{1 + b}{2} = 2 \Rightarrow 3 + a = 9 and 1 + b = 4 \Rightarrow a = 6 and b = 3$

Hence, the values of a and b is 6 and 3, respectively.

Question 40:

Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.

Diagonals of a parallelogram bisect each other.

∴ Mid point of AC = Mid point of BD

$Mid point of (x_{1}, y_{1}) and (x_{2}, y_{2}) is (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) . Mid point of A C = (\frac{3 + a}{2}, \frac{1 + b}{2}) Mid point of B D = (\frac{5 + 4}{2}, \frac{1 + 3}{2}) = (\frac{9}{2}, \frac{4}{2}) = (\frac{9}{2}, 2) ∴ (\frac{3 + a}{2}, \frac{1 + b}{2}) = (\frac{9}{2}, 2) \Rightarrow \frac{3 + a}{2} = \frac{9}{2} and \frac{1 + b}{2} = 2 \Rightarrow 3 + a = 9 and 1 + b = 4 \Rightarrow a = 6 and b = 3$

Hence, the values of a and b is 6 and 3, respectively.

Answer:

Let the points A(2, 1) and B(5, –8) is trisected at the points P(x, y) and Q(a, b).

Thus, AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2.

Section formula: if the point (x, y) divides the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m : n, then the coordinates (x, y) = $(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$

Therefore, using section formula, the coordinates of P are:

$(x, y) = (\frac{1 (5) + 2 (2)}{1 + 2}, \frac{1 (- 8) + 2 (1)}{1 + 2}) \Rightarrow (x, y) = (\frac{5 + 4}{3}, \frac{- 8 + 2}{3}) \Rightarrow (x, y) = (\frac{9}{3}, \frac{- 6}{3}) \Rightarrow (x, y) = (3, - 2)$

Hence, the coordinates of P are (3, –2).

Since, P also lies on the line given by 2x – y + k = 0,
Therefore, (3, –2) satisfies the equation 2x – y + k = 0
$2 (3) - (- 2) + k = 0 \Rightarrow 6 + 2 + k = 0 \Rightarrow k = - 8$

Hence, the values of k is –8.

Question 41:

Let the points A(2, 1) and B(5, –8) is trisected at the points P(x, y) and Q(a, b).

Thus, AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2.

Section formula: if the point (x, y) divides the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio m : n, then the coordinates (x, y) = $(\frac{m x_{2} + n x_{1}}{m + n}, \frac{m y_{2} + n y_{1}}{m + n})$

Therefore, using section formula, the coordinates of P are:

$(x, y) = (\frac{1 (5) + 2 (2)}{1 + 2}, \frac{1 (- 8) + 2 (1)}{1 + 2}) \Rightarrow (x, y) = (\frac{5 + 4}{3}, \frac{- 8 + 2}{3}) \Rightarrow (x, y) = (\frac{9}{3}, \frac{- 6}{3}) \Rightarrow (x, y) = (3, - 2)$

Hence, the coordinates of P are (3, –2).

Since, P also lies on the line given by 2x – y + k = 0,
Therefore, (3, –2) satisfies the equation 2x – y + k = 0
$2 (3) - (- 2) + k = 0 \Rightarrow 6 + 2 + k = 0 \Rightarrow k = - 8$

Hence, the values of k is –8.

Answer:

Section formula: if the point (x, y) divides the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio k : 1, then the coordinates (x, y) = $(\frac{k x_{2} + x_{1}}{k + 1}, \frac{k y_{2} + y_{1}}{k + 1})$

Let the point P(0, y) divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio k : 1.

Therefore, using section formula, the coordinates of P are:

$(0, y) = (\frac{k (- 1) + 1 (5)}{k + 1}, \frac{k (4) + 1 (- 6)}{k + 1}) \Rightarrow (0, y) = (\frac{- k + 5}{k + 1}, \frac{4 k - 6}{k + 1}) \Rightarrow 0 = \frac{- k + 5}{k + 1} and y = \frac{4 k - 6}{k + 1} \Rightarrow 0 = \frac{- k + 5}{k + 1} \Rightarrow - k + 5 = 0 \Rightarrow k = 5 Now, y = \frac{4 k - 6}{k + 1} \Rightarrow y = \frac{4 (5) - 6}{5 + 1} (∵ k = 5) \Rightarrow y = \frac{20 - 6}{6} \Rightarrow y = \frac{14}{6} \Rightarrow y = \frac{7}{3}$

Hence, the y-axis divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio 5 : 1.
and the coordinates of the point of division are $(0, \frac{7}{3})$ .

Question 1:

Section formula: if the point (x, y) divides the line segment joining the points (x₁, y₁) and (x₂, y₂) internally in the ratio k : 1, then the coordinates (x, y) = $(\frac{k x_{2} + x_{1}}{k + 1}, \frac{k y_{2} + y_{1}}{k + 1})$

Let the point P(0, y) divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio k : 1.

Therefore, using section formula, the coordinates of P are:

$(0, y) = (\frac{k (- 1) + 1 (5)}{k + 1}, \frac{k (4) + 1 (- 6)}{k + 1}) \Rightarrow (0, y) = (\frac{- k + 5}{k + 1}, \frac{4 k - 6}{k + 1}) \Rightarrow 0 = \frac{- k + 5}{k + 1} and y = \frac{4 k - 6}{k + 1} \Rightarrow 0 = \frac{- k + 5}{k + 1} \Rightarrow - k + 5 = 0 \Rightarrow k = 5 Now, y = \frac{4 k - 6}{k + 1} \Rightarrow y = \frac{4 (5) - 6}{5 + 1} (∵ k = 5) \Rightarrow y = \frac{20 - 6}{6} \Rightarrow y = \frac{14}{6} \Rightarrow y = \frac{7}{3}$

Hence, the y-axis divides the line segment joining the points A(5, –6) and B(–1, 4) in the ratio 5 : 1.
and the coordinates of the point of division are $(0, \frac{7}{3})$ .

Answer:

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of âˆ†ABC. Then,
(x₁ = 1, y₁ = 2), (x₂ = −2, y₂ = 3) and (x₃ = −3, y₃ = -4)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{1 (3 - (- 4)) + (- 2) (- 4 - 2) + (- 3) (2 - 3)\} = \frac{1}{2} \{1 (3 + 4) - 2 (- 6) - 3 (- 1)\} = \frac{1}{2} \{7 + 12 + 3\} = \frac{1}{2} (22) = 11 sq . units$

(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of âˆ†ABC. Then,
(x₁ = −5, y₁ = 7), (x₂ = −4, y₂ = −5) and (x₃ = 4, y₃ = 5)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{(- 5) (- 5 - 5) + (- 4) (5 - 7) + 4 (7 - (- 5))\} = \frac{1}{2} \{(- 5) (- 10) - 4 (- 2) + 4 (12)\} = \frac{1}{2} \{50 + 8 + 48\} = \frac{1}{2} (106) = 53 sq . units$

(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of âˆ†ABC. Then,
(x₁ = 3, y₁ = 8), (x₂ = −4, y₂ = 2) and (x₃ = 5, y₃ = −1)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{3 (2 - (- 1)) + (- 4) (- 1 - 8) + 5 (8 - 2)\} = \frac{1}{2} \{3 (2 + 1) - 4 (- 9) + 5 (6)\} = \frac{1}{2} \{9 + 36 + 30\} = \frac{1}{2} (75) = 37.5 sq . units$

(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of âˆ†ABC. Then,
(x₁ = 10, y₁ = −6), (x₂ = 2, y₂ = 5) and (x₃ = −1, y₃ = 3)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{10 (5 - 3) + 2 (3 - (- 6)) + (- 1) (- 6 - 5)\} = \frac{1}{2} \{10 (2) + 2 (9) - 1 (- 11)\} = \frac{1}{2} \{20 + 18 + 11\} = \frac{1}{2} (49) = 24.5 sq . units$

Question 2:

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of âˆ†ABC. Then,
(x₁ = 1, y₁ = 2), (x₂ = −2, y₂ = 3) and (x₃ = −3, y₃ = -4)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{1 (3 - (- 4)) + (- 2) (- 4 - 2) + (- 3) (2 - 3)\} = \frac{1}{2} \{1 (3 + 4) - 2 (- 6) - 3 (- 1)\} = \frac{1}{2} \{7 + 12 + 3\} = \frac{1}{2} (22) = 11 sq . units$

(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of âˆ†ABC. Then,
(x₁ = −5, y₁ = 7), (x₂ = −4, y₂ = −5) and (x₃ = 4, y₃ = 5)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{(- 5) (- 5 - 5) + (- 4) (5 - 7) + 4 (7 - (- 5))\} = \frac{1}{2} \{(- 5) (- 10) - 4 (- 2) + 4 (12)\} = \frac{1}{2} \{50 + 8 + 48\} = \frac{1}{2} (106) = 53 sq . units$

(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of âˆ†ABC. Then,
(x₁ = 3, y₁ = 8), (x₂ = −4, y₂ = 2) and (x₃ = 5, y₃ = −1)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{3 (2 - (- 1)) + (- 4) (- 1 - 8) + 5 (8 - 2)\} = \frac{1}{2} \{3 (2 + 1) - 4 (- 9) + 5 (6)\} = \frac{1}{2} \{9 + 36 + 30\} = \frac{1}{2} (75) = 37.5 sq . units$

(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of âˆ†ABC. Then,
(x₁ = 10, y₁ = −6), (x₂ = 2, y₂ = 5) and (x₃ = −1, y₃ = 3)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{10 (5 - 3) + 2 (3 - (- 6)) + (- 1) (- 6 - 5)\} = \frac{1}{2} \{10 (2) + 2 (9) - 1 (- 11)\} = \frac{1}{2} \{20 + 18 + 11\} = \frac{1}{2} (49) = 24.5 sq . units$

Answer:

By joining A and C, we get two triangles ABC and ACD.
Let $A (x_{1}, y_{1}) = A (3, - 1), B (x_{2}, y_{2}) = B (9, - 5), C (x_{3}, y_{3}) = C (14, 0) and D (x_{4}, y_{4}) = D (9, 19)$ . Then
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [3 (- 5 - 0) + 9 (0 + 1) + 14 (- 1 + 5)] = \frac{1}{2} [- 15 + 9 + 56] = 25 sq . units$
$Area of ∆ A C D = \frac{1}{2} [x_{1} (y_{3} - y_{4}) + x_{3} (y_{4} - y_{1}) + x_{4} (y_{1} - y_{3})] = \frac{1}{2} [3 (0 - 19) + 14 (19 + 1) + 9 (- 1 - 0)] = \frac{1}{2} [- 57 + 280 - 9] = 107 sq . units$
So, the area of the quadrilateral is 25 + 107 = 132 sq. units.

Question 3:

By joining A and C, we get two triangles ABC and ACD.
Let $A (x_{1}, y_{1}) = A (3, - 1), B (x_{2}, y_{2}) = B (9, - 5), C (x_{3}, y_{3}) = C (14, 0) and D (x_{4}, y_{4}) = D (9, 19)$ . Then
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [3 (- 5 - 0) + 9 (0 + 1) + 14 (- 1 + 5)] = \frac{1}{2} [- 15 + 9 + 56] = 25 sq . units$
$Area of ∆ A C D = \frac{1}{2} [x_{1} (y_{3} - y_{4}) + x_{3} (y_{4} - y_{1}) + x_{4} (y_{1} - y_{3})] = \frac{1}{2} [3 (0 - 19) + 14 (19 + 1) + 9 (- 1 - 0)] = \frac{1}{2} [- 57 + 280 - 9] = 107 sq . units$
So, the area of the quadrilateral is 25 + 107 = 132 sq. units.

Answer:

By joining P and R, we get two triangles PQR and PRS.
Let $P (x_{1}, y_{1}) = P (- 5, - 3), Q (x_{2}, y_{2}) = Q (- 4, - 6), R (x_{3}, y_{3}) = R (2, - 3) and S (x_{4}, y_{4}) = S (1, 2)$ . Then
$Area of ∆ P Q R = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [- 5 (- 6 + 3) - 4 (- 3 + 3) + 2 (- 3 + 6)] = \frac{1}{2} [15 - 0 + 6] = \frac{21}{2} sq . units$
$Area of ∆ P R S = \frac{1}{2} [x_{1} (y_{3} - y_{4}) + x_{3} (y_{4} - y_{1}) + x_{4} (y_{1} - y_{3})] = \frac{1}{2} [- 5 (- 3 - 2) + 2 (2 + 3) + 1 (- 3 + 3)] = \frac{1}{2} [25 + 10 + 0] = \frac{35}{2} sq . units$
So, the area of the quadrilateral PQRS is $\frac{21}{2} + \frac{35}{2} = 28 sq . units$ sq. units.

Question 4:

By joining P and R, we get two triangles PQR and PRS.
Let $P (x_{1}, y_{1}) = P (- 5, - 3), Q (x_{2}, y_{2}) = Q (- 4, - 6), R (x_{3}, y_{3}) = R (2, - 3) and S (x_{4}, y_{4}) = S (1, 2)$ . Then
$Area of ∆ P Q R = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [- 5 (- 6 + 3) - 4 (- 3 + 3) + 2 (- 3 + 6)] = \frac{1}{2} [15 - 0 + 6] = \frac{21}{2} sq . units$
$Area of ∆ P R S = \frac{1}{2} [x_{1} (y_{3} - y_{4}) + x_{3} (y_{4} - y_{1}) + x_{4} (y_{1} - y_{3})] = \frac{1}{2} [- 5 (- 3 - 2) + 2 (2 + 3) + 1 (- 3 + 3)] = \frac{1}{2} [25 + 10 + 0] = \frac{35}{2} sq . units$
So, the area of the quadrilateral PQRS is $\frac{21}{2} + \frac{35}{2} = 28 sq . units$ sq. units.

Answer:

By joining A and C, we get two triangles ABC and ACD.
Let $A (x_{1}, y_{1}) = A (- 3, - 1), B (x_{2}, y_{2}) = B (- 2, - 4), C (x_{3}, y_{3}) = C (4, - 1) and D (x_{4}, y_{4}) = D (3, 4)$ . Then
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [- 3 (- 4 + 1) - 2 (- 1 + 1) + 4 (- 1 + 4)] = \frac{1}{2} [9 - 0 + 12] = \frac{21}{2} sq . units$
$Area of ∆ A C D = \frac{1}{2} [x_{1} (y_{3} - y_{4}) + x_{3} (y_{4} - y_{1}) + x_{4} (y_{1} - y_{3})] = \frac{1}{2} [- 3 (- 1 - 4) + 4 (4 + 1) + 3 (- 1 + 1)] = \frac{1}{2} [15 + 20 + 0] = \frac{35}{2} sq . units$
So, the area of the quadrilateral ABCD is $\frac{21}{2} + \frac{35}{2} = 28 sq . units$ sq. units.

Question 5:

By joining A and C, we get two triangles ABC and ACD.
Let $A (x_{1}, y_{1}) = A (- 3, - 1), B (x_{2}, y_{2}) = B (- 2, - 4), C (x_{3}, y_{3}) = C (4, - 1) and D (x_{4}, y_{4}) = D (3, 4)$ . Then
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [- 3 (- 4 + 1) - 2 (- 1 + 1) + 4 (- 1 + 4)] = \frac{1}{2} [9 - 0 + 12] = \frac{21}{2} sq . units$
$Area of ∆ A C D = \frac{1}{2} [x_{1} (y_{3} - y_{4}) + x_{3} (y_{4} - y_{1}) + x_{4} (y_{1} - y_{3})] = \frac{1}{2} [- 3 (- 1 - 4) + 4 (4 + 1) + 3 (- 1 + 1)] = \frac{1}{2} [15 + 20 + 0] = \frac{35}{2} sq . units$
So, the area of the quadrilateral ABCD is $\frac{21}{2} + \frac{35}{2} = 28 sq . units$ sq. units.

Answer:

Consider the figure.
Construction: Produce AC by joining points A to C to form two triangles, $∆ ABC and ∆ ADC$ .
In $∆ ABC,$
$x_{1} = - 7, x_{2} = - 6 and x_{3} = - 3; y_{1} = 5, y_{2} = - 7 and y_{3} = - 8$
We know that,
$a r (∆ ABC) = |\frac{1}{2} \{(x_{1}) (y_{2} - y_{3}) + (x_{2}) (y_{3} - y_{1}) + (x_{3}) (y_{1} - y_{2})\}| \Rightarrow a r (∆ ABC) = |\frac{1}{2} \{(- 7) (- 7 + 8) + (- 6) (- 8 - 5) + (- 3) (5 + 7)\}| \Rightarrow a r (∆ ABC) = |\frac{1}{2} \{(- 7) (1) + (- 6) (- 13) + (- 3) (12)\}| \Rightarrow a r (∆ ABC) = |\frac{1}{2} \{(- 7) + 78 + (- 36)\}| \Rightarrow a r (∆ ABC) = |\frac{1}{2} \{35\}| \Rightarrow a r (∆ ABC) = \frac{35}{2} sq . units$
Similarly, in $∆ ADC,$
$x_{1} = - 7, x_{2} = 2, x_{3} = - 3, y_{1} = 5, y_{2} = 3 and y_{3} = - 8$
$\Rightarrow a r (∆ ADC) = |\frac{1}{2} \{(- 7) (3 + 8) + (2) (- 8 - 5) + (- 3) (5 - 3)\}| \Rightarrow a r (∆ ADC) = |\frac{1}{2} \{(- 7) (11) + (2) (- 13) + (- 3) (2)\}| \Rightarrow a r (∆ ADC) = |\frac{1}{2} \{(- 77) - 26 - 6\}| \Rightarrow a r (∆ ADC) = |\frac{1}{2} \{- 109\}| \Rightarrow a r (∆ ADC) = \frac{109}{2} sq . units$
Now, ar(quad. ABCD) = $a r (∆ ABC) + a r (∆ ADC)$

ar(quad. ABCD) = $\frac{35}{2} + \frac{109}{2} = \frac{144}{2} = 72$
Therefore, area of quadrilateral ABCD is 72 sq. units

Disclaimer: The answer thus calculated does not match with the answer given in the book.
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Question 6:

Consider the figure.
Construction: Produce AC by joining points A to C to form two triangles, $∆ ABC and ∆ ADC$ .
In $∆ ABC,$
$x_{1} = - 7, x_{2} = - 6 and x_{3} = - 3; y_{1} = 5, y_{2} = - 7 and y_{3} = - 8$
We know that,
$a r (∆ ABC) = |\frac{1}{2} \{(x_{1}) (y_{2} - y_{3}) + (x_{2}) (y_{3} - y_{1}) + (x_{3}) (y_{1} - y_{2})\}| \Rightarrow a r (∆ ABC) = |\frac{1}{2} \{(- 7) (- 7 + 8) + (- 6) (- 8 - 5) + (- 3) (5 + 7)\}| \Rightarrow a r (∆ ABC) = |\frac{1}{2} \{(- 7) (1) + (- 6) (- 13) + (- 3) (12)\}| \Rightarrow a r (∆ ABC) = |\frac{1}{2} \{(- 7) + 78 + (- 36)\}| \Rightarrow a r (∆ ABC) = |\frac{1}{2} \{35\}| \Rightarrow a r (∆ ABC) = \frac{35}{2} sq . units$
Similarly, in $∆ ADC,$
$x_{1} = - 7, x_{2} = 2, x_{3} = - 3, y_{1} = 5, y_{2} = 3 and y_{3} = - 8$
$\Rightarrow a r (∆ ADC) = |\frac{1}{2} \{(- 7) (3 + 8) + (2) (- 8 - 5) + (- 3) (5 - 3)\}| \Rightarrow a r (∆ ADC) = |\frac{1}{2} \{(- 7) (11) + (2) (- 13) + (- 3) (2)\}| \Rightarrow a r (∆ ADC) = |\frac{1}{2} \{(- 77) - 26 - 6\}| \Rightarrow a r (∆ ADC) = |\frac{1}{2} \{- 109\}| \Rightarrow a r (∆ ADC) = \frac{109}{2} sq . units$
Now, ar(quad. ABCD) = $a r (∆ ABC) + a r (∆ ADC)$

ar(quad. ABCD) = $\frac{35}{2} + \frac{109}{2} = \frac{144}{2} = 72$
Therefore, area of quadrilateral ABCD is 72 sq. units

Disclaimer: The answer thus calculated does not match with the answer given in the book.
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Answer:

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).
$Coordinates of midpoint of A B = P (x_{1}, y_{1}) = (\frac{2 + 4}{2}, \frac{1 + 3}{2}) = (3, 2) Coordinates of midpoint of B C = Q (x_{2}, y_{2}) = (\frac{4 + 2}{2}, \frac{3 + 5}{2}) = (3, 4) Coordinates of midpoint of A C = R (x_{3}, y_{3}) = (\frac{2 + 2}{2}, \frac{1 + 5}{2}) = (2, 3)$
Now
$Area of ∆ P Q R = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [3 (4 - 3) + 3 (3 - 2) + 2 (2 - 4)] = \frac{1}{2} [3 + 3 - 4] = 1 sq . unit$
Hence, the area of the required triangle is 1 sq. unit.

Question 7:

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).
$Coordinates of midpoint of A B = P (x_{1}, y_{1}) = (\frac{2 + 4}{2}, \frac{1 + 3}{2}) = (3, 2) Coordinates of midpoint of B C = Q (x_{2}, y_{2}) = (\frac{4 + 2}{2}, \frac{3 + 5}{2}) = (3, 4) Coordinates of midpoint of A C = R (x_{3}, y_{3}) = (\frac{2 + 2}{2}, \frac{1 + 5}{2}) = (2, 3)$
Now
$Area of ∆ P Q R = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [3 (4 - 3) + 3 (3 - 2) + 2 (2 - 4)] = \frac{1}{2} [3 + 3 - 4] = 1 sq . unit$
Hence, the area of the required triangle is 1 sq. unit.

Answer:

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).
$Coordinates of D = (\frac{5 + 3}{2}, \frac{3 - 1}{2}) = (4, 1)$
For the area of the triangle ADC, let $A (x_{1}, y_{1}) = A (7, - 3), D (x_{2}, y_{2}) = D (4, 1) and C (x_{3}, y_{3}) = C (3, - 1)$ . Then
$Area of ∆ A D C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [7 (1 + 1) + 4 (- 1 + 3) + 3 (- 3 - 1)] = \frac{1}{2} [14 + 8 - 12] = 5 sq . unit$
Now, for the area of triangle ABD, let $A (x_{1}, y_{1}) = A (7, - 3), B (x_{2}, y_{2}) = B (5, 3) and D (x_{3}, y_{3}) = D (4, 1)$ . Then
$Area of ∆ A B D = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [7 (3 - 1) + 5 (1 + 3) + 4 (- 3 - 3)] = \frac{1}{2} [14 + 20 - 24] = 5 sq . unit$
Thus, $Area (∆ A D C) = Area (∆ A B D) = 5 sq . units$ .
Hence, AD divides $∆ A B C$ into two triangles of equal areas.

Question 8:

The vertices of the triangle are A(7, −3), B(5, 3), C(3, −1).
$Coordinates of D = (\frac{5 + 3}{2}, \frac{3 - 1}{2}) = (4, 1)$
For the area of the triangle ADC, let $A (x_{1}, y_{1}) = A (7, - 3), D (x_{2}, y_{2}) = D (4, 1) and C (x_{3}, y_{3}) = C (3, - 1)$ . Then
$Area of ∆ A D C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [7 (1 + 1) + 4 (- 1 + 3) + 3 (- 3 - 1)] = \frac{1}{2} [14 + 8 - 12] = 5 sq . unit$
Now, for the area of triangle ABD, let $A (x_{1}, y_{1}) = A (7, - 3), B (x_{2}, y_{2}) = B (5, 3) and D (x_{3}, y_{3}) = D (4, 1)$ . Then
$Area of ∆ A B D = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [7 (3 - 1) + 5 (1 + 3) + 4 (- 3 - 3)] = \frac{1}{2} [14 + 20 - 24] = 5 sq . unit$
Thus, $Area (∆ A D C) = Area (∆ A B D) = 5 sq . units$ .
Hence, AD divides $∆ A B C$ into two triangles of equal areas.

Answer:

Let $(x_{2}, y_{2}) and (x_{3}, y_{3})$ be the coordinates of B and C respectively. Since, the coordinates of A are (1, −4), therefore
$\frac{1 + x_{2}}{2} = 2 \Rightarrow x_{2} = 3 \frac{- 4 + y_{2}}{2} = - 1 \Rightarrow y_{2} = 2 \frac{1 + x_{3}}{2} = 0 \Rightarrow x_{3} = - 1 \frac{- 4 + y_{3}}{2} = - 1 \Rightarrow y_{3} = 2$
Let $A (x_{1}, y_{1}) = A (1, - 4), B (x_{2}, y_{2}) = B (3, 2) and C (x_{3}, y_{3}) = C (- 1, 2)$ . Now
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [1 (2 - 2) + 3 (2 + 4) - 1 (- 4 - 2)] = \frac{1}{2} [0 + 18 + 6] = 12 sq . units$
Hence, the area of the triangle $∆ A B C$ is 12 sq. units.

Question 9:

Let $(x_{2}, y_{2}) and (x_{3}, y_{3})$ be the coordinates of B and C respectively. Since, the coordinates of A are (1, −4), therefore
$\frac{1 + x_{2}}{2} = 2 \Rightarrow x_{2} = 3 \frac{- 4 + y_{2}}{2} = - 1 \Rightarrow y_{2} = 2 \frac{1 + x_{3}}{2} = 0 \Rightarrow x_{3} = - 1 \frac{- 4 + y_{3}}{2} = - 1 \Rightarrow y_{3} = 2$
Let $A (x_{1}, y_{1}) = A (1, - 4), B (x_{2}, y_{2}) = B (3, 2) and C (x_{3}, y_{3}) = C (- 1, 2)$ . Now
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [1 (2 - 2) + 3 (2 + 4) - 1 (- 4 - 2)] = \frac{1}{2} [0 + 18 + 6] = 12 sq . units$
Hence, the area of the triangle $∆ A B C$ is 12 sq. units.

Answer:

Let (x, y) be the coordinates of D and $(x', y')$ be the coordinates of E. Since, the diagonals of a parallelogram bisect
each other at the same point, therefore
$\frac{x + 8}{2} = \frac{6 + 9}{2} \Rightarrow x = 7 \frac{y + 2}{2} = \frac{1 + 4}{2} \Rightarrow y = 3$
Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore
$x' = \frac{7 + 9}{2} \Rightarrow x' = 8 y' = \frac{3 + 4}{2} \Rightarrow y' = \frac{7}{2}$
Thus, the coordinates of E are $(8, \frac{7}{2})$ .
Let $A (x_{1}, y_{1}) = A (6, 1), E (x_{2}, y_{2}) = E (8, \frac{7}{2}) and D (x_{3}, y_{3}) = D (7, 3)$ . Now
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [6 (\frac{7}{2} - 3) + 8 (3 - 1) + 7 (1 - \frac{7}{2})] = \frac{1}{2} [\frac{3}{2}] = \frac{3}{4} sq . unit$
Hence, the area of the triangle $∆ A D E$ is $\frac{3}{4} sq . units$ .

Question 10:

Let (x, y) be the coordinates of D and $(x', y')$ be the coordinates of E. Since, the diagonals of a parallelogram bisect
each other at the same point, therefore
$\frac{x + 8}{2} = \frac{6 + 9}{2} \Rightarrow x = 7 \frac{y + 2}{2} = \frac{1 + 4}{2} \Rightarrow y = 3$
Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore
$x' = \frac{7 + 9}{2} \Rightarrow x' = 8 y' = \frac{3 + 4}{2} \Rightarrow y' = \frac{7}{2}$
Thus, the coordinates of E are $(8, \frac{7}{2})$ .
Let $A (x_{1}, y_{1}) = A (6, 1), E (x_{2}, y_{2}) = E (8, \frac{7}{2}) and D (x_{3}, y_{3}) = D (7, 3)$ . Now
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [6 (\frac{7}{2} - 3) + 8 (3 - 1) + 7 (1 - \frac{7}{2})] = \frac{1}{2} [\frac{3}{2}] = \frac{3}{4} sq . unit$
Hence, the area of the triangle $∆ A D E$ is $\frac{3}{4} sq . units$ .

Answer:

(i) Let $A (x_{1}, y_{1}) = A (1, - 3), B (x_{2}, y_{2}) = B (4, p) and C (x_{3}, y_{3}) = C (- 9, 7)$ . Now
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \Rightarrow 15 = \frac{1}{2} [1 (p - 7) + 4 (7 + 3) - 9 (- 3 - p)] \Rightarrow 15 = \frac{1}{2} [10 p + 60] \Rightarrow |10 p + 60| = 30$
Therefore
$\Rightarrow 10 p + 60 = - 30 or 30 \Rightarrow 10 p = - 90 or - 30 \Rightarrow p = - 9 or - 3$
Hence, $p = - 9 or p = - 3$ .

(ii)
Let $A (x_{1}, y_{1}) = A (2, 1), B (x_{2}, y_{2}) = B (3, - 2) and C (x_{3}, y_{3}) = C (\frac{7}{2}, y) .$
Now
$Area (∆ A B C) = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| \Rightarrow 5 = \frac{1}{2} |2 (- 2 - y) + 3 (y - 1) + \frac{7}{2} (1 + 2)| \Rightarrow 10 = |- 4 - 2 y + 3 y - 3 + \frac{21}{2}| \Rightarrow 10 = |y + \frac{7}{2}| \Rightarrow 10 = y + \frac{7}{2} or - 10 = y + \frac{7}{2} \Rightarrow y = \frac{13}{2} or y = \frac{- 27}{2}$
Hence, y = $\frac{13}{2} or \frac{- 27}{2} .$

Question 11:

(i) Let $A (x_{1}, y_{1}) = A (1, - 3), B (x_{2}, y_{2}) = B (4, p) and C (x_{3}, y_{3}) = C (- 9, 7)$ . Now
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \Rightarrow 15 = \frac{1}{2} [1 (p - 7) + 4 (7 + 3) - 9 (- 3 - p)] \Rightarrow 15 = \frac{1}{2} [10 p + 60] \Rightarrow |10 p + 60| = 30$
Therefore
$\Rightarrow 10 p + 60 = - 30 or 30 \Rightarrow 10 p = - 90 or - 30 \Rightarrow p = - 9 or - 3$
Hence, $p = - 9 or p = - 3$ .

(ii)
Let $A (x_{1}, y_{1}) = A (2, 1), B (x_{2}, y_{2}) = B (3, - 2) and C (x_{3}, y_{3}) = C (\frac{7}{2}, y) .$
Now
$Area (∆ A B C) = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| \Rightarrow 5 = \frac{1}{2} |2 (- 2 - y) + 3 (y - 1) + \frac{7}{2} (1 + 2)| \Rightarrow 10 = |- 4 - 2 y + 3 y - 3 + \frac{21}{2}| \Rightarrow 10 = |y + \frac{7}{2}| \Rightarrow 10 = y + \frac{7}{2} or - 10 = y + \frac{7}{2} \Rightarrow y = \frac{13}{2} or y = \frac{- 27}{2}$
Hence, y = $\frac{13}{2} or \frac{- 27}{2} .$

Answer:

Let $A (x_{1}, y_{1}) = A (k + 1, 1), B (x_{2}, y_{2}) = B (4, - 3) and C (x_{3}, y_{3}) = C (7, - k)$ . Now
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \Rightarrow 6 = \frac{1}{2} [(k + 1) (- 3 + k) + 4 (- k - 1) + 7 (1 + 3)] \Rightarrow 6 = \frac{1}{2} [k^{2} - 2 k - 3 - 4 k - 4 + 28] \Rightarrow k^{2} - 6 k + 9 = 0$
$\Rightarrow {(k - 3)}^{2} = 0 \Rightarrow k = 3$
Hence, k = 3.

Question 12:

Let $A (x_{1}, y_{1}) = A (k + 1, 1), B (x_{2}, y_{2}) = B (4, - 3) and C (x_{3}, y_{3}) = C (7, - k)$ . Now
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \Rightarrow 6 = \frac{1}{2} [(k + 1) (- 3 + k) + 4 (- k - 1) + 7 (1 + 3)] \Rightarrow 6 = \frac{1}{2} [k^{2} - 2 k - 3 - 4 k - 4 + 28] \Rightarrow k^{2} - 6 k + 9 = 0$
$\Rightarrow {(k - 3)}^{2} = 0 \Rightarrow k = 3$
Hence, k = 3.

Answer:

Let $A (x_{1} = - 2, y_{1} = 5), B (x_{2} = k, y_{2} = - 4) and C (x_{3} = 2 k + 1, y_{3} = 10)$ be the vertices of
the triangle. So
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \Rightarrow 53 = \frac{1}{2} [(- 2) (- 4 - 10) + k (10 - 5) + (2 k + 1) (5 + 4)] \Rightarrow 53 = \frac{1}{2} [28 + 5 k + 9 (2 k + 1)] \Rightarrow 28 + 5 k + 18 k + 9 = 106$

$\Rightarrow 37 + 23 k = 106 \Rightarrow 23 k = 106 - 37 = 69 \Rightarrow k = \frac{69}{23} = 3$
Hence, k = 3.

Question 13:

Let $A (x_{1} = - 2, y_{1} = 5), B (x_{2} = k, y_{2} = - 4) and C (x_{3} = 2 k + 1, y_{3} = 10)$ be the vertices of
the triangle. So
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] \Rightarrow 53 = \frac{1}{2} [(- 2) (- 4 - 10) + k (10 - 5) + (2 k + 1) (5 + 4)] \Rightarrow 53 = \frac{1}{2} [28 + 5 k + 9 (2 k + 1)] \Rightarrow 28 + 5 k + 18 k + 9 = 106$

$\Rightarrow 37 + 23 k = 106 \Rightarrow 23 k = 106 - 37 = 69 \Rightarrow k = \frac{69}{23} = 3$
Hence, k = 3.

Answer:

(i)
Let A(x₁ = 2, y₁ = −2), B(x₂ = −3, y₂ = 8) and C(x₃ = −1, y₃ = 4) be the given points. Now
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 2 (8 - 4) + (- 3) (4 + 2) + (- 1) (- 2 - 8) = 8 - 18 + 10 = 0$
Hence, the given points are collinear.

(ii)
Let A(x₁ = −5, y₁ = 1), B(x₂ = 5, y₂ = 5) and C(x₃ = 10, y₃ = 7) be the given points. Now
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = (- 5) (5 - 7) + 5 (7 - 1) + 10 (1 - 5) = - 5 (- 2) + 5 (6) + 10 (- 4) = 10 + 30 - 40 = 0$
Hence, the given points are collinear.

(iii)
Let A(x₁ = 5, y₁ = 1), B(x₂ = 1, y₂ = −1) and C(x₃ = 11, y₃ = 4) be the given points. Now
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 5 (- 1 - 4) + 1 (4 - 1) + 11 (1 + 1) = - 25 + 3 + 22 = 0$
Hence, the given points are collinear.

(iv)
Let A(x₁ = 8, y₁ = 1), B(x₂ = 3, y₂ = −4) and C(x₃ = 2, y₃ = −5) be the given points. Now
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 8 (- 4 + 5) + 3 (- 5 - 1) + 2 (1 + 4) = 8 - 18 + 10 = 0$
Hence, the given points are collinear.

Question 14:

(i)
Let A(x₁ = 2, y₁ = −2), B(x₂ = −3, y₂ = 8) and C(x₃ = −1, y₃ = 4) be the given points. Now
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 2 (8 - 4) + (- 3) (4 + 2) + (- 1) (- 2 - 8) = 8 - 18 + 10 = 0$
Hence, the given points are collinear.

(ii)
Let A(x₁ = −5, y₁ = 1), B(x₂ = 5, y₂ = 5) and C(x₃ = 10, y₃ = 7) be the given points. Now
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = (- 5) (5 - 7) + 5 (7 - 1) + 10 (1 - 5) = - 5 (- 2) + 5 (6) + 10 (- 4) = 10 + 30 - 40 = 0$
Hence, the given points are collinear.

(iii)
Let A(x₁ = 5, y₁ = 1), B(x₂ = 1, y₂ = −1) and C(x₃ = 11, y₃ = 4) be the given points. Now
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 5 (- 1 - 4) + 1 (4 - 1) + 11 (1 + 1) = - 25 + 3 + 22 = 0$
Hence, the given points are collinear.

(iv)
Let A(x₁ = 8, y₁ = 1), B(x₂ = 3, y₂ = −4) and C(x₃ = 2, y₃ = −5) be the given points. Now
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 8 (- 4 + 5) + 3 (- 5 - 1) + 2 (1 + 4) = 8 - 18 + 10 = 0$
Hence, the given points are collinear.

Answer:

Let $A (x_{1}, y_{1}) = A (x, 2), B (x_{2}, y_{2}) = B (- 3, - 4) and C (x_{3}, y_{3}) = C (7, - 5)$ . So, the condition for three collinear points is
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow x (- 4 + 5) - 3 (- 5 - 2) + 7 (2 + 4) = 0 \Rightarrow x + 21 + 42 = 0 \Rightarrow x = - 63$
Hence, x = − 63.

Question 15:

Let $A (x_{1}, y_{1}) = A (x, 2), B (x_{2}, y_{2}) = B (- 3, - 4) and C (x_{3}, y_{3}) = C (7, - 5)$ . So, the condition for three collinear points is
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow x (- 4 + 5) - 3 (- 5 - 2) + 7 (2 + 4) = 0 \Rightarrow x + 21 + 42 = 0 \Rightarrow x = - 63$
Hence, x = − 63.

Answer:

A(−3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x₁ = −3, y₁ = 12), (x₂ = 7, y₂ = 6) and (x₃ = x, y₃ = 9)
It is given that the points A, B and C are collinear. Therefore,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow (- 3) (6 - 9) + 7 (9 - 12) + x (12 - 6) = 0 \Rightarrow (- 3) (- 3) + 7 (- 3) + x (6) = 0 \Rightarrow 9 - 21 + 6 x = 0 \Rightarrow 6 x - 12 = 0 \Rightarrow 6 x = 12 \Rightarrow x = \frac{12}{6} = 2$
Therefore, when x= 2, the given points are collinear.

Question 16:

A(−3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x₁ = −3, y₁ = 12), (x₂ = 7, y₂ = 6) and (x₃ = x, y₃ = 9)
It is given that the points A, B and C are collinear. Therefore,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow (- 3) (6 - 9) + 7 (9 - 12) + x (12 - 6) = 0 \Rightarrow (- 3) (- 3) + 7 (- 3) + x (6) = 0 \Rightarrow 9 - 21 + 6 x = 0 \Rightarrow 6 x - 12 = 0 \Rightarrow 6 x = 12 \Rightarrow x = \frac{12}{6} = 2$
Therefore, when x= 2, the given points are collinear.

Answer:

If the area of the triangle formed by three points is equal to zero, then the points are collinear.

Area of the triangle formed by the vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is$ $\frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})|$ .

Now, the given points A(–5, 1), B(1, p) and C(4, –2) are collinear.

Therefore, Area of triangle formed by them is equal to zero.

$Area of triangle = 0 \Rightarrow \frac{1}{2} |(- 5) (p - (- 2)) + 1 (- 2 - 1) + 4 (1 - p)| = 0 \Rightarrow \frac{1}{2} |(- 5) (p + 2) + 1 (- 3) + 4 (1 - p)| = 0 \Rightarrow |- 5 p - 10 - 3 + 4 - 4 p| = 0 \Rightarrow |- 9 p - 9| = 0 \Rightarrow - 9 p - 9 = 0 \Rightarrow - 9 p = 9 \Rightarrow p = - 1$

Hence, the value of p is –1.

Question 17:

If the area of the triangle formed by three points is equal to zero, then the points are collinear.

Area of the triangle formed by the vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is$ $\frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})|$ .

Now, the given points A(–5, 1), B(1, p) and C(4, –2) are collinear.

Therefore, Area of triangle formed by them is equal to zero.

$Area of triangle = 0 \Rightarrow \frac{1}{2} |(- 5) (p - (- 2)) + 1 (- 2 - 1) + 4 (1 - p)| = 0 \Rightarrow \frac{1}{2} |(- 5) (p + 2) + 1 (- 3) + 4 (1 - p)| = 0 \Rightarrow |- 5 p - 10 - 3 + 4 - 4 p| = 0 \Rightarrow |- 9 p - 9| = 0 \Rightarrow - 9 p - 9 = 0 \Rightarrow - 9 p = 9 \Rightarrow p = - 1$

Hence, the value of p is –1.

Answer:

Let A(x₁ = −3, y₁ = 9), B(x₂ = 2, y₂ = y) and C(x₃ = 4, y₃ = −5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow (- 3) (y + 5) + 2 (- 5 - 9) + 4 (9 - y) = 0 \Rightarrow - 3 y - 15 - 28 + 36 - 4 y = 0 \Rightarrow 7 y = 36 - 43$
$\Rightarrow y = - 1$

Question 18:

Let A(x₁ = −3, y₁ = 9), B(x₂ = 2, y₂ = y) and C(x₃ = 4, y₃ = −5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow (- 3) (y + 5) + 2 (- 5 - 9) + 4 (9 - y) = 0 \Rightarrow - 3 y - 15 - 28 + 36 - 4 y = 0 \Rightarrow 7 y = 36 - 43$
$\Rightarrow y = - 1$

Answer:

Let A(x₁ = 8, y₁ = 1), B(x₂ = 3, y₂ = −2k) and C(x₃ = k, y₃ = −5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 8 (- 2 k + 5) + 3 (- 5 - 1) + k (1 + 2 k) = 0 \Rightarrow - 16 k + 40 - 18 + k + 2 k^{2} = 0 \Rightarrow 2 k^{2} - 15 k + 22 = 0$
$\Rightarrow 2 k^{2} - 11 k - 4 k + 22 = 0 \Rightarrow k (2 k - 11) - 2 (2 k - 11) = 0 \Rightarrow (k - 2) (2 k - 11) = 0 \Rightarrow k = 2 or k = \frac{11}{2}$
Hence, $k = 2 or k = \frac{11}{2}$ .

Question 19:

Let A(x₁ = 8, y₁ = 1), B(x₂ = 3, y₂ = −2k) and C(x₃ = k, y₃ = −5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 8 (- 2 k + 5) + 3 (- 5 - 1) + k (1 + 2 k) = 0 \Rightarrow - 16 k + 40 - 18 + k + 2 k^{2} = 0 \Rightarrow 2 k^{2} - 15 k + 22 = 0$
$\Rightarrow 2 k^{2} - 11 k - 4 k + 22 = 0 \Rightarrow k (2 k - 11) - 2 (2 k - 11) = 0 \Rightarrow (k - 2) (2 k - 11) = 0 \Rightarrow k = 2 or k = \frac{11}{2}$
Hence, $k = 2 or k = \frac{11}{2}$ .

Answer:

Let A(x₁ = 2, y₁ = 1), B(x₂ = x, y₂ = y) and C(x₃ = 7, y₃ = 5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 2 (y - 5) + x (5 - 1) + 7 (1 - y) = 0 \Rightarrow 2 y - 10 + 4 x + 7 - 7 y = 0 \Rightarrow 4 x - 5 y - 3 = 0$
Hence, the required relation is 4x − 5y − 3 = 0.

Question 20:

Let A(x₁ = 2, y₁ = 1), B(x₂ = x, y₂ = y) and C(x₃ = 7, y₃ = 5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 2 (y - 5) + x (5 - 1) + 7 (1 - y) = 0 \Rightarrow 2 y - 10 + 4 x + 7 - 7 y = 0 \Rightarrow 4 x - 5 y - 3 = 0$
Hence, the required relation is 4x − 5y − 3 = 0.

Answer:

Let A(x₁ = x, y₁ = y), B(x₂ = −5, y₂ = 7) and C(x₃ = −4, y₃ = 5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow x (7 - 5) + (- 5) (5 - y) + (- 4) (y - 7) = 0 \Rightarrow 7 x - 5 x - 25 + 5 y - 4 y + 28 = 0 \Rightarrow 2 x + y + 3 = 0$
Hence, the required relation is 2x + y + 3 = 0.

Question 21:

Let A(x₁ = x, y₁ = y), B(x₂ = −5, y₂ = 7) and C(x₃ = −4, y₃ = 5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow x (7 - 5) + (- 5) (5 - y) + (- 4) (y - 7) = 0 \Rightarrow 7 x - 5 x - 25 + 5 y - 4 y + 28 = 0 \Rightarrow 2 x + y + 3 = 0$
Hence, the required relation is 2x + y + 3 = 0.

Answer:

Consider the points A(a, 0), B(0, b) and C(1, 1).
Here, (x₁ = a, y₁ = 0), (x₂ = 0, y₂ = b) and (x₃ = 1, y₃ = 1).
It is given that the points are collinear. So,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow a (b - 1) + 0 (1 - 0) + 1 (0 - b) = 0 \Rightarrow a b - a - b = o Divid ing the equation by a b : \Rightarrow 1 - \frac{1}{b} - \frac{1}{a} = 0 \Rightarrow 1 - (\frac{1}{a} + \frac{1}{b}) = 0 \Rightarrow (\frac{1}{a} + \frac{1}{b}) = 1$
Therefore, the given points are collinear if $(\frac{1}{a} + \frac{1}{b})$ = 1.

Question 22:

Consider the points A(a, 0), B(0, b) and C(1, 1).
Here, (x₁ = a, y₁ = 0), (x₂ = 0, y₂ = b) and (x₃ = 1, y₃ = 1).
It is given that the points are collinear. So,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow a (b - 1) + 0 (1 - 0) + 1 (0 - b) = 0 \Rightarrow a b - a - b = o Divid ing the equation by a b : \Rightarrow 1 - \frac{1}{b} - \frac{1}{a} = 0 \Rightarrow 1 - (\frac{1}{a} + \frac{1}{b}) = 0 \Rightarrow (\frac{1}{a} + \frac{1}{b}) = 1$
Therefore, the given points are collinear if $(\frac{1}{a} + \frac{1}{b})$ = 1.

Answer:

Let A(x₁ = −3, y₁ = 9), B(x₂ = a, y₂ = b) and C(x₃ = 4, y₃ = −5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow (- 3) (b + 5) + a (- 5 - 9) + 4 (9 - b) = 0 \Rightarrow - 3 b - 15 - 14 a + 36 - 4 b = 0 \Rightarrow 2 a + b = 3$
Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = −1.
Hence, a = 2 and b = −1.

Question 23:

Let A(x₁ = −3, y₁ = 9), B(x₂ = a, y₂ = b) and C(x₃ = 4, y₃ = −5) be the given points.
The given points are collinear if
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow (- 3) (b + 5) + a (- 5 - 9) + 4 (9 - b) = 0 \Rightarrow - 3 b - 15 - 14 a + 36 - 4 b = 0 \Rightarrow 2 a + b = 3$
Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = −1.
Hence, a = 2 and b = −1.

Answer:

Let A(x₁ = 0, y₁ = −1), B(x₂ = 2, y₂ = 1) and C(x₃ = 0, y₃ = 3) be the given points. Then
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [0 (1 - 3) + 2 (3 + 1) + 0 (- 1 - 1)] = \frac{1}{2} \times 8 = 4 sq . units$
So, the area of the triangle $∆ A B C$ is 4 sq. units.
Let D(a₁, b₁), E(a₂, b₂) and F(a₃, b₃) be the midpoints of AB, BC and AC respectively. Then
$a_{1} = \frac{0 + 2}{2} = 1 b_{1} = \frac{- 1 + 1}{2} = 0 a_{2} = \frac{2 + 0}{2} = 1 b_{2} = \frac{1 + 3}{2} = 2 a_{3} = \frac{0 + 0}{2} = 0 b_{3} = \frac{- 1 + 3}{2} = 1$
Thus, the coordinates of D, E and F are D(a₁ = 1, b₁ = 0), E(a₂ = 1, b₂ = 2) and F(a₃ = 0, b₃ = 1). Now
$Area (∆ D E F) = \frac{1}{2} [a_{1} (b_{2} - b_{3}) + a_{2} (b_{3} - b_{1}) + a_{3} (b_{1} - b_{2})] = \frac{1}{2} [1 (2 - 1) + 1 (1 - 0) + 0 (0 - 2)] = \frac{1}{2} [1 + 1 + 0] = 1 sq . unit$
So, the area of the triangle $∆ D E F$ is 1 sq. unit.
Hence, $∆ A B C : ∆ D E F = 4 : 1$ .

Question 24:

Let A(x₁ = 0, y₁ = −1), B(x₂ = 2, y₂ = 1) and C(x₃ = 0, y₃ = 3) be the given points. Then
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [0 (1 - 3) + 2 (3 + 1) + 0 (- 1 - 1)] = \frac{1}{2} \times 8 = 4 sq . units$
So, the area of the triangle $∆ A B C$ is 4 sq. units.
Let D(a₁, b₁), E(a₂, b₂) and F(a₃, b₃) be the midpoints of AB, BC and AC respectively. Then
$a_{1} = \frac{0 + 2}{2} = 1 b_{1} = \frac{- 1 + 1}{2} = 0 a_{2} = \frac{2 + 0}{2} = 1 b_{2} = \frac{1 + 3}{2} = 2 a_{3} = \frac{0 + 0}{2} = 0 b_{3} = \frac{- 1 + 3}{2} = 1$
Thus, the coordinates of D, E and F are D(a₁ = 1, b₁ = 0), E(a₂ = 1, b₂ = 2) and F(a₃ = 0, b₃ = 1). Now
$Area (∆ D E F) = \frac{1}{2} [a_{1} (b_{2} - b_{3}) + a_{2} (b_{3} - b_{1}) + a_{3} (b_{1} - b_{2})] = \frac{1}{2} [1 (2 - 1) + 1 (1 - 0) + 0 (0 - 2)] = \frac{1}{2} [1 + 1 + 0] = 1 sq . unit$
So, the area of the triangle $∆ D E F$ is 1 sq. unit.
Hence, $∆ A B C : ∆ D E F = 4 : 1$ .

Answer:

Let A(a, a²), B(b, b²) and C(0, 0) be the coordinates of the given points.
We know that the area of triangle having vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})$ is $|\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]|$ square units.
So,
Area of âˆ†ABC
$= |\frac{1}{2} [a (b^{2} - 0) + b (0 - a^{2}) + 0 (a^{2} - b^{2})]| = |\frac{1}{2} (a b^{2} - a^{2} b)| = \frac{1}{2} |a b (b - a)| \neq 0 (∵ a \neq b \neq 0)$
Since the area of the triangle formed by the points (a, a²), (b, b²) and (0, 0) is not zero, so the given points are not collinear.

Question 25:

Let A(a, a²), B(b, b²) and C(0, 0) be the coordinates of the given points.
We know that the area of triangle having vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})$ is $|\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]|$ square units.
So,
Area of âˆ†ABC
$= |\frac{1}{2} [a (b^{2} - 0) + b (0 - a^{2}) + 0 (a^{2} - b^{2})]| = |\frac{1}{2} (a b^{2} - a^{2} b)| = \frac{1}{2} |a b (b - a)| \neq 0 (∵ a \neq b \neq 0)$
Since the area of the triangle formed by the points (a, a²), (b, b²) and (0, 0) is not zero, so the given points are not collinear.

Answer:

Area of the triangle formed by the vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is$ $\frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})|$ .

Now, the given vertices are (x, 3), (4, 4) and (3, 5)
and the given area is 4 square units.

Therefore,
$Area of triangle = \frac{1}{2} |x (4 - 5) + 4 (5 - 3) + 3 (3 - 4)| \Rightarrow 4 = \frac{1}{2} |x (- 1) + (4) (2) + (3) (- 1)| \Rightarrow 4 = \frac{1}{2} |- x + 8 - 3| \Rightarrow 4 = \frac{1}{2} |- x + 5| \Rightarrow 8 = |- x + 5| \Rightarrow - x + 5 = 8 or - x + 5 = - 8 \Rightarrow - x = 3 or - x = - 13 \Rightarrow x = - 3 or x = 13$

Hence, â€‹the value of x is 13 and −3.

Question 1:

Area of the triangle formed by the vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is$ $\frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})|$ .

Now, the given vertices are (x, 3), (4, 4) and (3, 5)
and the given area is 4 square units.

Therefore,
$Area of triangle = \frac{1}{2} |x (4 - 5) + 4 (5 - 3) + 3 (3 - 4)| \Rightarrow 4 = \frac{1}{2} |x (- 1) + (4) (2) + (3) (- 1)| \Rightarrow 4 = \frac{1}{2} |- x + 8 - 3| \Rightarrow 4 = \frac{1}{2} |- x + 5| \Rightarrow 8 = |- x + 5| \Rightarrow - x + 5 = 8 or - x + 5 = - 8 \Rightarrow - x = 3 or - x = - 13 \Rightarrow x = - 3 or x = 13$

Hence, â€‹the value of x is 13 and −3.

Answer:

The given points are A(−1, y), B(5, 7) and O(2, −3y).
Here, AO and BO are the radii of the circle. So
$A O = B O \Rightarrow A O^{2} = B O^{2} \Rightarrow {(2 + 1)}^{2} + {(- 3 y - y)}^{2} = {(2 - 5)}^{2} + {(- 3 y - 7)}^{2} \Rightarrow 9 + {(4 y)}^{2} = {(- 3)}^{2} + {(3 y + 7)}^{2} \Rightarrow 9 + 16 y^{2} = 9 + 9 y^{2} + 49 + 42 y$
$\Rightarrow 7 y^{2} - 42 y - 49 = 0 \Rightarrow y^{2} - 6 y - 7 = 0 \Rightarrow y^{2} - 7 y + y - 7 = 0 \Rightarrow y (y - 7) + 1 (y - 7) = 0$
$\Rightarrow (y - 7) (y + 1) = 0 \Rightarrow y = - 1 or y = 7$
Hence, y = 7 or y = −1.

Question 2:

The given points are A(−1, y), B(5, 7) and O(2, −3y).
Here, AO and BO are the radii of the circle. So
$A O = B O \Rightarrow A O^{2} = B O^{2} \Rightarrow {(2 + 1)}^{2} + {(- 3 y - y)}^{2} = {(2 - 5)}^{2} + {(- 3 y - 7)}^{2} \Rightarrow 9 + {(4 y)}^{2} = {(- 3)}^{2} + {(3 y + 7)}^{2} \Rightarrow 9 + 16 y^{2} = 9 + 9 y^{2} + 49 + 42 y$
$\Rightarrow 7 y^{2} - 42 y - 49 = 0 \Rightarrow y^{2} - 6 y - 7 = 0 \Rightarrow y^{2} - 7 y + y - 7 = 0 \Rightarrow y (y - 7) + 1 (y - 7) = 0$
$\Rightarrow (y - 7) (y + 1) = 0 \Rightarrow y = - 1 or y = 7$
Hence, y = 7 or y = −1.

Answer:

The given points are A(0, 2), B(3, p) and C(p, 5).
$A B = A C \Rightarrow A B^{2} = A C^{2} \Rightarrow {(3 - 0)}^{2} + {(p - 2)}^{2} = {(p - 0)}^{2} + {(5 - 2)}^{2} \Rightarrow 9 + p^{2} - 4 p + 4 = p^{2} + 9 \Rightarrow 4 p = 4 \Rightarrow p = 1$
Hence, p = 1.

Question 3:

The given points are A(0, 2), B(3, p) and C(p, 5).
$A B = A C \Rightarrow A B^{2} = A C^{2} \Rightarrow {(3 - 0)}^{2} + {(p - 2)}^{2} = {(p - 0)}^{2} + {(5 - 2)}^{2} \Rightarrow 9 + p^{2} - 4 p + 4 = p^{2} + 9 \Rightarrow 4 p = 4 \Rightarrow p = 1$
Hence, p = 1.

Answer:

The given vertices are B(4, 0), C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So
$B D = \sqrt{{(4 - 0)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$
Hence, the length of the diagonal is 5 units..

Question 4:

The given vertices are B(4, 0), C(4, 3) and D(0, 3). Here, BD is one of the diagonals. So
$B D = \sqrt{{(4 - 0)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5$
Hence, the length of the diagonal is 5 units..

Answer:

The given points are P(k − 1, 2), A(3, k) and B(k, 5).
$∵ A P = B P ∴ A P^{2} = B P^{2} \Rightarrow {(k - 1 - 3)}^{2} + {(2 - k)}^{2} = {(k - 1 - k)}^{2} + {(2 - 5)}^{2} \Rightarrow {(k - 4)}^{2} + {(2 - k)}^{2} = {(- 1)}^{2} + {(- 3)}^{2}$
$\Rightarrow k^{2} - 8 y + 16 + 4 + k^{2} - 4 k = 1 + 9 \Rightarrow k^{2} - 6 y + 5 = 0 \Rightarrow (k - 1) (k - 5) = 0 \Rightarrow k = 1 or k = 5$
Hence, k = 1 or k = 5.

Question 5:

The given points are P(k − 1, 2), A(3, k) and B(k, 5).
$∵ A P = B P ∴ A P^{2} = B P^{2} \Rightarrow {(k - 1 - 3)}^{2} + {(2 - k)}^{2} = {(k - 1 - k)}^{2} + {(2 - 5)}^{2} \Rightarrow {(k - 4)}^{2} + {(2 - k)}^{2} = {(- 1)}^{2} + {(- 3)}^{2}$
$\Rightarrow k^{2} - 8 y + 16 + 4 + k^{2} - 4 k = 1 + 9 \Rightarrow k^{2} - 6 y + 5 = 0 \Rightarrow (k - 1) (k - 5) = 0 \Rightarrow k = 1 or k = 5$
Hence, k = 1 or k = 5.

Answer:

Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x₁ = 12, y₁ = 5) and B(x₂ = 4, y₂ = −3). Then
$x = \frac{k \times 4 + 12}{k + 1} and 2 = \frac{k \times (- 3) + 5}{k + 1}$
Now
$2 = \frac{k \times (- 3) + 5}{k + 1} \Rightarrow 2 k + 2 = - 3 k + 5 \Rightarrow k = \frac{3}{5}$
Hence, the required ratio is 3 : 5.

Question 6:

Let k be the ratio in which the point P(x, 2) divides the line joining the points A(x₁ = 12, y₁ = 5) and B(x₂ = 4, y₂ = −3). Then
$x = \frac{k \times 4 + 12}{k + 1} and 2 = \frac{k \times (- 3) + 5}{k + 1}$
Now
$2 = \frac{k \times (- 3) + 5}{k + 1} \Rightarrow 2 k + 2 = - 3 k + 5 \Rightarrow k = \frac{3}{5}$
Hence, the required ratio is 3 : 5.

Answer:

The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). Now
$Coordinates of midpoint of A C = (\frac{2 + 5}{2}, \frac{- 1 + 6}{2}) = (\frac{7}{2}, \frac{5}{2}) Coordinates of midpoint of B D = (\frac{5 + 2}{2}, \frac{- 1 + 6}{2}) = (\frac{7}{2}, \frac{5}{2})$
Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

Question 7:

The vertices of the rectangle ABCD are A(2, −1), B(5, −1), C(5, 6) and D(2, 6). Now
$Coordinates of midpoint of A C = (\frac{2 + 5}{2}, \frac{- 1 + 6}{2}) = (\frac{7}{2}, \frac{5}{2}) Coordinates of midpoint of B D = (\frac{5 + 2}{2}, \frac{- 1 + 6}{2}) = (\frac{7}{2}, \frac{5}{2})$
Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

Answer:

The given vertices are A(7, −3), B(5, 3) and C(3, −1).
Since D and E are the midpoints of BC and AC respectively, therefore
$Coordinates of D = (\frac{5 + 3}{2}, \frac{3 - 1}{2}) = (4, 1) Coordinates of E = (\frac{7 + 3}{2}, \frac{- 3 - 1}{2}) = (5, - 2)$
Now
$A D = \sqrt{{(7 - 4)}^{2} + {(- 3 - 1)}^{2}} = \sqrt{9 + 16} = 5 B E = \sqrt{{(5 - 5)}^{2} + {(3 + 2)}^{2}} = \sqrt{0 + 25} = 5$
Hence, AD = BE = 5 units.

Question 8:

The given vertices are A(7, −3), B(5, 3) and C(3, −1).
Since D and E are the midpoints of BC and AC respectively, therefore
$Coordinates of D = (\frac{5 + 3}{2}, \frac{3 - 1}{2}) = (4, 1) Coordinates of E = (\frac{7 + 3}{2}, \frac{- 3 - 1}{2}) = (5, - 2)$
Now
$A D = \sqrt{{(7 - 4)}^{2} + {(- 3 - 1)}^{2}} = \sqrt{9 + 16} = 5 B E = \sqrt{{(5 - 5)}^{2} + {(3 + 2)}^{2}} = \sqrt{0 + 25} = 5$
Hence, AD = BE = 5 units.

Answer:

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3. So
$k = \frac{2 \times 5 + 3 \times 2}{2 + 3} = \frac{10 + 6}{5} = \frac{16}{5}$
Hence, $k = \frac{16}{5}$ .

Question 9:

Here, the point C(k, 4) divides the join of A(2, 6) and B(5, 1) in the ratio 2 : 3. So
$k = \frac{2 \times 5 + 3 \times 2}{2 + 3} = \frac{10 + 6}{5} = \frac{16}{5}$
Hence, $k = \frac{16}{5}$ .

Answer:

Let P(x, 0) be the point on x-axis. Then
$A P = B P \Rightarrow A P^{2} = B P^{2} \Rightarrow {(x + 1)}^{2} + {(0 - 0)}^{2} = {(x - 5)}^{2} + {(0 - 0)}^{2} \Rightarrow x^{2} + 2 x + 1 = x^{2} - 10 x + 25 \Rightarrow 12 x = 24 \Rightarrow x = 2$
Hence, x = 2.

Question 10:

Let P(x, 0) be the point on x-axis. Then
$A P = B P \Rightarrow A P^{2} = B P^{2} \Rightarrow {(x + 1)}^{2} + {(0 - 0)}^{2} = {(x - 5)}^{2} + {(0 - 0)}^{2} \Rightarrow x^{2} + 2 x + 1 = x^{2} - 10 x + 25 \Rightarrow 12 x = 24 \Rightarrow x = 2$
Hence, x = 2.

Answer:

The given points are $A (\frac{- 8}{5}, 2) and B (\frac{2}{5}, 2)$ .
Then, $(x_{1} = \frac{- 8}{5}, y_{1} = 2) and (x_{2} = \frac{2}{5}, y_{2} = 2)$
Therefore,
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{\{\frac{2}{5} - (\frac{- 8}{5})\}}^{2} + {(2 - 2)}^{2}} = \sqrt{{(2)}^{2} + {(0)}^{2}} = \sqrt{4 + 0} = \sqrt{4} = 2 units$

Question 11:

The given points are $A (\frac{- 8}{5}, 2) and B (\frac{2}{5}, 2)$ .
Then, $(x_{1} = \frac{- 8}{5}, y_{1} = 2) and (x_{2} = \frac{2}{5}, y_{2} = 2)$
Therefore,
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{\{\frac{2}{5} - (\frac{- 8}{5})\}}^{2} + {(2 - 2)}^{2}} = \sqrt{{(2)}^{2} + {(0)}^{2}} = \sqrt{4 + 0} = \sqrt{4} = 2 units$

Answer:

$The point (3, a) lies on the line 2 x - 3 y = 5 . If point (3, a) lies on the line 2 x - 3 y = 5, then 2 x - 3 y = 5 \Rightarrow (2 \times 3) - (3 \times a) = 5$
$\Rightarrow 6 - 3 a = 5 \Rightarrow 3 a = 1 \Rightarrow a = \frac{1}{3}$
$Hence, the value of a is \frac{1}{3}$ .

Question 12:

$The point (3, a) lies on the line 2 x - 3 y = 5 . If point (3, a) lies on the line 2 x - 3 y = 5, then 2 x - 3 y = 5 \Rightarrow (2 \times 3) - (3 \times a) = 5$
$\Rightarrow 6 - 3 a = 5 \Rightarrow 3 a = 1 \Rightarrow a = \frac{1}{3}$
$Hence, the value of a is \frac{1}{3}$ .

Answer:

$The given points A (4, 3) and B (x, 5) lie on the circle with centre O (2, 3) . Then, OA = OB \Rightarrow \sqrt{{(x - 2)}^{2} + {(5 - 3)}^{2}} = \sqrt{{(4 - 2)}^{2} + {(3 - 3)}^{2}} \Rightarrow {(x - 2)}^{2} + 2^{2} = 2^{2} + 0^{2} \Rightarrow {(x - 2)}^{2} = (2^{2} - 2^{2}) \Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$
$Hence, the value of x = 2$ .

Question 13:

$The given points A (4, 3) and B (x, 5) lie on the circle with centre O (2, 3) . Then, OA = OB \Rightarrow \sqrt{{(x - 2)}^{2} + {(5 - 3)}^{2}} = \sqrt{{(4 - 2)}^{2} + {(3 - 3)}^{2}} \Rightarrow {(x - 2)}^{2} + 2^{2} = 2^{2} + 0^{2} \Rightarrow {(x - 2)}^{2} = (2^{2} - 2^{2}) \Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$
$Hence, the value of x = 2$ .

Answer:

Let the point $P (x, y)$ be equidistant from the points A(7, 1) and B(3, 5).
Then,
$P A = P B \Rightarrow P A^{2} = P B^{2} \Rightarrow {(x - 7)}^{2} + {(y - 1)}^{2} = {(x - 3)}^{2} + {(y - 5)}^{2} \Rightarrow x^{2} + y^{2} - 14 x - 2 y + 50 = x^{2} + y^{2} - 6 x - 10 y + 34 \Rightarrow 8 x - 8 y = 16 \Rightarrow x - y = 2$

Question 14:

Let the point $P (x, y)$ be equidistant from the points A(7, 1) and B(3, 5).
Then,
$P A = P B \Rightarrow P A^{2} = P B^{2} \Rightarrow {(x - 7)}^{2} + {(y - 1)}^{2} = {(x - 3)}^{2} + {(y - 5)}^{2} \Rightarrow x^{2} + y^{2} - 14 x - 2 y + 50 = x^{2} + y^{2} - 6 x - 10 y + 34 \Rightarrow 8 x - 8 y = 16 \Rightarrow x - y = 2$

Answer:

The given points are A(a, b), B(b, c) and C(c, a).
Here,
$(x_{1} = a, y_{1} = b), (x_{2} = b, y_{2} = c) and (x_{3} = c, y_{3} = a)$
Let the centroid be (x, y).
Then,
$x = \frac{1}{3} (x_{1} + x_{2} + x_{3}) = \frac{1}{3} (a + b + c) = \frac{a + b + c}{3} y = \frac{1}{3} (y_{1} + y_{2} + y_{3}) = \frac{1}{3} (b + c + a) = \frac{a + b + c}{3}$
But it is given that the centroid of the triangle is the origin.
Then, we have:
$\frac{a + b + c}{3} = 0 \Rightarrow a + b + c = 0$

Question 15:

The given points are A(a, b), B(b, c) and C(c, a).
Here,
$(x_{1} = a, y_{1} = b), (x_{2} = b, y_{2} = c) and (x_{3} = c, y_{3} = a)$
Let the centroid be (x, y).
Then,
$x = \frac{1}{3} (x_{1} + x_{2} + x_{3}) = \frac{1}{3} (a + b + c) = \frac{a + b + c}{3} y = \frac{1}{3} (y_{1} + y_{2} + y_{3}) = \frac{1}{3} (b + c + a) = \frac{a + b + c}{3}$
But it is given that the centroid of the triangle is the origin.
Then, we have:
$\frac{a + b + c}{3} = 0 \Rightarrow a + b + c = 0$

Answer:

The given points are A(2, 2), B(−4, −4) and C(5, −8).
Here, $(x_{1} = 2, y_{1} = 2), (x_{2} = - 4, y_{2} = - 4) and (x_{3} = 5, y_{3} = - 8)$
Let G(x, y) be the centroid of $∆ A B C$ . Then,
$x = \frac{1}{3} (x_{1} + x_{2} + x_{3}) = \frac{1}{3} (2 - 4 + 5) = 1$
$y = \frac{1}{3} (y_{1} + y_{2} + y_{3}) = \frac{1}{3} (2 - 4 - 8) = \frac{- 10}{3}$
Hence, the centroid of $∆ A B C is G (1, \frac{- 10}{3})$ .

Question 16:

The given points are A(2, 2), B(−4, −4) and C(5, −8).
Here, $(x_{1} = 2, y_{1} = 2), (x_{2} = - 4, y_{2} = - 4) and (x_{3} = 5, y_{3} = - 8)$
Let G(x, y) be the centroid of $∆ A B C$ . Then,
$x = \frac{1}{3} (x_{1} + x_{2} + x_{3}) = \frac{1}{3} (2 - 4 + 5) = 1$
$y = \frac{1}{3} (y_{1} + y_{2} + y_{3}) = \frac{1}{3} (2 - 4 - 8) = \frac{- 10}{3}$
Hence, the centroid of $∆ A B C is G (1, \frac{- 10}{3})$ .

Answer:

Let the required ratio be $k : 1$ .
Then, by section formula, the coordinates of C are
$C (\frac{7 k + 2}{k + 1}, \frac{8 k + 3}{k + 1})$
Therefore,
$\frac{7 k + 2}{k + 1} = 4 and \frac{8 k + 3}{k + 1} = 5 [∵ C (4, 5) is given] \Rightarrow 7 k + 2 = 4 k + 4 and 8 k + 3 = 5 k + 5 \Rightarrow 3 k = 2 and 3 k = 2$
$\Rightarrow k = \frac{2}{3} in each case$
$So, the required ratio is \frac{2}{3} : 1, which is same as 2:3.$

Question 17:

Let the required ratio be $k : 1$ .
Then, by section formula, the coordinates of C are
$C (\frac{7 k + 2}{k + 1}, \frac{8 k + 3}{k + 1})$
Therefore,
$\frac{7 k + 2}{k + 1} = 4 and \frac{8 k + 3}{k + 1} = 5 [∵ C (4, 5) is given] \Rightarrow 7 k + 2 = 4 k + 4 and 8 k + 3 = 5 k + 5 \Rightarrow 3 k = 2 and 3 k = 2$
$\Rightarrow k = \frac{2}{3} in each case$
$So, the required ratio is \frac{2}{3} : 1, which is same as 2:3.$

Answer:

The given points are $A (2, 3), B (4, k) and C (6, - 3)$ .
Here, $(x_{1} = 2, y_{1} = 3), (x_{2} = 4, y_{2} = k) and (x_{3} = 6, y_{3} = - 3)$
It is given that the points A, B and Câ€‹ are collinear. Then,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 2 (k + 3) + 4 (- 3 - 3) + 6 (3 - k) = 0 \Rightarrow 2 k + 6 - 24 + 18 - 6 k = 0 \Rightarrow - 4 k = 0 \Rightarrow k = 0$

Question 1:

The given points are $A (2, 3), B (4, k) and C (6, - 3)$ .
Here, $(x_{1} = 2, y_{1} = 3), (x_{2} = 4, y_{2} = k) and (x_{3} = 6, y_{3} = - 3)$
It is given that the points A, B and Câ€‹ are collinear. Then,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 2 (k + 3) + 4 (- 3 - 3) + 6 (3 - k) = 0 \Rightarrow 2 k + 6 - 24 + 18 - 6 k = 0 \Rightarrow - 4 k = 0 \Rightarrow k = 0$

Answer:

The distance of a point (x, y) from the origin O(0, 0) is $\sqrt{x^{2} + y^{2}}$ .
Let P(x = −6, y = 8) be the given point. Then
$O P = \sqrt{x^{2} + y^{2}} = \sqrt{{(- 6)}^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$
Hence, the correct answer is option (d).

Question 2:

The distance of a point (x, y) from the origin O(0, 0) is $\sqrt{x^{2} + y^{2}}$ .
Let P(x = −6, y = 8) be the given point. Then
$O P = \sqrt{x^{2} + y^{2}} = \sqrt{{(- 6)}^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10$
Hence, the correct answer is option (d).

Answer:

The distance of a point (x, y) from x-axis is $|y|$ .
Here, the point is (−3, 4). So, its distance from x-axis is $|4| = 4$ .
Hence, the correct answer is option (c).

Question 3:

The distance of a point (x, y) from x-axis is $|y|$ .
Here, the point is (−3, 4). So, its distance from x-axis is $|4| = 4$ .
Hence, the correct answer is option (c).

Answer:

Let P(x, 0) the point on x-axis, then
$A P = B P \Rightarrow A P^{2} = B P^{2} \Rightarrow {(x + 1)}^{2} + {(0 - 0)}^{2} = {(x - 5)}^{2} + {(0 - 0)}^{2} \Rightarrow x^{2} + 2 x + 1 = x^{2} - 10 x + 25 \Rightarrow 12 x = 24 \Rightarrow x = 2$
Thus, the required point is (2, 0).
Hence, the correct answer is option (b).

Question 4:

Let P(x, 0) the point on x-axis, then
$A P = B P \Rightarrow A P^{2} = B P^{2} \Rightarrow {(x + 1)}^{2} + {(0 - 0)}^{2} = {(x - 5)}^{2} + {(0 - 0)}^{2} \Rightarrow x^{2} + 2 x + 1 = x^{2} - 10 x + 25 \Rightarrow 12 x = 24 \Rightarrow x = 2$
Thus, the required point is (2, 0).
Hence, the correct answer is option (b).

Answer:

Since R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y), therefore
$\frac{5 + y}{2} = 6 \Rightarrow 5 + y = 12 \Rightarrow y = 12 - 5 = 7$
Hence, the correct option is (b).

Question 5:

Since R(5, 6) is the midpoint of the line segment AB joining the points A(6, 5) and B(4, y), therefore
$\frac{5 + y}{2} = 6 \Rightarrow 5 + y = 12 \Rightarrow y = 12 - 5 = 7$
Hence, the correct option is (b).

Answer:

The point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3. So
$k = \frac{2 \times 5 + 3 \times 2}{2 + 3} = \frac{10 + 6}{5} = \frac{16}{5}$
Hence, the correct answer is option (c).

Question 6:

The point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3. So
$k = \frac{2 \times 5 + 3 \times 2}{2 + 3} = \frac{10 + 6}{5} = \frac{16}{5}$
Hence, the correct answer is option (c).

Answer:

Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
$A B = \sqrt{{(0 - 0)}^{2} + {(4 - 0)}^{2}} = \sqrt{16} = 4 B C = \sqrt{{(0 - 3)}^{2} + {(0 - 0)}^{2}} = \sqrt{9} = 3 A C = \sqrt{{(0 - 3)}^{2} + {(4 - 0)}^{2}} = \sqrt{9 + 16} = 5$
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).

Question 7:

Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
$A B = \sqrt{{(0 - 0)}^{2} + {(4 - 0)}^{2}} = \sqrt{16} = 4 B C = \sqrt{{(0 - 3)}^{2} + {(0 - 0)}^{2}} = \sqrt{9} = 3 A C = \sqrt{{(0 - 3)}^{2} + {(4 - 0)}^{2}} = \sqrt{9 + 16} = 5$
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).

Answer:

The diagonals of a parallelogram bisect each other. The vertices of the ||gm ABCD are A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4).
Here, AC and BD are the diagonals. So
$\frac{1 + 2}{2} = \frac{- 1 + x}{2} \Rightarrow x - 1 = 3 \Rightarrow x = 1 + 3 = 4$
Hence, the correct answer is option (b).

Question 8:

The diagonals of a parallelogram bisect each other. The vertices of the ||gm ABCD are A(1, 3), B(−1, 2) and C(2, 5) and D(x, 4).
Here, AC and BD are the diagonals. So
$\frac{1 + 2}{2} = \frac{- 1 + x}{2} \Rightarrow x - 1 = 3 \Rightarrow x = 1 + 3 = 4$
Hence, the correct answer is option (b).

Answer:

Let A(x₁ = x, y₁ = 2), B(x₂ = −3, y₂ = −4) and C(x₃ = 7, y₃ = −5) be collinear points. Then
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow x (- 4 + 5) + (- 3) (- 5 - 2) + 7 (2 + 4) = 0 \Rightarrow x + 21 + 42 = 0 \Rightarrow x = - 63$
Hence, the correct answer is option (a).

Question 9:

Let A(x₁ = x, y₁ = 2), B(x₂ = −3, y₂ = −4) and C(x₃ = 7, y₃ = −5) be collinear points. Then
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow x (- 4 + 5) + (- 3) (- 5 - 2) + 7 (2 + 4) = 0 \Rightarrow x + 21 + 42 = 0 \Rightarrow x = - 63$
Hence, the correct answer is option (a).

Answer:

Let A(x₁ = 5, y₁ = 0), B(x₂ = 8, y₂ = 0) and C(x₃ = 8, y₃ = 4) be the vertices of the triangle. Then
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [5 (0 - 4) + 8 (4 - 0) + 8 (0 - 0)] = \frac{1}{2} [- 20 + 32 + 0] = 6 sq . units$
Hence, the correct answer is option (c).

Question 10:

Let A(x₁ = 5, y₁ = 0), B(x₂ = 8, y₂ = 0) and C(x₃ = 8, y₃ = 4) be the vertices of the triangle. Then
$Area (∆ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [5 (0 - 4) + 8 (4 - 0) + 8 (0 - 0)] = \frac{1}{2} [- 20 + 32 + 0] = 6 sq . units$
Hence, the correct answer is option (c).

Answer:

Let A(x₁ = a, y₁ = 0), O(x₂ = 0, y₂ = 0) and B(x₃ = 0, y₃ = b) be the given vertices. So
$Area (∆ A B O) = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| = \frac{1}{2} |a (0 - b) + 0 (b - 0) + 0 (0 - 0)| = \frac{1}{2} |- a b| = = \frac{1}{2} a b$
Hence, the correct answer is (b).

Question 11:

Let A(x₁ = a, y₁ = 0), O(x₂ = 0, y₂ = 0) and B(x₃ = 0, y₃ = b) be the given vertices. So
$Area (∆ A B O) = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})| = \frac{1}{2} |a (0 - b) + 0 (b - 0) + 0 (0 - 0)| = \frac{1}{2} |- a b| = = \frac{1}{2} a b$
Hence, the correct answer is (b).

Answer:

The point $P (\frac{a}{2}, 4)$ is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3). So
$\frac{a}{2} = \frac{- 6 - 2}{2} \Rightarrow \frac{a}{2} = - 4 \Rightarrow a = - 8$
Hence, the correct answer is option (a).

Question 12:

The point $P (\frac{a}{2}, 4)$ is the midpoint of the line segment joining the points A(−6, 5) and B(−2, 3). So
$\frac{a}{2} = \frac{- 6 - 2}{2} \Rightarrow \frac{a}{2} = - 4 \Rightarrow a = - 8$
Hence, the correct answer is option (a).

Answer:

Here, AC and BD are two diagonals of the rectangle ABCD. So
$B D = \sqrt{{(4 - 0)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$
Hence, the correct answer is option (a).

Question 13:

Here, AC and BD are two diagonals of the rectangle ABCD. So
$B D = \sqrt{{(4 - 0)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$
Hence, the correct answer is option (a).

Answer:

Here, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Then
$Coordinates of P = (\frac{2 \times 4 + 1 \times 1}{2 + 1}, \frac{2 \times 6 + 1 \times 3}{2 + 1}) = (\frac{8 + 1}{3}, \frac{12 + 3}{3}) = (\frac{9}{3}, \frac{15}{3}) = (3, 5)$
Hence, the correct answer is option (b).

Question 14:

Here, the point P divides the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1. Then
$Coordinates of P = (\frac{2 \times 4 + 1 \times 1}{2 + 1}, \frac{2 \times 6 + 1 \times 3}{2 + 1}) = (\frac{8 + 1}{3}, \frac{12 + 3}{3}) = (\frac{9}{3}, \frac{15}{3}) = (3, 5)$
Hence, the correct answer is option (b).

Answer:

Let (x, y) be the coordinates of the other end of the diameter. Then
$- 2 = \frac{2 + x}{2} \Rightarrow x = - 6 5 = \frac{3 + y}{2} \Rightarrow y = 7$
Hence, the correct answer is option (a).

Question 15:

Let (x, y) be the coordinates of the other end of the diameter. Then
$- 2 = \frac{2 + x}{2} \Rightarrow x = - 6 5 = \frac{3 + y}{2} \Rightarrow y = 7$
Hence, the correct answer is option (a).

Answer:

Here, AQ : BQ = 2 : 1. Then
$y = \frac{2 \times (- 5) + 1 \times (- 2)}{2 + 1} = \frac{- 10 - 2}{3} = - 4$
Hence, the correct answer is option (c).

Question 16:

Here, AQ : BQ = 2 : 1. Then
$y = \frac{2 \times (- 5) + 1 \times (- 2)}{2 + 1} = \frac{- 10 - 2}{3} = - 4$
Hence, the correct answer is option (c).

Answer:

Let (x, y) be the coordinates of A. Then
$0 = \frac{- 2 + x}{2} \Rightarrow x = 2 4 = \frac{3 + y}{2} \Rightarrow y = 8 - 3 = 5$
Thus, the coordinates of A are (2, 5).
Hence, the correct answer is option (a).

Question 17:

Let (x, y) be the coordinates of A. Then
$0 = \frac{- 2 + x}{2} \Rightarrow x = 2 4 = \frac{3 + y}{2} \Rightarrow y = 8 - 3 = 5$
Thus, the coordinates of A are (2, 5).
Hence, the correct answer is option (a).

Answer:

Let (x, y) be the coordinates of P. Then
$x = \frac{2 \times 5 + 3 \times 2}{2 + 3} = \frac{10 + 6}{5} = \frac{16}{5} y = \frac{2 \times 2 + 3 \times (- 5)}{2 + 3} = \frac{4 - 15}{5} = \frac{- 11}{5}$
Thus, the coordinates of point P are $(\frac{16}{5}, \frac{- 11}{5})$ and so it lies in the fourth quadrant.
Hence, the correct answer is option (d).

Question 18:

Let (x, y) be the coordinates of P. Then
$x = \frac{2 \times 5 + 3 \times 2}{2 + 3} = \frac{10 + 6}{5} = \frac{16}{5} y = \frac{2 \times 2 + 3 \times (- 5)}{2 + 3} = \frac{4 - 15}{5} = \frac{- 11}{5}$
Thus, the coordinates of point P are $(\frac{16}{5}, \frac{- 11}{5})$ and so it lies in the fourth quadrant.
Hence, the correct answer is option (d).

Answer:

The given points are A(−6, 7) and B(−1, −5). So
$A B = \sqrt{{(- 6 + 1)}^{2} + {(7 + 5)}^{2}} = \sqrt{{(- 5)}^{2} + {(12)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$
Thus, 2AB = 26.
Hence, the correct answer is option (b).

Question 19:

The given points are A(−6, 7) and B(−1, −5). So
$A B = \sqrt{{(- 6 + 1)}^{2} + {(7 + 5)}^{2}} = \sqrt{{(- 5)}^{2} + {(12)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$
Thus, 2AB = 26.
Hence, the correct answer is option (b).

Answer:

Let P(x, 0) be the point on x-axis. Then as per the question
$A P = B P \Rightarrow A P^{2} = B P^{2} \Rightarrow {(x - 7)}^{2} + {(0 - 6)}^{2} = {(x + 3)}^{2} + {(0 - 4)}^{2} \Rightarrow x^{2} - 14 x + 49 + 36 = x^{2} + 6 x + 9 + 16 \Rightarrow 60 = 20 x \Rightarrow x = \frac{60}{20} = 3$
Thus, the required point is (3, 0).
Hence, the correct answer is option (c).

Question 20:

Let P(x, 0) be the point on x-axis. Then as per the question
$A P = B P \Rightarrow A P^{2} = B P^{2} \Rightarrow {(x - 7)}^{2} + {(0 - 6)}^{2} = {(x + 3)}^{2} + {(0 - 4)}^{2} \Rightarrow x^{2} - 14 x + 49 + 36 = x^{2} + 6 x + 9 + 16 \Rightarrow 60 = 20 x \Rightarrow x = \frac{60}{20} = 3$
Thus, the required point is (3, 0).
Hence, the correct answer is option (c).

Answer:

(b) 4 units
The y-coordinate is the distance of the point from the x-axis.
Here, the y-coordinate is 4.

Question 21:

(b) 4 units
The y-coordinate is the distance of the point from the x-axis.
Here, the y-coordinate is 4.

Answer:

(c) 1 : 2
Let AB be divided by the x axis in the ratio $k : 1$ at the point P.
Then, by section formula, the coordinates of P are
$P (\frac{5 k + 2}{k + 1}, \frac{6 k - 3}{k + 1})$
But P lies on the x axis: so, its ordinate is 0.
$\frac{6 k - 3}{k + 1} = 0$
$\Rightarrow 6 k - 3 = 0$
$\Rightarrow 6 k = 3$
$\Rightarrow k = \frac{1}{2}$
Hence, the required ratio is $\frac{1}{2} : 1$ , which is same as 1 : 2.

Question 22:

(c) 1 : 2
Let AB be divided by the x axis in the ratio $k : 1$ at the point P.
Then, by section formula, the coordinates of P are
$P (\frac{5 k + 2}{k + 1}, \frac{6 k - 3}{k + 1})$
But P lies on the x axis: so, its ordinate is 0.
$\frac{6 k - 3}{k + 1} = 0$
$\Rightarrow 6 k - 3 = 0$
$\Rightarrow 6 k = 3$
$\Rightarrow k = \frac{1}{2}$
Hence, the required ratio is $\frac{1}{2} : 1$ , which is same as 1 : 2.

Answer:

(d) 1 : 2
Let AB be divided by the y axis in the ratio $k : 1$ at the point P.
Then, by section formula, the coordinates of P are
$P (\frac{8 k - 4}{k + 1}, \frac{3 k + 2}{k + 1})$
But, P lies on the y axis; so, its abscissa is 0.
$\Rightarrow \frac{8 k - 4}{k + 1} = 0$
$\Rightarrow 8 k - 4 = 0$
$\Rightarrow 8 k = 4$
$\Rightarrow k = \frac{1}{2}$
Hence, the required ratio is $\frac{1}{2} : 1$ , which is same as 1 : 2.

Question 23:

(d) 1 : 2
Let AB be divided by the y axis in the ratio $k : 1$ at the point P.
Then, by section formula, the coordinates of P are
$P (\frac{8 k - 4}{k + 1}, \frac{3 k + 2}{k + 1})$
But, P lies on the y axis; so, its abscissa is 0.
$\Rightarrow \frac{8 k - 4}{k + 1} = 0$
$\Rightarrow 8 k - 4 = 0$
$\Rightarrow 8 k = 4$
$\Rightarrow k = \frac{1}{2}$
Hence, the required ratio is $\frac{1}{2} : 1$ , which is same as 1 : 2.

Answer:

(b) −1
The given points are A(−3, b) and B(1, b+4).
Then, $(x_{1} = - 3, y_{1} = b) and (x_{2} = 1, y_{2} = b + 4)$
Therefore,
$x = \frac{[(- 3) + 1]}{2} = \frac{- 2}{2} = - 1$
and
$y = \frac{[b + (b + 4)]}{2} = \frac{2 b + 4}{2} = b + 2$
But the midpoint is $P (- 1, 1)$ .
Therefore,
$b + 2 = 1 \Rightarrow b = - 1$

Question 24:

(b) −1
The given points are A(−3, b) and B(1, b+4).
Then, $(x_{1} = - 3, y_{1} = b) and (x_{2} = 1, y_{2} = b + 4)$
Therefore,
$x = \frac{[(- 3) + 1]}{2} = \frac{- 2}{2} = - 1$
and
$y = \frac{[b + (b + 4)]}{2} = \frac{2 b + 4}{2} = b + 2$
But the midpoint is $P (- 1, 1)$ .
Therefore,
$b + 2 = 1 \Rightarrow b = - 1$

Answer:

(b) 2 : 9
Let the lineâ€‹ $2 x + y - 4 = 0$ divide the line segment in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
$P (\frac{3 k + 2}{k + 1}, \frac{7 k - 2}{k + 1})$
Since P lies on the line $2 x + y - 4 = 0$ , we have:
$\frac{2 (3 k + 2)}{k + 1} + \frac{7 k - 2}{k + 1} - 4 = 0 \Rightarrow (6 k + 4) + (7 k - 2) - (4 k + 4) = 0 \Rightarrow 9 k = 2 \Rightarrow k = \frac{2}{9}$
Hence, the required ratio is $\frac{2}{9} : 1$ , which is same as 2 : 9.

Question 25:

(b) 2 : 9
Let the lineâ€‹ $2 x + y - 4 = 0$ divide the line segment in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
$P (\frac{3 k + 2}{k + 1}, \frac{7 k - 2}{k + 1})$
Since P lies on the line $2 x + y - 4 = 0$ , we have:
$\frac{2 (3 k + 2)}{k + 1} + \frac{7 k - 2}{k + 1} - 4 = 0 \Rightarrow (6 k + 4) + (7 k - 2) - (4 k + 4) = 0 \Rightarrow 9 k = 2 \Rightarrow k = \frac{2}{9}$
Hence, the required ratio is $\frac{2}{9} : 1$ , which is same as 2 : 9.

Answer:

(c) $(\frac{7}{2}, \frac{9}{2})$
D is the midpoint of BC.
So, the coordinates of D are
$D (\frac{6 + 1}{2}, \frac{5 + 4}{2}) [B (6, 5) and C (1, 4) \Rightarrow (x_{1} = 6, y_{1} = 5) and (x_{2} = 1, y_{2} = 4)] i.e. D (\frac{7}{2}, \frac{9}{2})$

Question 26:

(c) $(\frac{7}{2}, \frac{9}{2})$
D is the midpoint of BC.
So, the coordinates of D are
$D (\frac{6 + 1}{2}, \frac{5 + 4}{2}) [B (6, 5) and C (1, 4) \Rightarrow (x_{1} = 6, y_{1} = 5) and (x_{2} = 1, y_{2} = 4)] i.e. D (\frac{7}{2}, \frac{9}{2})$

Answer:

(d) (4, 0)
The given points are $A (- 1, 0), B (5, - 2) and C (8, 2)$ .
Here, $(x_{1} = - 1, y_{1} = 0), (x_{2} = 5, y_{2} = - 2) and (x_{3} = 8, y_{3} = 2)$
Let G(x, y) be the centroid of $∆ A B C$ . Then,
$x = \frac{1}{3} (x_{1} + x_{2} + x_{3}) = \frac{1}{3} (- 1 + 5 + 8) = 4$
and
$y = \frac{1}{3} (y_{1} + y_{2} + y_{3}) = \frac{1}{3} (0 - 2 + 2) = 0$
Hence, the centroid of $∆ A B C$ is G(4, 0).

Question 27:

(d) (4, 0)
The given points are $A (- 1, 0), B (5, - 2) and C (8, 2)$ .
Here, $(x_{1} = - 1, y_{1} = 0), (x_{2} = 5, y_{2} = - 2) and (x_{3} = 8, y_{3} = 2)$
Let G(x, y) be the centroid of $∆ A B C$ . Then,
$x = \frac{1}{3} (x_{1} + x_{2} + x_{3}) = \frac{1}{3} (- 1 + 5 + 8) = 4$
and
$y = \frac{1}{3} (y_{1} + y_{2} + y_{3}) = \frac{1}{3} (0 - 2 + 2) = 0$
Hence, the centroid of $∆ A B C$ is G(4, 0).

Answer:

(c) (−4, −15)
Two vertices of $∆ A B C are A (- 1, 4) and B (5, 2)$ .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
$G (\frac{- 1 + 5 + a}{3}, \frac{4 + 2 + b}{3}) i . e . G (\frac{4 + a}{3}, \frac{6 + b}{3})$
But it is given that the centroid is $G (0, - 3)$ .
Therefore,
$\frac{4 + a}{3} = 0 and \frac{6 + b}{3} = - 3$
$\Rightarrow 4 + a = 0 and 6 + b = - 9$
$\Rightarrow a = - 4 and b = - 15$
Hence, the third vertex of $∆ A B C is C (- 4, - 15)$ .

Question 28:

(c) (−4, −15)
Two vertices of $∆ A B C are A (- 1, 4) and B (5, 2)$ .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
$G (\frac{- 1 + 5 + a}{3}, \frac{4 + 2 + b}{3}) i . e . G (\frac{4 + a}{3}, \frac{6 + b}{3})$
But it is given that the centroid is $G (0, - 3)$ .
Therefore,
$\frac{4 + a}{3} = 0 and \frac{6 + b}{3} = - 3$
$\Rightarrow 4 + a = 0 and 6 + b = - 9$
$\Rightarrow a = - 4 and b = - 15$
Hence, the third vertex of $∆ A B C is C (- 4, - 15)$ .

Answer:

(a) isosceles
Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then,
$A B = \sqrt{{(4 + 4)}^{2} + {(0 - 0)}^{2}} = \sqrt{{(8)}^{2} + {(0)}^{2}} = \sqrt{64 + 0} = \sqrt{64} = 8 units B C = \sqrt{{(0 - 4)}^{2} + {(3 - 0)}^{2}} = \sqrt{{(- 4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units A C = \sqrt{{(0 + 4)}^{2} + {(3 - 0)}^{2}} = \sqrt{{(4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$
BC = AC = 5 units
Therefore, $∆ A B C$ is isosceles.

Question 29:

(a) isosceles
Let A(−4, 0), B(4, 0) and C(0, 3) be the given points. Then,
$A B = \sqrt{{(4 + 4)}^{2} + {(0 - 0)}^{2}} = \sqrt{{(8)}^{2} + {(0)}^{2}} = \sqrt{64 + 0} = \sqrt{64} = 8 units B C = \sqrt{{(0 - 4)}^{2} + {(3 - 0)}^{2}} = \sqrt{{(- 4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units A C = \sqrt{{(0 + 4)}^{2} + {(3 - 0)}^{2}} = \sqrt{{(4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$
BC = AC = 5 units
Therefore, $∆ A B C$ is isosceles.

Answer:

(d) right-angled

Let P(0, 6), Q(−5, 3) and R(3, 1) be the given points. Then,
$P Q = \sqrt{{(- 5 - 0)}^{2} + {(3 - 6)}^{2}} = \sqrt{}$

RS Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 6 - Coordinate Geometry

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