Rs Aggarwal 2015 for Class 10 Math Chapter 16 - Coordinate Geometry

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Page No 636:

Question 1:

Find the distance between the points:

(i) A(9, 3) and B(15, 11)
(ii) A(7, −4) and B(−5, 1)
(iii) A(−6, −4) and B(9, −12)
(iv) A(1, −3) and B(4, −6)
(v) P(a + b, a − b) and Q(a −b, a + b)
(vi) P(a sin α, a cos α) and Q(a cos α, −a sin α)

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)
$P Q = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(a \cos α - a \sin α)}^{2} + {(- a \sin α - a \cos α)}^{2}} = \sqrt{(a^{2} \cos^{2} α + a^{2} \sin^{2} α - 2 a^{2} \cos α \times \sin α) + (a^{2} \sin^{2} α + a^{2} \cos^{2} α + 2 a^{2} \cos α \times \sin α)} = \sqrt{2 a^{2} \cos^{2} α + 2 a^{2} \sin^{2} α} = \sqrt{2 a^{2} (\cos^{2} α + \sin^{2} α)} = \sqrt{2 a^{2} (1)} (From the identity \cos^{2} α + \sin^{2} α = 1) = \sqrt{2 a^{2}} = \sqrt{2} a units$

Page No 637:

Question 2:

Find the distance of each of the following points from the origin:

(i) A(5, −12)
(ii) B(−5, 5)
(iii) C(−, −6)4

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.
$O C = \sqrt{{(- 4 - 0)}^{2} + {(- 6 - 0)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 6)}^{2}} = \sqrt{16 + 36} = \sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13} units$

Page No 637:

Question 3:

Find all possible values of a for which the distance between the points A(a, −1) and B(5, 3) is 5 units.

Answer:

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

Page No 637:

Question 4:

Find the ordinate of a point whose abscissa is 10 and which is at a distance of 10 units from the points P(2, −3).

Answer:

Let Q be the other point whose coordinates are Q(10, y).
Given, PQ = 10 units
Therefore, (PQ)² = 100
$\Rightarrow {(10 - 2)}^{2} + {\{y - (- 3)\}}^{2} = 100 \Rightarrow {(8)}^{2} + {(y + 3)}^{2} = 100 \Rightarrow 64 + {(y + 3)}^{2} = 100 \Rightarrow {(y + 3)}^{2} = 100 - 64 \Rightarrow {(y + 3)}^{2} = 36 \Rightarrow (y + 3) = \pm \sqrt{36} \Rightarrow y + 3 = \pm 6 \Rightarrow y + 3 = 6 or y + 3 = - 6 \Rightarrow y = 3 or - 9$
Therefore, y = 3 or −9.

Page No 637:

Question 5:

If A(6, −1), B(1, 3) and C(k, 8) are three points such that AB = BC, find the value of k.

Answer:

The given points are A(6, −1), B(1, 3) and C(k, 8).
Also, it is given that AB = BC.
Therefore, (AB)² = (BC)²
$\Rightarrow {(1 - 6)}^{2} + {\{3 - (- 1)\}}^{2} = {(k - 1)}^{2} + {(8 - 3)}^{2} \Rightarrow {(1 - 6)}^{2} + {(3 + 1)}^{2} = {(k - 1)}^{2} + {(8 - 3)}^{2} \Rightarrow {(- 5)}^{2} + {(4)}^{2} = {(k - 1)}^{2} + {(5)}^{2} \Rightarrow 25 + 16 = {(k - 1)}^{2} + 25 \Rightarrow 25 + 16 - 25 = {(k - 1)}^{2} \Rightarrow {(k - 1)}^{2} = 16 \Rightarrow (k - 1) = \pm \sqrt{16} \Rightarrow k - 1 = \pm 4 \Rightarrow k - 1 = 4 or k - 1 = - 4 \Rightarrow k = 5 or - 3$
Therefore, the value of k will be either 5 or −3.

Page No 637:

Question 6:

If the point A(a, 2) is equidistant from the points B(8, −2) and C(2, −2), find the value of a.

Answer:

The given points are A(a, 2), B(8, −2) and C(2, −2).
Also, it is given that AB = AC.
Therefore, (AB)² = (AC)²
$\Rightarrow {(8 - a)}^{2} + {(- 2 - 2)}^{2} = {(2 - a)}^{2} + {(- 2 - 2)}^{2} \Rightarrow {(8 - a)}^{2} + {(- 4)}^{2} = {(2 - a)}^{2} + {(- 4)}^{2} \Rightarrow {(8 - a)}^{2} + {(4)}^{2} = {(2 - a)}^{2} + {(- 4)}^{2} \Rightarrow {(8 - a)}^{2} + 16 = {(2 - a)}^{2} + 16 \Rightarrow {(8 - a)}^{2} = {(2 - a)}^{2} \Rightarrow 64 + a^{2} - 16 a = 4 + a^{2} - 4 a \Rightarrow - 12 a = - 60 \Rightarrow a = \frac{- 60}{- 12} = 5$
Therefore, the value of a is 5.

Page No 637:

Question 7:

Find the point on the x-axis that is equidistant from the points (−2, 5) and (−2, 9).

Answer:

Let the given points be A(−2, 5) and B(−2, 9) and the required point be P(x,0). Then, AP = PB.
That is, (AP)² = (PB)²
$\Rightarrow {\{x - (- 2)\}}^{2} + {(0 - 5)}^{2} = {(- 2 - x)}^{2} + {(9 - 0)}^{2} \Rightarrow {(x + 2)}^{2} + {(0 - 5)}^{2} = {(- 2 - x)}^{2} + {(9)}^{2} \Rightarrow {(x + 2)}^{2} + {(- 5)}^{2} = {(- 2 - x)}^{2} + {(9)}^{2} \Rightarrow {(x + 2)}^{2} + 25 = {(- 2 - x)}^{2} + 81 \Rightarrow x^{2} + 4 x + 4 + 25 = x^{2} + 4 x + 4 + 81 \Rightarrow x^{2} + 4 x + 29 = x^{2} + 4 x + 85 \Rightarrow x^{2} + 4 x - x^{2} - 4 x = 85 - 29 \Rightarrow 0 = 56$

Disclaimer : As it is never possible that 0 will become equal to 56
Therefore, we can say that there is no point on the x-axis which is equidistance from A(-2,5) and B(-2,9)

Page No 637:

Question 8:

Find the point on the y-axis that is equidistant from the points (5, −2) and (−3, 2).

Answer:

Let the given points be A(5, −2) and B(−3, 2) and the required point be P(0, y). Then, AP = BP.
Also, (AP)² = (BP)²
$\Rightarrow {(0 - 5)}^{2} + {(y + 2)}^{2} = {(0 + 3)}^{2} + {(y - 2)}^{2} \Rightarrow {(- 5)}^{2} + {(y + 2)}^{2} = {(3)}^{2} + {(y - 2)}^{2} \Rightarrow 25 + (y^{2} + 4 + 4 y) = 9 + (y^{2} + 4 - 4 y) \Rightarrow y^{2} + 4 + 4 y + 25 = y^{2} + 4 - 4 y + 9 \Rightarrow y^{2} + 4 y - y^{2} + 4 y = 4 + 9 - 4 - 25 \Rightarrow 8 y = - 16 \Rightarrow y = \frac{- 16}{8} = - 2$
So, the point is (0, −2).

Page No 637:

Question 9:

If the points A(4, 3) and B(x, 5) lie on a circle with centre O(2, 3), find the value of x.

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.

Page No 637:

Question 10:

If P(x, y) is a point equidistant from the points A(6, −1) and B(2, 3), show that x − y = 3.

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.

Page No 637:

Question 11:

Find points on the x-axis, each of which is at a distance of 10 units from the point A(11, −8).

Answer:

A(11,−8) is the given point. Let P(x,0) be the required point on the x-axis; then, AP = 10 units.
Also (AP)² = (10)² = 100
$\Rightarrow {(x - 11)}^{2} + {(0 + 8)}^{2} = 100 \Rightarrow {(x - 11)}^{2} + {(8)}^{2} = 100 \Rightarrow {(x - 11)}^{2} + 64 = 100 \Rightarrow {(x - 11)}^{2} = 100 - 64 \Rightarrow {(x - 11)}^{2} = 36 \Rightarrow (x - 11) = \pm \sqrt{36} \Rightarrow x - 11 = \pm 6 If x - 11 = 6, x = 6 + 11 \Rightarrow x = 17 Or if x - 11 = - 6, x = 11 - 6 = 5$
Therefore, the coordinates of the point are (17, 0) or (5, 0). It is 10 units away from the point A(11, −8).

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Question 12:

Find the coordinates of the point equidistant from the three points A(5, 3), B(5, −5) and C(1, −5).

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).

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Question 13:

Using the distance formula, show that the points (1, −1), (5, 2) and (9, 5) are collinear.

Answer:

Let A(1, −1), B (5, 2) and C(9, 5) be the given points. Then:
$A B = \sqrt{{(5 - 1)}^{2} + {(2 + 1)}^{2}} = \sqrt{{(4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units B C = \sqrt{{(9 - 5)}^{2} + {(5 - 2)}^{2}} = \sqrt{{(4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units A C = \sqrt{{(9 - 1)}^{2} + {(5 + 1)}^{2}} = \sqrt{{(8)}^{2} + {(6)}^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 units A B + B C = (5 + 5) = 10 units = A C$

Hence, the given points are collinear.

Page No 637:

Question 14:

Show that the following points are collinear:

(i) A(6, 9) B(0, 1) and C(−6, −7)
(ii) A(−1, −1) B(2, 3) and C(8, 11)
(iii) P(1, 1) Q(−2, 7) and R(3, −3)
(iv) P(2, 0) Q(11, 6) and R(−4, −4)

Answer:

(i) A(6, 9) B(0, 1) and C(−6, −7)
$A B = \sqrt{{(0 - 6)}^{2} + {(1 - 9)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 units B C = \sqrt{{(- 6 - 0)}^{2} + {(- 7 - 1)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 units A C = \sqrt{{(- 6 - 6)}^{2} + {(- 7 - 9)}^{2}} = \sqrt{{(- 12)}^{2} + {(- 16)}^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20 units A B + B C = (10 + 10) = 20 units = A C$

Hence, the points A, B and C are collinear.

(ii) A(−1, −1) B(2, 3) and C(8, 11)
$A B = \sqrt{{(2 + 1)}^{2} + {(3 + 1)}^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 units B C = \sqrt{{(8 - 2)}^{2} + {(11 - 3)}^{2}} = \sqrt{{(6)}^{2} + {(8)}^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 units A C = \sqrt{{(8 + 1)}^{2} + {(11 + 1)}^{2}} = \sqrt{{(9)}^{2} + {(12)}^{2}} = \sqrt{81 + 144} = \sqrt{225} = 15 units A B + B C = (5 + 10) = 15 units = A C$
Hence, the points A, B and C are collinear.

(iii) P(1, 1) Q(−2, 7) and R(3, −3)
$P Q = \sqrt{{(- 2 - 1)}^{2} + {(7 - 1)}^{2}} = \sqrt{{(- 3)}^{2} + {(6)}^{2}} = \sqrt{9 + 36} = \sqrt{45} = \sqrt{9 \times 5} = 3 \sqrt{5} units Q R = \sqrt{{(3 + 2)}^{2} + {(- 3 - 7)}^{2}} = \sqrt{{(5)}^{2} + {(- 10)}^{2}} = \sqrt{25 + 100} = \sqrt{125} = \sqrt{25 \times 5} = 5 \sqrt{5} units P R = \sqrt{{(3 - 1)}^{2} + {(- 3 - 1)}^{2}} = \sqrt{{(2)}^{2} + {(- 4)}^{2}} = \sqrt{4 + 16} = \sqrt{20} = \sqrt{4 \times 5} = 2 \sqrt{5} units P Q + P R = (3 \sqrt{5} + 2 \sqrt{5}) = 5 \sqrt{5} units = Q R$

Hence, the points Q, P and R are collinear.

(iv) P(2, 0) Q(11, 6) and R(−4, −4)

$P Q = \sqrt{{(11 - 2)}^{2} + {(6 - 0)}^{2}} = \sqrt{{(9)}^{2} + {(6)}^{2}} = \sqrt{81 + 36} = \sqrt{117} = \sqrt{9 \times 13} = 3 \sqrt{13} units Q R = \sqrt{{(- 4 - 11)}^{2} + {(- 4 - 6)}^{2}} = \sqrt{{(- 15)}^{2} + {(- 10)}^{2}} = \sqrt{225 + 100} = \sqrt{325} = \sqrt{25 \times 13} = 5 \sqrt{13} units P R = \sqrt{{(- 4 - 2)}^{2} + {(- 4 - 0)}^{2}} = \sqrt{{(- 6)}^{2} + {(- 4)}^{2}} = \sqrt{36 + 16} = \sqrt{52} = \sqrt{4 \times 13} = 2 \sqrt{13} units P Q + P R = (3 \sqrt{13} + 2 \sqrt{13}) = 5 \sqrt{13} units = Q R$

Hence, the points Q, P and R are collinear.

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Question 15:

Show that the points A(3, 0), B(6, 4) and C(−1, 3) are the vertices of a right-angled triangle. Also, prove that these are the vertices of an isosceles triangle.

Answer:

A(3, 0), B (6, 4) and C(−1, 3) are the given points. Then:
$A B = \sqrt{{(6 - 3)}^{2} + {(4 - 0)}^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 units B C = \sqrt{{(- 1 - 6)}^{2} + {(3 - 4)}^{2}} = \sqrt{{(- 7)}^{2} + {(- 1)}^{2}} = \sqrt{49 + 1} = \sqrt{50} = \sqrt{25 \times 2} = 5 \sqrt{2} units A C = \sqrt{{(- 1 - 3)}^{2} + {(3 - 0)}^{2}} = \sqrt{{(- 4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$
Thus, AB = BC = 5 units
Also, (AB)²+(AC)² = (5)²+(5)² = 50
and (BC)² = (5 $\sqrt{2}$ )² = 25 $\times$ 2 = 50
Thus, (AB)²+(AC)² = (BC)²
This show that $∆ A B C$ is right- angled at A.
This proves that the the points A(3, 0), B(6, 4) and C(−1, 3) are the vertices of an isosceles right- angled triangle.

Page No 637:

Question 16:

Prove that the points A(7, 10), B(−2, 5) and C(3, −4) are the vertices of an isosceles right triangle.

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.

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Question 17:

Show that the points A(−5, 6) B(3, 0) and C(9, 8) are the vertices of an isosceles right-angled triangle. Calculate its area.

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, $a rea of a triangle = \frac{1}{2} \times base \times height$
$If A B is the height and B C is the base, Area = \frac{1}{2} \times 10 \times 10 = 50 square units$

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Question 18:

Show that the points O(0, 0) A(3, $\sqrt{3}$ ) and B(3, − $\sqrt{3}$ ) are the vertices of an equilateral triangle. Find the area of this triangle.

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 = $\frac{\sqrt{3}}{4} \times {(side)}^{2}$
$\frac{\sqrt{3}}{4} \times {(2 \sqrt{3})}^{2} = \frac{\sqrt{3}}{4} \times 12 = 3 \sqrt{3} square units$

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Question 19:

Show that the points A(2, 1), B(5, 2), C(6, 4) and D(3, 3) are the angular points of a parallelogram. Is this figure a rectangle?

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.

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Question 20:

Show that the following points are the vertices of a rectangle.

(i) A(0, −4), B(6, 2) C(3, 5) and D(−3, −1)
(ii) A(2, −2), B(14, 10) C(11, 13) and D(−1, 1)
(iii) A(−4, −1), B(−2, −4) C(−3, 2) and D(0, −1)

Answer:

(i) 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.

(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(−4, −1), B(−2, −4) C(4, 0) and D(2, 3).
$A B = \sqrt{{\{- 2 - (- 4)\}}^{2} + {\{- 4 - (- 1)\}}^{2}} = \sqrt{{(2)}^{2} + {(- 3)}^{2}} = \sqrt{4 + 9} = \sqrt{13} units B C = \sqrt{{\{4 - (- 2)\}}^{2} + {\{0 - (- 4)\}}^{2}} = \sqrt{{(6)}^{2} + {(4)}^{2}} = \sqrt{36 + 16} = \sqrt{52} = 2 \sqrt{13} units C D = \sqrt{{(2 - 4)}^{2} + {(3 - 0)}^{2}} = \sqrt{{(- 2)}^{2} + {(3)}^{2}} = \sqrt{4 + 9} = \sqrt{13} units A D = \sqrt{{\{2 - (- 4)\}}^{2} + {\{3 - (- 1)\}}^{2}} = \sqrt{{(6)}^{2} + {(4)}^{2}} = \sqrt{36 + 16} = \sqrt{52} = 2 \sqrt{13} units Thus, A B = C D = \sqrt{13} units and B C = A D = 2 \sqrt{13} units Also, A C = \sqrt{{\{4 - (- 4)\}}^{2} + {\{0 - (- 1)\}}^{2}} = \sqrt{{(8)}^{2} + {(1)}^{2}} = \sqrt{64 + 1} = \sqrt{65} units B D = \sqrt{{\{2 - (- 2)\}}^{2} + {\{3 - (- 4)\}}^{2}} = \sqrt{{(4)}^{2} + {(7)}^{2}} = \sqrt{16 + 49} = \sqrt{65} units$
Also, diagonal AC = diagonal BD
Hence, the given points form a rectangle.

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Question 21:

Show that the following points are the vertices of a square.

(i) A(6, 2), B(2, 1), C(1, 5) and D(5, 6)
(ii) P(0, −2), Q(3, 1), R(0, 4) and S(−3, 1)
(iii) A(3, 2), B(0, 5), C(−3, 2) and D(0, −1)

Answer:

(i) 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.

(ii) 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.

(iii) 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.

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Question 22:

Show that the points A(−3, 2), B(−5, −5), C(2, −3) and D(4, 4) are the vertices of a rhombus. Find the area of this rhombus.

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.
$Area of a rhombus = \frac{1}{2} \times (product of diagonals) = \frac{1}{2} \times (5 \sqrt{2}) \times (9 \sqrt{2}) = \frac{45 (2)}{2} = 45 square units$

Page No 649:

Question 1:

Find the coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3.

Answer:

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).

Page No 649:

Question 2:

Find the coordinates of the points that divide the join of A(−5, 11) and B(4, −7) in the ratio 7 : 2.

Answer:

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).

Page No 649:

Question 3:

Find the coordinates of the points of trisection of the line segment AB, whose end points are A(2, 1) and B(5, −8).

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
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} x = \frac{(1 \times 5 + 2 \times 2)}{1 + 2}, y = \frac{\{1 \times (- 8) + 2 \times 1\}}{1 + 2} x = \frac{5 + 4}{3}, y = \frac{- 8 + 2}{3} x = \frac{9}{3}, y = - \frac{6}{3} So, x = 3 and y = - 2$
Therefore, the coordinates of P are (3, −2).
Also, Q divides AB in the ratio 2:1.
So, the coordinates of Q are
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} x = \frac{(2 \times 5 + 1 \times 2)}{2 + 1}, y = \frac{\{2 \times (- 8) + 1 \times 1\}}{2 + 1} x = \frac{10 + 2}{3}, y = \frac{- 16 + 1}{3} x = \frac{12}{3}, y = - \frac{15}{3} So, x = 4 and y = - 5$
Therefore, the coordinates of Q are (4, −5).
Hence, the points are (3, −2) and (4, −5).

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Question 4:

Find the coordinates of the points that divide the join of A(−4, 0) and B(0, 6) in three equal parts.

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
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{\{1 \times 0 + 2 \times (- 4)\}}{1 + 2}, y = \frac{(1 \times 6 + 2 \times 0)}{1 + 2} \Rightarrow x = - \frac{8}{3}, y = \frac{6}{3} \Rightarrow x = - \frac{8}{3}, y = 2 Therefore, coordinates of P are (- \frac{8}{3}, 2) .$
Also Q divides AB in the ratio 2:1
So the coordinates of Q are
$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} x = \frac{\{2 \times 0 + 1 \times (- 4)\}}{2 + 1}, y = \frac{(2 \times 6 + 1 \times 0)}{2 + 1} x = - \frac{4}{3}, y = \frac{12}{3} x = - \frac{4}{3}, y = 4 Therefore, coordinates of Q are (- \frac{4}{3}, 4) .$
Hence, the points are $(- \frac{8}{3}, 2) and (- \frac{4}{3}, 4)$ .

Page No 649:

Question 5:

The line segment joining the points A(3, −4) and B(1, 2) is trisected at the points P(p, −2) and $Q (\frac{5}{3}, q)$ . Find the values of p and q.

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

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Question 6:

The line segment joining the points A(4, −5) and B(4, 5) is divided by the point P such that $\frac{A P}{A B} = \frac{2}{5}$ . Find the coordinates of P.

Answer:

The given points are A(4, −5) and B(4, 5).
P divides AB, such that $\frac{A P}{A B} = \frac{2}{5}$ .
$Here, A B = A P + P B Therefore, \frac{A P}{A P + P B} = \frac{2}{5} \Rightarrow 5 A P = 2 A P + 2 P B \Rightarrow 5 A P - 2 A P = 2 P B \Rightarrow 3 A P = 2 P B Therefore, \frac{A P}{P B} = \frac{2}{3} So, m = 2 and n = 3 .$

$x = \frac{(m x_{2} + n x_{1})}{(m + n)}, y = \frac{(m y_{2} + n y_{1})}{(m + n)} \Rightarrow x = \frac{(2 \times 4 + 3 \times 4)}{2 + 3}, y = \frac{(2 \times 5 + 3 \times (- 5))}{2 + 3} \Rightarrow x = \frac{8 + 12}{5}, y = \frac{10 - 15}{5} \Rightarrow x = \frac{20}{5}, y = \frac{- 5}{5} x = 4 and y = - 1 Hence, the coordinates of P are (4, - 1) .$

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Question 7:

Find the coordinates of the midpoints of the line segment joining:

(i) A(3, 0) and B(−5, 4)
(ii) P(−11, −8) and Q(8, −2)

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.

Page No 649:

Question 8:

If (2, p) is the midpoint of the line segment joining the points A(6, −5) and B(−2, 11), find the value of p.

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.

Page No 649:

Question 9:

The midpoint of the line segment joining A(2a, 4) and B(−2, 3b) is C(1, 2a + 1). Find the values of a and b.

Answer:

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

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Question 10:

Points P, Q and R, in that order, divide a line segment joining A(1, 6) and B(5, −2) in four equal parts. Find the coordinates of P, Q and R.

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.

Page No 650:

Question 11:

The line segment joining A(−2, 9) and B(6, 3) is a diameter of a circle with centre C. Find the coordinates of C.

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).

Page No 650:

Question 12:

Find the coordinates of a point A, where AB is the diameter of a circle with centre C(2, −3) and the other end of the diameter is B(1, 4).

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).

Page No 650:

Question 13:

In what ratio does the point P(2, 5) divide the join of A(8, 2) and B(−6, 9)?

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.

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Question 14:

Find the ratio in which the point P(−6, a) divides the join of A(−3, −1) and B(−8, 9), Also, find the value of a.

Answer:

Let the point P(−6, a) divide the line AB in the ratio k : 1.
Then, by the section formula:
$x = \frac{- 8 k - 3}{k + 1}, y = \frac{9 k - 1}{k + 1} The coordinates of P are P (- 6, a) . - 6 = \frac{- 8 k - 3}{k + 1}, a = \frac{9 k - 1}{k + 1} \Rightarrow - 6 k - 6 = - 8 k - 3, a (k + 1) = 9 k - 1 \Rightarrow - 6 k + 8 k = - 3 + 6, a (k + 1) = 9 k - 1 \Rightarrow 2 k = 3, a (k + 1) = 9 k - 1 \Rightarrow k = \frac{3}{2}, a (k + 1) = 9 k - 1 Therefore, the point P divides the line A B in the ratio 3 : 2 . Now, put ting the value of k in the equation a (k + 1) = 9 k - 1, we get: a (\frac{3}{2} + 1) = 9 (\frac{3}{2}) - 1 \Rightarrow a (\frac{3 + 2}{2}) = \frac{27 - 2}{2} \Rightarrow \frac{5 a}{2} = \frac{25}{2} \Rightarrow 5 a = 25 \Rightarrow a = \frac{25}{5} = 5 Therefore, the value of a = 5 Th at is, the coordinates of P are (- 6, 5) . P oint P divides the line A B in the ratio 3 : 2 .$

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Question 15:

Find the ratio in which the point P(m, 6) divides the join of A(−4, 3) and B(2, 8). Also, find the value of m.

Answer:

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

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Question 16:

Find the ratio in which the point (−3, k) divides the join of A(−5, −4) and B(−2, 3). Also, find the value of k.

Answer:

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

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Question 17:

In what ratio is the line segment joining A(2, −3) and B(5, 6) divided by the x-axis? Also, find the coordinates of the point of division.

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).

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Question 18:

In what ratio is the line segment joining the points A(−2, −3) and B(3, 7) divided by the y-axis? Also, find the coordinates of the points of division.

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).

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Question 19:

In what ratio does the line x − y − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9)?

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.

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Question 20:

Find the lengths of the median of âˆ†ABC whose vertices are A(0, −1), B(2, 1) and C(0, 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

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Question 21:

Find the centroid of âˆ†ABC whose vertices are A(−1, 0), B(5, −2) and C(8, 2).

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).

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Question 22:

If G(−2, 1) is the centroid of a âˆ†ABC and two of its vertices are A(1, −6) and B(−5, 2), find the third vertex of the triangle.

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).

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Question 23:

Find the third vertex of âˆ†ABC if two of its vertices are B(−3, 1) and C(0, −2) and its centroid is at the origin.

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).

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Question 24:

Show that the points A(3, 1), B(0, −2), C(1, 1) and D(4, 4) are the vertices of a parallelogram ABCD.

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.

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Question 25:

If the points P(a, −11), Q(5, b), R(2, 15) and S(1, 1) are the vertices of a parallelogram PQRS, find the values of a and b.

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.
$Midpoint of P R = (\frac{a + 2}{2}, \frac{- 11 + 15}{2}) = (\frac{a + 2}{2}, \frac{4}{2}) = (\frac{a + 2}{2}, 2) Midpoint of Q S = (\frac{5 + 1}{2}, \frac{b + 1}{2}) = (\frac{6}{2}, \frac{b + 1}{2}) = (3, \frac{b + 1}{2}) Therefore, \frac{a + 2}{2} = 3, \frac{b + 1}{2} = 2 \Rightarrow a + 2 = 6, b + 1 = 4 \Rightarrow a = 6 - 2, b = 4 - 1 \Rightarrow a = 4 and b = 3$

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Question 26:

If three consecutive vertices of a parallelogram ABCD are A(1, −2), B(3, 6) and C(5, 10), find its fourth vertex D.

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).

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Question 1:

Find the area of âˆ†ABC whose vertices are:

(i) A(1, 2), B(−2, 3) and C(−3, −4)
(ii) A(−5, 7), B(−4, −5) and C(4, 5)
(iii) A(3, 8), B(−4, 2) and C(5, −1)
(iv) A(10, −6), B(2, 5) and C(−1, 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)
$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{10 (5 - 3) + 2 (3 - (- 6)) + (- 1) (- 6 - 5)\} = \frac{1}{2} \{10 (2) + 2 (9) - 1 (- 11)\} = \frac{1}{2} \{20 + 18 + 11\} = \frac{1}{2} (49) = 24.5 sq . units$

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Question 2:

(i) Find the area of a quadrilateral ABCD whose vertices are
A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3)

(ii) Find the area of the quadrilateral ABCD whose vertices are
A(0, 0), B(6, 0), C(4, 3) and D(0, 3)

(iii) Find the area of the quadrilateral ABCD whose vertices are
A(1, 0), B(5, 3), C(2, 7) and D(−2, 4).

Answer:

(i) The vertices are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{(- 4) (- 5 - (- 2)) + (- 3) ((- 2) - (- 2)) + 3 (- 2 - (- 5))\} = \frac{1}{2} \{- 4 (- 3) - 3 (0) + 3 (3)\} = \frac{1}{2} \{12 + 9\} = \frac{1}{2} (21) = 10.5 sq . units$
Now,
$Area of triangle A C D = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{(- 4) (- 2 - 3) + 3 (3 - (- 2)) + 2 (- 2 - (- 2))\} = \frac{1}{2} \{- 4 (- 5) + 3 (5) + 2 (0)\} = \frac{1}{2} \{20 + 15\} = \frac{1}{2} (35) = 17.5 sq . units$
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 10.5 sq. units + 17.5 sq. units = 28 sq. units

(ii) The vertices are A(0, 0), B(6, 0), C(4, 3) and D(0, 3).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{0 (0 - 3) + 6 (3 - 0) + 4 (0 - 0)\} = \frac{1}{2} \{0 + 18 + 0\} = \frac{1}{2} \{18\} = 9 sq . units$
Now,
$Area of triangle A C D = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{0 (3 - 3) + 4 (3 - 0) + 0 (0 - 3)\} = \frac{1}{2} \{0 + 12 + 0\} = \frac{1}{2} \{12\} = 6 sq . units$
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 9 sq. units + 6 sq. units = 15 sq. units

(iii) The vertices are A(1, 0), B(5, 3), C(2, 7) and D(−2, 4).
Join AC. Then,
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD

$Area of triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{1 (3 - 7) + 5 (7 - 0) + 2 (0 - 3)\} = \frac{1}{2} \{1 (- 4) + 5 (7) + 2 (- 3)\} = \frac{1}{2} \{- 4 + 35 - 6\} = \frac{1}{2} (25) = 12.5 sq . units$
Now,
$Area of triangle A C D = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{1 (7 - 4) + 2 (4 - 0) - 2 (0 - 7)\} = \frac{1}{2} \{1 (3) + 2 (4) - 2 (- 7)\} = \frac{1}{2} \{3 + 8 + 14\} = \frac{1}{2} (25) Area of triangle A C D = 12.5 sq . units$
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of quadrilateral ABCD = 12.5 sq. units + 12.5 sq. units = 25 sq. units

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Question 3:

Show that the following points are collinear:

(i) A(0, 1), B(1, 2) and C(−2, −1)
(ii) A(−5, 1), B(5, 5) and C(10, 7)
(iii) P(a, b + c), Q(b, c + a) and R(c, a + b)

Answer:

(i) A(0, 1), B(1, 2) and C(−2, −1) are the given points. Then:
(x₁ = 0, y₁ = 1), (x₂ = 1, y₂ = 2) and (x₃ =−2, y₃ = −1)
$∴ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 (2 - (- 1)) + 1 (- 1 - 1) + (- 2) (1 - 2) = 0 (3) + 1 (- 2) - 2 (- 1) = 0 - 2 + 2 = 0$
Hence, the given points are collinear.

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

(iii) P(a, b + c), Q(b, c + a) and R(c, a + b) are the given points. Then:
(x₁ = a, y₁ = b + c), (x₂ = b, y₂ = c + a) and (x₃ = c, y₃ = a + b)
$∴ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = a (c + a - a - b) + b (a + b - b - c) + c (b + c - c - a) = a (c - b) + b (a - c) + c (b - a) = a c - a b + b a - b c + c b - c a = 0$
Hence, the given points are collinear.

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Question 4:

Find the values of p for which the given points are collinear:

(i) A(−1, 3) B(2, p) and C(5, −1)
(ii) A(3, 2) B(4, p) and C(5, 3)
(iii) A(−3, 9) B(2, p) and C(4, −5)

Answer:

(i) A(−1, 3), B(2, p) and C(5, −1) are the given points. Then:
(x₁ = −1, y₁ = 3), (x₂ = 2, y₂ = p) and (x₃ = 5, y₃ = −1)
It is given that points A, B and C are collinear.
$\Rightarrow x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow (- 1) (p - (- 1)) + 2 (- 1 - 3) + 5 (3 - p) = 0 \Rightarrow (- 1) (p + 1) + 2 (- 4) + 5 (3 - p) = 0 \Rightarrow - p - 1 - 8 + 15 - 5 p = 0 \Rightarrow 6 - 6 p = 0 \Rightarrow 6 p = 6 \Rightarrow p = \frac{6}{6} = 1$
Therefore, the value of p is 1.

(ii) A(3, 2) B(4, p) and C(5, 3) are the given points. Then:
(x₁ = 3, y₁ = 2), (x₂ = 4, y₂ = p) and (x₃ = 5, y₃ = 3)
It is given that points A, B and C are collinear.
$\Rightarrow x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 3 (p - 3) + 4 (3 - 2) + 5 (2 - p) = 0 \Rightarrow 3 (p - 3) + 4 (1) + 5 (2 - p) = 0 \Rightarrow 3 p - 9 + 4 + 10 - 5 p = 0 \Rightarrow 5 - 2 p = 0 \Rightarrow 2 p = 5 \Rightarrow p = \frac{5}{2}$
Therefore, the value of p is $\frac{5}{2}$ .

(iii) A(−3, 9) B(2, p) and C(4, −5) are the given points. Then:
(x₁ = −3, y₁ = 9), (x₂ = 2, y₂ = p) and (x₃ = 4, y₃ = −5)
It is given that points A, B and C are collinear.
$\Rightarrow x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow (- 3) (p - (- 5)) + 2 (- 5 - 9) + 4 (9 - p) = 0 \Rightarrow (- 3) (p + 5) + 2 (- 14) + 4 (9 - p) = 0 \Rightarrow - 3 p - 15 - 28 + 36 - 4 p = 0 \Rightarrow - 7 - 7 p = 0 \Rightarrow 7 p = - 7 \Rightarrow p = \frac{- 7}{7} = - 1$
Therefore, the value of p is −1.

Page No 655:

Question 5:

For what value of x are the points A(−3, 12), B(7, 6) and C(x, 9) collinear?

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.

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Question 6:

For what value of y, are the points P(1, 4), Q(3, y) and R(−3, 16) collinear?

Answer:

P(1, 4), Q(3, y) and R(−3, 16) are the given points. Then:
(x₁ = 1, y₁ = 4), (x₂ = 3, y₂ = y) and (x₃ = −3, y₃ = 16)
It is given that the points P, Q and R are collinear.
Therefore,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 1 (y - 16) + 3 (16 - 4) + (- 3) (4 - y) = 0 \Rightarrow 1 (y - 16) + 3 (12) - 3 (4 - y) = 0 \Rightarrow y - 16 + 36 - 12 + 3 y = 0 \Rightarrow 8 + 4 y = 0 \Rightarrow 4 y = - 8 \Rightarrow y = - \frac{8}{4} = - 2$
When y = −2, the given points are collinear.

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Question 7:

If the points A(x, y), B(−5, 7) and C(−4, 5) are collinear, then show that 2x + y + 3 = 0.

Answer:

The given points are A(x, y), B(−5, 7) and C(−4, 5).
So, (x₁ = x, y₁ = y), (x₂ = −5, y₂ = 7) and (x₃ = −4, y₃ = 5).
If the given points are collinear, then:
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow x (7 - 5) + (- 5) (5 - y) + (- 4) (y - 7) = 0 \Rightarrow x (2) - 5 (5 - y) - 4 (y - 7) = 0 \Rightarrow 2 x - 25 + 5 y - 4 y + 28 = 0 \Rightarrow 2 x + y + 3 = 0$

Page No 655:

Question 8:

A(−4, −2), B(−3, −5) C(3, −2) and D(2, k) are the vertices of a quadrilateral ABCD. Find the value of k, if the area of the quadrilateral is 28 sq units.

Answer:

Let A(−4, −2), B(−3, −5), C(3, −2) and D(2, k) be the vertices of quadrilateral ABCD.
Join AC. Then, area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD.

$Area of the triangle A B C = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{(- 4) (- 5 - (- 2)) + (- 3) ((- 2) - (- 2)) + 3 (- 2 - (- 5))\} = \frac{1}{2} \{- 4 (- 3) - 3 (0) + 3 (3)\} = \frac{1}{2} \{12 + 9\} = \frac{1}{2} (21) Area of the triangle A B C = 10.5 sq . units .$
$Therefore, area of triangle A C D = Area of the quadrilateral A B C D - Area of the triangle A B C Area of triangle A C D = 28 sq units - 10.5 sq units = 17.5 sq . units .$
To find the value of k, let us consider the area of triangle ACD.
$Area of the triangle A C D = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} 17.5 = \frac{1}{2} \{(- 4) (- 2 - k) + 3 (k - (- 2)) + 2 (- 2 - (- 2))\} 17.5 = \frac{1}{2} \{- 4 (- 2 - k) + 3 (k + 2) + 2 (0)\} 17.5 = \frac{1}{2} \{8 + 4 k + 3 k + 6\} 17.5 = \frac{1}{2} (7 k + 14) 35 = 7 k + 14 7 k = 35 - 14 k = \frac{21}{7} = 3$
Therefore, the value of k = 3.

Page No 655:

Question 9:

For âˆ†ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2), verify that a median of the triangle divides it into two triangles of equal areas.

Answer:

To prove : Area of triangle ABD = Area of triangle ACD
Given âˆ†ABC, whose vertices are A(4, −6), B(3, −2) and C(5, 2).
Join AD. Let AD be the median that divides the triangle into two triangles.

So, D is the midpoint of BC.
Therefore, the coordinates of D are
$D (\frac{3 + 5}{2}, \frac{- 2 + 2}{2}) = D (\frac{8}{2}, \frac{0}{2}) = D (4, 0)$
Now,
$Area of triangle A B D = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} [4 (- 2 - 0) + 3 \{0 - (- 6)\} + 4 \{- 6 - (- 2)\}] = \frac{1}{2} \{4 (- 2) + 3 (6) + 4 (- 4)\} = \frac{1}{2} (- 8 + 18 - 16) = \frac{1}{2} (- 6) = 3 sq . units . [Area cannot be - ve]$
and
$Area of triangle A C D = \frac{1}{2} \{x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})\} = \frac{1}{2} \{4 (2 - 0) + 5 (0 - (- 6)) + 4 (- 6 - 2)\} = \frac{1}{2} \{4 (2) + 5 (6) + 4 (- 8)\} = \frac{1}{2} (8 + 30 - 32) = \frac{1}{2} (6) = 3 sq . units .$
Therefore, area of triangle ABD = Area of triangle ACD
Hence, the median of âˆ†ABC divides it into two triangles of equal areas.

Page No 655:

Question 10:

Find a relation between x and y if the points A(2, 1), B(x, y) and C(7, 5) are collinear.

Answer:

Consider the points A(2, 1), B(x, y) and C(7, 5) .
Here, (x₁ = 2, y₁ = 1), (x₂ = x, y₂ = y) and (x₃ = 7, y₃ = 5).
It is given that the points are collinear. So,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 2 (y - 5) + x (5 - 1) + 7 (1 - y) = 0 \Rightarrow 2 y - 10 + 5 x - x + 7 - 7 y = 0 \Rightarrow 4 x - 5 y - 3 = 0$

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Question 11:

Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear if $(\frac{1}{a} + \frac{1}{b}) = 1 .$

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.

Page No 656:

Question 1:

Find the distance between the points $(\frac{- 8}{5}, 2) and (\frac{2}{5}, 2) .$

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,
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{\{\frac{2}{5} - (\frac{- 8}{5})\}}^{2} + {(2 - 2)}^{2}} = \sqrt{{(2)}^{2} + {(0)}^{2}} = \sqrt{4 + 0} = \sqrt{4} = 2 units$

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Question 2:

Find the value of a, so that the point (3, a) lies on the line 2x − 3y = 5.

Answer:

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

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Question 3:

If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.

Answer:

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

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Question 4:

If P(x, y) is equidistant from the points A(7, 1) and B(3, 5), find the relation between x and y.

Answer:

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

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Question 5:

If the centroid of âˆ†ABC, which has vertices A(a, b), B(b, c) and C(c, a), is the origin, find the value of (a + b + c).

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:
$\frac{a + b + c}{3} = 0 \Rightarrow a + b + c = 0$

Page No 656:

Question 6:

Find the centroid of âˆ†ABC whose vertices are A(2, 2), B(−4, −4) and C(5, −8).

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 $∆ A B C is G (1, \frac{- 10}{3})$ .

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Question 7:

In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)?

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,
$\frac{7 k + 2}{k + 1} = 4 and \frac{8 k + 3}{k + 1} = 5 [∵ C (4, 5) is given] \Rightarrow 7 k + 2 = 4 k + 4 and 8 k + 3 = 5 k + 5 \Rightarrow 3 k = 2 and 3 k = 2$
$\Rightarrow k = \frac{2}{3} in each case$
$So, the required ratio is \frac{2}{3} : 1, which is same as 2:3.$

Page No 656:

Question 8:

If the points A(2, 3) B(4, k) and C(6, −3) are collinear, find the value of k.

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,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 2 (k + 3) + 4 (- 3 - 3) + 6 (3 - k) = 0 \Rightarrow 2 k + 6 - 24 + 18 - 6 k = 0 \Rightarrow - 4 k = 0 \Rightarrow k = 0$

Page No 658:

Question 1:

In which quadrant does the point (−3, 5) lie?

(a) I
(b) II
(c) III
(d) IV

Answer:

(b) II
The point (−3, 5) lies in quadrant II, since the signs of the coordinates are (−, +).

Page No 658:

Question 2:

In which quadrant does the point (2, −4) lie?

(a) I
(b) II
(c) III
(d) IV

Answer:

(d) IV
The point (2, −4) lies in quadrant IV since the signs of the coordinates are (+, −).

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Question 3:

The distance of the point P(3, 4) from the x-axis is

(a) 3 units
(b) 4 units
(c) 7 units
(d) 1 unit

Answer:

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

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Question 4:

P is a point on x-axis at a distance of 3 units from y-axis, to its right. The coordinates of P are

(a) (3, 0)
(b) (0, 3)
(c) (3, 3)
(d) (−3, 3)

Answer:

(a) (3, 0)
$It is given that point P is on the x - axis at a distance of 3 units from the y - axis, to its right . The refore, the coordinates of P are (3, 0) .$

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Question 5:

A is a point on the y-axis at a distance of 4 units from the x-axis and lying below the x-axis. The coordinates of A are

(a) (4, 0)
(b) (0, 4)
(c) (−4, 0)
(d) (0, −4)

Answer:

(d) (0, −4)
$It is given that point A is on the y - axis at a distance of 4 units from the x - axis and lies below the x - axis . Therefore, the coordinates of A are (0, - 4) .$

Page No 658:

Question 6:

The distance of the point P(4, −3) from the origin is

(a) 1 unit
(b) 3 units
(c) 5 units
(d) 7 units

Answer:

(c) 5 units
The given point is Pâ€‹(4, −3) and the origin is (0, 0).
Then, $(x_{1} = 4, y_{1} = - 3) and (x_{2} = 0, y_{2} = 0)$
$O P = \sqrt{{(0 - 4)}^{2} + {(0 - (- 3))}^{2}} = \sqrt{{(- 4)}^{2} + {(3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$

Page No 658:

Question 7:

The distance between the points A(2, −3) and B(2, 2) is

(a) 4 units
(b) 5 units
(c) 3 units
(d) 2 units

Answer:

(b) 5 units
The given points are A(2, −3) and B(2, 2).
Then, $(x_{1} = 2, y_{1} = - 3) and (x_{2} = 2, y_{2} = 2)$
Therefore,
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(2 - 2)}^{2} + {(2 + 3)}^{2}} = \sqrt{{(0)}^{2} + {(5)}^{2}} = \sqrt{0 + 25} = \sqrt{25} = 5 units$

Page No 658:

Question 8:

What point on the x-axis is equidistant from the points A(7, 6) and B(−3, 4)?

(a) (0, 4)
(b) (−4, 0)
(c) (3, 0)
(d) (0, 3)

Answer:

(c) (3, 0)
Let P(x,0) be equidistant from the points A(7,6) and B(−3,4).
Then,
$P A = P B \Rightarrow P A^{2} = P B^{2} \Rightarrow {(x - 7)}^{2} + {(0 - 6)}^{2} = {(x + 3)}^{2} + {(0 - 4)}^{2} \Rightarrow x^{2} - 14 x + 49 + 36 = x^{2} + 6 x + 9 + 16 \Rightarrow x^{2} - 14 x + 85 = x^{2} + 6 x + 25 \Rightarrow 20 x = 60 \Rightarrow x = 3 units$
Hence,
$(x = 3, y = 0) \Rightarrow (3, 0)$

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Question 9:

The distance between the points A(0, 5) and B(−5, 0) is

(a) 5 units
(b) $2 \sqrt{5}$ units
(c) $\sqrt{10}$ units
(d) $5 \sqrt{2}$ units

Answer:

(d) $5 \sqrt{2}$ units
The given points are A(0, 5) and B(−5, 0).
Then, $(x_{1} = 0, y_{1} = 5) and (x_{2} = - 5, y_{2} = 0)$
Therefore,
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(- 5 - 0)}^{2} + {(0 - 5)}^{2}} = \sqrt{{(- 5)}^{2} + {(- 5)}^{2}} = \sqrt{25 + 25} = \sqrt{50} = 5 \sqrt{2} units$

Page No 658:

Question 10:

A point P divides the join of A(5, −2) and B(9, 6) in the ratio 3 : 1. The coordinates of P are

(a) (4, 7)
(b) (8, 4)
(c) (12, 8)
(d) $(\frac{11}{2}, 5)$

Answer:

(b) (8, 4)
The given points are A(5, −2) and B(9, 6)
Then, $(x_{1} = 5, y_{1} = - 2) and (x_{2} = 9, y_{2} = 6)$
Also, m â€‹= 3 and n = 1
Let the required point be P(x, y).
By section formula, we have
$x = \frac{m x_{2} + n x_{1}}{m + n}, y = \frac{m y_{2} + n y_{1}}{m + n} \Rightarrow x = \frac{(3 \times 9 + 1 \times 5)}{(3 + 1)}, y = \frac{(3 \times 6 + 1 \times (- 2))}{(3 + 1)} \Rightarrow x = \frac{32}{4}, y = \frac{16}{4} \Rightarrow x = 8, y = 4$
Hence, the coordinates of P are (8, 4).

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Question 11:

In what ratio does the point P(1, 2) divide the join of A(−2, 1) and B(7, 4)?

(a) 1 : 2
(b) 2 : 1
(c) 3 : 2
(d) 2 : 3

Answer:

(a) 1 : 2
Let the required ratio be k : 1.
Then, by section formula, the coordinates of P are
$P (\frac{7 k - 2}{k + 1}, \frac{4 k + 1}{k + 1})$
Therefore,
$\frac{7 k - 2}{k + 1} = 1 and \frac{4 k + 1}{k + 1} = 2 [Since P (1, 2) is given]$
$\Rightarrow 7 k - 2 = k + 1 and 4 k + 1 = 2 k + 2$
$\Rightarrow 6 k = 3 and 2 k = 1$
$\Rightarrow k = \frac{1}{2}$ in each case
So, the required ratio is $\frac{1}{2} : 1$ , which is same as 1 : 2.

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Question 12:

The point which divides the line segment joining the points A(7, −6) and B(3, 4) in the ratio 1 : 2 lies in

(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant

Answer:

(d) IV quadrant
The given points are $A (7, - 6)$ and $B (3, 4)$ .
Then, $(x_{1} = 7, y_{1} = - 6) and (x_{2} = 3, y_{2} = 4)$
Also, m = 1 and n = 2
Let the required point be $P (x, y)$ .
By section formula, we have
$x = \frac{(1 \times 3 + 2 \times 7)}{(1 + 2)}, y = \frac{(1 \times 4 + 2 \times (- 6))}{(1 + 2)}$
$\Rightarrow x = \frac{17}{3}, y = \frac{- 8}{3}$
Hence, the coordinates of P are $(\frac{17}{3}, \frac{- 8}{3})$ , which lies in the IV quadrant.

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Question 13:

In what ratio does the x-axis divide the join of A(2, −3) and B(5, 6)?

(a) 2 : 3
(b) 3 : 5
(c) 1 : 2
(d) 2 : 1

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.

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Question 14:

In what ratio does the y-axis divide the join of P(−4, 2) and Q(8, 3)?

(a) 3 : 1
(b) 1 : 3
(c) 2 : 1
(d) 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.

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Question 15:

The midpoint of the line segment joining the points A(−2, 8) and B(−6, −4) is

(a) (4, 2)
(b) (−4, 2)
(c) (2, 6)
(d) (−4, −6)

Answer:

(b) (−4, 2)
The given points are A(−2, 8) and B(−6, −4).
Then, $(x_{1} = - 2, y_{1} = 8) and (x_{2} = - 6, y_{2} = - 4)$
Let M(x,y) be the midpoint of AB. Then,
$x = \frac{[(- 2) + (- 6)]}{2} = \frac{- 8}{2} = - 4$
and
$y = \frac{[8 + (- 4)]}{2} = \frac{4}{2} = 2$
Hence, the required point is $M (- 4, 2)$ .

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Question 16:

If P(−1, 1) is the midpoint of the line segment joining A(−3, b) and B(1, b + 4), then b = ?

(a) 1
(b) −1
(c) 2
(d) 0

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,
$b + 2 = 1 \Rightarrow b = - 1$

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Question 17:

If $P (\frac{a}{3}, 4)$ is the midpoint of the line segment joining A(−6, 5) and B(−2, 3), then a = ?

(a) −4
(b) −12
(c) 12
(d) −6

Answer:

(b) −12
The given points are A(−6, 5) and B(−2, 3).
Then, $(x_{1} = - 6, y_{1} = 5) and (x_{2} = - 2, y_{2} = 3)$
Therefore,
$x = \frac{[(- 6) + (- 2)]}{2} = \frac{- 8}{2} = - 4$
and
$y = \frac{[5 + 3]}{2} = \frac{8}{2} = 4$
But the given midpoint is $P (\frac{a}{3}, 4)$ .
Therefore,
$- 4 = \frac{a}{3} \Rightarrow a = - 12$

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Question 18:

If the distance between the points A(2, −2) and B(−1, x) is 5, then

(a) x = −3 or x = 4
(b) x = 3 or x = −4
(c) x = −6 or x = 2
(d) x = 6 or x = −2

Answer:

(c) x = −6 or x = 2
The given points are A(2, −2) and B(−1, x) and AB = 5.
Then, $(x_{1} = 2, y_{1} = - 2) and (x_{2} = - 1, y_{2} = x)$
Therefore,
$A B = 5 \Rightarrow \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = 5 \Rightarrow \sqrt{{(- 1 - 2)}^{2} + {(x + 2)}^{2}} = 5 \Rightarrow {(- 3)}^{2} + {(x + 2)}^{2} = 25 \Rightarrow 9 + x^{2} + 4 x + 4 = 25 \Rightarrow x^{2} + 4 x + 13 = 25 \Rightarrow x^{2} + 4 x - 12 = 0 \Rightarrow (x + 6) (x - 2) = 0 \Rightarrow x = - 6 or x = 2$

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Question 19:

The line 2x + y −4 = 0 divides the line segment joining A(2, −2) and B(3, 7) in the ratio

(a) 2 : 5
(b) 2 : 9
(c) 2 : 7
(d) 2 : 3

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.

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Question 20:

If A(4, 2), B(6, 5) and C(1, 4) be the vertices of âˆ†ABC and AD is median, then the coordinates of D are

(a) $(\frac{5}{2}, 3)$
(b) $(5, \frac{7}{2})$
(c) $(\frac{7}{2}, \frac{9}{2})$
(d) None of these

Answer:

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

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Question 21:

If A(−1, 0), B(5, −2) and C(8, 2) are the vertices of a âˆ†ABC, then its centroid is

(a) (12, 0)
(b) (6, 0)
(c) (0, 6)
(d) (4, 0)

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).

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Question 22:

Two vertices of âˆ†ABC are A(−1, 4) and B(5, 2) and its centroid is G(0, −3). Then, the coordinates of C are

(a) (4, 3)
(b) (4, 15)
(c) (−4, −15)
(d) (−15, −4)

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 $∆ A B C is C (- 4, - 15)$ .

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Question 23:

If the distance between the points A(4, p) and B(1, 0) is 5, then

(a) p = 4 only
(b) p = −4 only
(c) p = ±4 only
(d) p = 0

Answer:

(c) p = ±4 only
The given points are A(4, p) and B(1, 0) and AB = 5.
Then, $(x_{1} = 4, y_{1} = p) and (x_{2} = 1, y_{2} = 0)$
Therefore,
$A B = 5 \Rightarrow \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = 5 \Rightarrow \sqrt{{(1 - 4)}^{2} + {(0 - p)}^{2}} = 5 \Rightarrow {(- 3)}^{2} + {(- p)}^{2} = 25 \Rightarrow 9 + p^{2} = 25 \Rightarrow p^{2} = 16 \Rightarrow p = \pm \sqrt{16} \Rightarrow p = \pm 4$

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Question 24:

The three vertices of a parallelogram ABCD are A(−2, 3) B(6, 7) and C(8, 3). The fourth vertex D is

(a) (1, 0)
(b) (0, 1)
(c) (−1, 0)
(d) (0, −1)

Answer:

(d) (0, −1)
Let the fourth vertex be D(a, b).
Join AC and BD.
Let AC and BD intersect at the point O.
We know that the diagonals of a parallelogram bisect each other.
$Midpoint of A C is (\frac{- 2 + 8}{2}, \frac{3 + 3}{2}) \Rightarrow (3, 3) Midpoint of B D is (\frac{6 + a}{2}, \frac{7 + b}{2})$
Therefore,
$\frac{6 + a}{2} = 3 and \frac{7 + b}{2} = 3 \Rightarrow 6 + a = 6 and 7 + b = 6$
$\Rightarrow a = 0 and b = - 1$
Hence, the fourth vertex is D(0,−1).

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Question 25:

The points A(−4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is

(a) isosceles
(b) equilateral
(c) scalene
(d) right-angled

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.

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Question 26:

The points P(0, 6), Q(−5, 3) and R(3, 1) are the vertices of a triangle, which is

(a) equilateral
(b) isosceles
(c) scalene
(d) right-angled

Answer:

(d) right-angled

Let P(0, 6), Q(−5, 3) and R(3, 1) be the given points. Then,
$P Q = \sqrt{{(- 5 - 0)}^{2} + {(3 - 6)}^{2}} = \sqrt{{(- 5)}^{2} + {(- 3)}^{2}} = \sqrt{25 + 9} = \sqrt{34} units Q R = \sqrt{{(3 + 5)}^{2} + {(1 - 3)}^{2}} = \sqrt{{(8)}^{2} + {(- 2)}^{2}} = \sqrt{64 + 4} = \sqrt{68} = 2 \sqrt{17} units P R = \sqrt{{(3 - 0)}^{2} + {(1 - 6)}^{2}} = \sqrt{{(3)}^{2} + {(- 5)}^{2}} = \sqrt{9 + 25} = \sqrt{34} units P Q^{2} + P R^{2} \Rightarrow \{{(\sqrt{34})}^{2} + {(\sqrt{34})}^{2}\} = 68 Q R^{2} \Rightarrow {(2 \sqrt{17})}^{2} = 68$
$Thus, P Q^{2} + P R^{2} = Q R^{2}$
Therefore, âˆ†PQR is right-angled.

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Question 27:

The points (a, a), (−a, a) and $(- \sqrt{3} a, \sqrt{3} a)$ form the vertices of

(a) an equilateral triangle
(b) a scalene triangle
(c) an isosceles triangle
(d) a right triangle

Answer:

(a) an equilateral triangle
Let A(a, a), B(−a, −a) and $C (- \sqrt{3} a, \sqrt{3} a)$ be the given points. Then,
$A B = \sqrt{{(- a - a)}^{2} + {(- a - a)}^{2}} = \sqrt{{(- 2 a)}^{2} + {(- 2 a)}^{2}} = \sqrt{4 a^{2} + 4 a^{2}} = \sqrt{8 a^{2}} = 2 \sqrt{2} a units B C = \sqrt{{(- \sqrt{3} a + a)}^{2} + {(\sqrt{3} a + a)}^{2}} = a \sqrt{{(1 - \sqrt{3})}^{2} + {(1 + \sqrt{3})}^{2}} = a \sqrt{2 (1 + 3)} = a \sqrt{8} = 2 \sqrt{2} a units A C = \sqrt{{(- \sqrt{3} a - a)}^{2} + {(\sqrt{3} a - a)}^{2}} = a \sqrt{{(\sqrt{3} + 1)}^{2} + {(\sqrt{3} - 1)}^{2}} = a \sqrt{2 (3 + 1)} = a \sqrt{8} = 2 \sqrt{2} a units$
AB = BC = AC
Hence, the points form the vertices of an equilateral triangle.

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Question 28:

Three points A(1, −2) B(3, 4) and C(4, 7) form

(a) a straight line
(b) an equilateral triangle
(c) a right-angled triangle
(d) None of these

Answer:

(a) a straight line
The given points are A(1, −2), B(3, 4) and C(4, 7).
Here, $(x_{1} = 1, y_{1} = - 2), (x_{2} = 3, y_{2} = 4) and (x_{3} = 4, y_{3} = 7)$ .
Therefore,
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \frac{1}{2} [1 (4 - 7) + 3 (7 + 2) + 4 (- 2 - 4)] = \frac{1}{2} [- 3 + 27 - 24] = 0$
Thus, the given points are collinear, i.e. they form a straight line.

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Question 29:

If the points A(2, 3), B(5, k) and C(6, 7) are collinear, then

(a) k = 4
(b) k = 6
(c) $k = \frac{- 3}{2}$
(d) $k = \frac{11}{4}$

Answer:

(b) k = 6
The given points are A(2, 3), B(5, k) and C(6, 7).
Here, $(x_{1} = 2, y_{1} = 3), (x_{2} = 5, y_{2} = k) and (x_{3} = 6, y_{3} = 7) .$
Points A,B and C are collinear. Then,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 2 (k - 7) + 5 (7 - 3) + 6 (3 - k) = 0 \Rightarrow 2 k - 14 + 20 + 18 - 6 k = 0 \Rightarrow - 4 k = - 24 \Rightarrow k = 6$

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Question 30:

If the points A(1, 2), O(0, 0) and C(a, b) are collinear, then

(a) a = b
(b) a = 2b
(c) 2a = b
(d) a + b = 0

Answer:

(c) 2a = b
The given points are A(1, 2), O(0, 0) and C(a, b).
Here, $(x_{1} = 1, y_{1} = 2), (x_{2} = 0, y_{2} = 0) and (x_{3} = a, y_{3} = b)$ .
Points A, O and C are collinear.
$\Rightarrow x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 1 (0 - b) + 0 (b - 2) + a (2 - 0) = 0 \Rightarrow - b + 2 a = 0 \Rightarrow 2 a = b$

Page No 660:

Question 31:

The area of âˆ†ABC with vertices A(a, b + c), B(b, c + a) and C(c, a + b) is

(a) (a + b + c)²
(b) a + b + c
(c) abc
(d) 0

Answer:

(d) 0
The given points are $A (a, b + c), B (b, c + a) and C (c, a + b)$ .
Here, $(x_{1} = a, y_{1} = b + c), (x_{2} = b, y_{2} = c + a) and (x_{3} = c, y_{3} = a + b)$
Therefore,
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \frac{1}{2} [a \{(c + a) - (a + b)\} + b \{(a + b) - (b + c)\} + c \{(b + c) - (c + a)\}] = \frac{1}{2} [a (c - b) + b (a - c) + c (b - a)] = \frac{1}{2} (a c - a b + a b - b c + b c - a c) = 0$

Page No 660:

Question 32:

The point which lies on the perpendicular bisector of the line segment joining the points A(−2, −5) and B(2, 5) is

(a) (0, 0)
(b) (0, 2)
(c) (2, 0)
(d) (−2, 0)

Answer:

(a) (0, 0)
The point which lies on the perpendicular bisector of AB is the midpoint of AB.
The given points are $A (- 2, - 5) and B (2, 5)$ .
Here, $(x_{1} = - 2, y_{1} = - 5) and (x_{2} = 2, y_{2} = 5)$
$Midpoint of A B = (\frac{- 2 + 2}{2}, \frac{- 5 + 5}{2}) i.e. (0, 0)$
Hence, the required point is (0,0).

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Question 33:

In the given figure, A(0, 2y) and B(2x, 0) are the end-points of line segment AB. The coordinates of point C, which is equidistant from the three vertices of âˆ†AOB, are

(a) (x , y)
(b) (y , x)
(c) $(\frac{x}{2}, \frac{y}{2})$
(d) $(\frac{y}{2}, \frac{x}{2})$

Answer:

(a) (x , y)
The midpoint of AB is equidistant from the three vertices of $∆ A O B .$
Therefore,
$Midpoint of A B = (\frac{0 + 2 x}{2}, \frac{2 y + 0}{2}) [A (0, 2 y) and B (2 x, 0) \Rightarrow (x_{1} = 0, y_{1} = 2 y) and (x_{2} = 2 x, y_{2} = 0)] i.e. (x, y)$
Hence, the coordinates of C are (x, y).

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Question 34:

The points A(9, 0), B(9, 6), C(−9, 6) and D(−9, 0) are the vertices of a

(a) square
(b) rectangle
(c) rhombus
(d) trapezium

Answer:

(b) rectangle
$A (9, 0), B (9, 6), C (- 9, 6) and D (- 9, 0)$ are the given vertices.
Then,
$A B^{2} = {(9 - 9)}^{2} + {(6 - 0)}^{2} = {(0)}^{2} + {(6)}^{2} = 0 + 36 = 36 units B C^{2} = {(- 9 - 9)}^{2} + {(6 - 6)}^{2} = {(- 18)}^{2} + {(0)}^{2} = 324 + 0 = 324 units C D^{2} = {(- 9 + 9)}^{2} + {(0 - 6)}^{2} = {(0)}^{2} + {(- 6)}^{2} = 0 + 36 = 36 units D A^{2} = {(- 9 - 9)}^{2} + {(0 - 0)}^{2} = {(- 18)}^{2} + {(0)}^{2} = 324 + 0 = 324 units$
Therefore, we have:
$A B^{2} = C D^{2} and B C^{2} = D A^{2}$
Now, the diagonals are:
$A C^{2} = {(- 9 - 9)}^{2} + {(6 - 0)}^{2} = {(- 18)}^{2} + {(6)}^{2} = 324 + 36 = 360 units B D^{2} = {(- 9 - 9)}^{2} + {(0 - 6)}^{2} = {(- 18)}^{2} + {(- 6)}^{2} = 324 + 36 = 360 units$
Therefore,
$A C^{2} = B D^{2}$
Hence, ABCD is a rectangle.

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Question 35:

The area of âˆ†ABC with vertices A(3, 0), B(7, 0) and C(8, 4) is

(a) 14 sq units
(b) 28 sq units
(c) 8 sq units
(d) 6 sq units

Answer:

(c) 8 sq units
The given points are $A (3, 0), B (7, 0) and C (8, 4)$ .
Here, $(x_{1} = 3, y_{1} = 0), (x_{2} = 7, y_{2} = 0) and (x_{3} = 8, y_{3} = 4)$
Therefore,
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \frac{1}{2} [3 (0 - 4) + 7 (4 - 0) + 8 (0 - 0)] = \frac{1}{2} [- 12 + 28 + 0] = (\frac{1}{2} \times 16) = 8 sq units$

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Question 36:

AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). Length of each of its diagonal is

(a) 5 units
(b) 3 units
(c) $\sqrt{34}$ units
(d) 4 units

Answer:

(c) $\sqrt{34}$ units
$A (0, 3), O (0, 0) and B (5, 0)$ are the three vertices of a rectangle; let C be the fourth vertex.
Then, the length of the diagonal,
$A B = \sqrt{{(5 - 0)}^{2} + {(0 - 3)}^{2}} = \sqrt{{(5)}^{2} + {(- 3)}^{2}} = \sqrt{25 + 9} = \sqrt{34} units$
Since, the diagonals of rectangle are equal .
Hence, the length of its diagonals is $\sqrt{34} units$ .

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Question 37:

A line intersects the y-axis and x-axis at the points P and Q, respectively. If (2, −5) is the midpoint of PQ, then the coordinates of P and Q, respectively, are

(a) (0, −5) and (2, 0)
(b) (0, −10) and (4, 0)
(c) (0, 10) and (−4, 0)
(d) (0, 4) and (−10, 0)

Answer:

(b) (0, −10) and (4, 0)
Let the line PQ intersect y axis at P(0, a) and x axis at Q(b, 0).
$Midpoint of P Q = (\frac{0 + b}{2}, \frac{a + 0}{2})$
But the midpoint of PQ is (2, −5).
Therefore,
$\frac{0 + b}{2} = 2 and \frac{a + 0}{2} = - 5 \Rightarrow b = 4 and a = - 10 .$
Therefore,
$P (0, a) \Rightarrow P (0, - 10) Q (b, 0) \Rightarrow Q (4, 0)$
Hence, the coordinates of P and Q are (0, −10) and (4, 0).

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Question 38:

A circle drawn with origin as the centre passes through the point $A (\frac{13}{2}, 0)$ . Which of the following points does not lie inside the circle?

(a) $(\frac{- 3}{4}, 1)$
(b) $(2, \frac{7}{3})$
(c) $(5, \frac{- 1}{3})$
(d) $(- 6, \frac{5}{2})$

Answer:

(d) $(- 6, \frac{5}{2})$
Let the given point be $A (\frac{13}{2}, 0)$ and the origin (0, 0). Then,
the radius of the circle,
$O A = \sqrt{(0 - \frac{13}{2}) + {(0 - 0)}^{2}}$
$\Rightarrow O A = \sqrt{{(\frac{- 13}{2})}^{2} + {(0)}^{2}} \Rightarrow O A = \sqrt{\frac{169}{4}} \Rightarrow O A = \frac{13}{2} units$
Now, the distance of the point $(\frac{- 3}{4}, 1)$ from the origin is
$\sqrt{(0 + \frac{3}{4}) + {(0 - 1)}^{2}} = \sqrt{{(\frac{3}{4})}^{2} + {(- 1)}^{2}} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4} units, which lies inside the circle.$
Now, the distance of the point $(2, \frac{7}{3})$ from the origin is
$\sqrt{{(0 - 2)}^{2} + {(0 - \frac{7}{3})}^{2}} = \sqrt{{(- 2)}^{2} + {(\frac{- 7}{3})}^{2}} = \sqrt{4 + \frac{49}{9}} = \sqrt{\frac{85}{9}} = \frac{\sqrt{85}}{3} units, which lies inside the circle.$
Now, the distance of the point $(5, \frac{- 1}{2})$ from the origin is
$\sqrt{{(0 - 5)}^{2} + {(0 + \frac{1}{2})}^{2}} = \sqrt{{(- 5)}^{2} + {(\frac{1}{2})}^{2}} = \sqrt{25 + \frac{1}{4}} = \sqrt{\frac{101}{4}} = \frac{\sqrt{101}}{2} units, which lies inside the circle .$
Now, the distance of the point $(- 6, \frac{5}{2})$ from the origin is
$\sqrt{{(0 + 6)}^{2} + {(0 - \frac{5}{2})}^{2}} = \sqrt{{(6)}^{2} + {(\frac{- 5}{2})}^{2}} = \sqrt{36 + \frac{25}{4}} = \sqrt{\frac{169}{4}} = \frac{13}{2} units=radius OA$
Hence, it does not lie inside the circle.

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Question 39:

The coordinates of one end point of a diameter AB of a circle are A(4, −1) and the coordinates of the centre of the circle are C(1, −3). Then, the coordinates of B are

(a) (2, −5)
(b) (−2, 5)
(c) (−2, −5)
(d) (2, 5)

Answer:

(c) (−2, −5)
Let the coordinates of B be (a, b). Then,
$Midpoint of A B = (\frac{4 + a}{2}, \frac{- 1 + b}{2})$
But the midpoint C(1, −3) is given.
Therefore,
$\frac{4 + a}{2} = 1 and \frac{- 1 + b}{2} = - 3 \Rightarrow 4 + a = 2 and - 1 + b = - 6 \Rightarrow a = - 2 and b = - 5 .$
Hence, the coordinates of B are (−2, −5).

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Question 40:

Match the following columns:

Column I	Column II
(a) The coordinates of the point that divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3 are .......	(p) (5, 6)
(b) Two vertices of âˆ†ABC are A(6, 4) and B(−2, 2) and its centroid is G(3, 4). The coordinates of C are .......	(q) $\sqrt{10}$
(c) If the points A(4, 3) and B(x, 5) lie on a circle with the centre O(2, 3), then x = ....... .	(r) (1, 3)
(d) If A(0, −1), B(2, 1) and C(0, 3) are the vertices of a âˆ†ABC, then the length of median AD is	(s) 2

Answer:

(a) ⇒ (r)
(b) ⇒ (p)
(c) ⇒ (s)
(d) ⇒ (q)

(a) The given points are $A (- 1, 7) and B (4, - 3)$ .
Then, $(x_{1} = - 1, y_{1} = 7) and (x_{2} = 4, y_{2} = - 3)$
Also, m = 2 and n = 3
Let the required point be P(x, y).
By, section formula, we have:
$x = \frac{(2 \times 4 + 3 \times (- 1))}{(2 + 3)}, y = \frac{(2 \times (- 3) + 3 \times 7)}{(2 + 3)} \Rightarrow x = \frac{5}{5}, y = \frac{15}{5} \Rightarrow x = 1, y = 3$
Therefore, the coordinates of P are (1, 3).
Hence, $(a) \Rightarrow (r)$

(b) The two vertices of $∆ A B C are A (6, 4) and B (- 2, 2)$ .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
$G (\frac{6 - 2 + a}{3}, \frac{4 + 2 + b}{3}) i.e. G (\frac{4 + a}{3}, \frac{6 + b}{3})$
But it is given that the centroid is G(3, 4).
Therefore,
$\frac{4 + a}{3} = 3 and \frac{6 + b}{3} = 4 \Rightarrow 4 + a = 9 and 6 + b = 12$
$\Rightarrow a = 5 and b = 6$
Therefore, the coordinates of C are (5, 6).
Hence, $(b) \Rightarrow (p)$

(c) The points $A (4, 3) and B (x, 5)$ lie on the circle with centre O(2, 3).
Then, we have:
${(x - 2)}^{2} + {(5 - 3)}^{2} = {(4 - 2)}^{2} + {(3 - 3)}^{2}$
$\Rightarrow {(x - 2)}^{2} + 2^{2} = 2^{2} + 0^{2} \Rightarrow {(x - 2)}^{2} = (2^{2} - 2^{2}) \Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x - 2 = 0 \Rightarrow x = 2$
Therefore, the value of x = 2
Hence, $(c) \Rightarrow (s)$

(d) D is the midpoint of BC.
So, the coordinates of D are
$D (\frac{2 + 0}{2}, \frac{1 + 3}{2}) [B (2, 1) and C (1, 4) \Rightarrow (x_{1} = 2, y_{1} = 1) and (x_{2} = 1, y_{2} = 4)] i.e. D (1, 2)$
Now, the length of the median
$A D = \sqrt{{(1 - 0)}^{2} + {(2 + 1)}^{2}} = \sqrt{{(1)}^{2} + {(3)}^{2}} = \sqrt{1 + 9} = \sqrt{10} units$
Hence, $(d) \Rightarrow (q)$

Page No 662:

Question 41:

Assertion (A)
If the points A(8, 1), B(k, −4) and C(2, −5) are collinear, then k = 4.

Reason (R)
Three points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are collinear only when x₁(y₂− y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂) = 0.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

The given points are $A (8, 1), B (k, - 4) and C (2, - 5)$
Here, $(x_{1} = 8, y_{1} = 1), (x_{2} = k, y_{2} = - 4) and (x_{3} = 2, y_{3} = - 5)$
Points A,B and C are collinear.
$\Rightarrow x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow 8 (- 4 + 5) + k (- 5 - 1) + 2 (1 + 4) = 0 \Rightarrow 8 (1) + k (- 6) + 2 (5) = 0 \Rightarrow - 6 k = - 18 \Rightarrow k = 3$
Hence, Assertion(A) is false.
Reason(R):
The given statement is true.
Hence, the correct answer is (d).

Page No 662:

Question 42:

Assertion (A)
The point P(−1, 6) divides the join of A(−3, 10) and B(6, −8) in the ratio 2 : 7.

Reason (R)
If the point C(x, y) divides the join of A(x₁, y₁) and B(x₂, y₂) in the ratio m : n, then $x = \frac{m x_{2} + n x_{1}}{m + n} and y = \frac{m y_{2} + n y_{1}}{m + n} .$

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

The correct answer is (a)
Let the required ratio be k : 1.
Then, by section formula, the coordinates of C are
$P (\frac{6 k - 3}{k + 1}, \frac{- 8 k + 10}{k + 1})$
Therefore,
$\frac{6 k - 3}{k + 1} = - 1 and \frac{- 8 k + 10}{k + 1} = 6 [Since P (- 1, 6) is given] \Rightarrow 6 k - 3 = - k - 1 and - 8 k + 10 = 6 k + 6$

$\Rightarrow 7 k = 2 and - 14 k = - 4$
$\Rightarrow k = \frac{2}{7} in each case.$
Therefore, the required ratio is $\frac{2}{7} : 1$ , which is same as 2 : 7.
Hence, Assertion(A) is true.
Reason(R):
The given statement is true.
Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

Page No 667:

Question 1:

The ratio in which the point P(6, −6) divides the line segment joining A(1, 4) and B(9, −12) is

(a) 2 : 3
(b) 3 : 2
(c) 3 : 5
(d) 5 : 3

Answer:

(d) 5 : 3
Let the required ratio be k : 1.
Then, by section formula, the coordinates of C are
$P (\frac{9 k + 1}{k + 1}, \frac{- 12 k + 4}{k + 1})$
Therefore,
$\frac{9 k + 1}{k + 1} = 6 and \frac{9 k + 1}{k + 1} = - 6 [Since P (6, - 6) is given] \Rightarrow 9 k + 1 = 6 k + 6 and 9 k + 1 = - 6 k - 6 \Rightarrow 3 k = 5 and - 6 k = - 10 \Rightarrow k = \frac{5}{3} in each case.$
Therefore, the required ratio is $\frac{5}{3} : 1$ , which is same as 5 : 3.

Page No 667:

Question 2:

In what ratio is the segment joining the points A(4, 6) and B(−7, −1) divided by x-axis?

(a) 3 : 1
(b) 6 : 1
(c) 1 : 2
(d) 2 : 3

Answer:

(b) 6 : 1
Let AB be divided by x-axis in the ratio k : 1 at the point P.
Then, by section formula, the coordinates of P are
$P (\frac{- 7 k + 4}{k + 1}, \frac{- k + 6}{k + 1})$
But P lies on the x-axis. So, its ordinate is 0.
Therefore,
$\frac{- k + 6}{k + 1} = 0 \Rightarrow - k + 6 = 0 \Rightarrow k = 6$
Hence, the required ratio is 6 : 1.

Page No 667:

Question 3:

The two vertices of a triangle are A(6, 3) and B(−1, 7) and its centroid is G(1, 5). The third vertex C of âˆ†ABC is

(a) (2, 5)
(b) (2, −5)
(c) (−2, 5)
(d) (−2, −5)

Answer:

(c) (−2, 5)
The two vertices of $∆ A B C are A (6, 3) and B (- 1, 7)$ .
Let the third vertex be C(a, b).
Then, the coordinates of its centroid are
$G (\frac{6 - 1 + a}{3}, \frac{3 + 7 + b}{3}) i.e. G (\frac{5 + a}{3}, \frac{10 + b}{3})$
But it is given that the centroid is G(1,5).
Therefore,
$\frac{5 + a}{3} = 1 and \frac{10 + b}{3} = 5 \Rightarrow 5 + a = 3 and 10 + b = 15 \Rightarrow a = - 2 and b = 5$
Hence, the third vertex of $∆ A B C is C (- 2, 5) .$

Page No 668:

Question 4:

The area of âˆ†ABC with vertices A(1, −1), B(−4, 6) and C(−3, −5) is

(a) 52 sq units
(b) 27 sq units
(c) 48 sq units
(d) 24 sq units

Answer:

(d) 24 sq units
The given points are $A (1, - 1), B (- 4, 6) and C (- 3, - 5)$ .
Here, $(x_{1} = 1, y_{1} = - 1), (x_{2} = - 4, y_{2} = 6) and (x_{3} = - 3, y_{3} = - 5)$
Therefore,
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
$= \frac{1}{2} [1 (6 + 5) + (- 4) (- 5 + 1) + (- 3) (- 1 - 6)] = \frac{1}{2} [11 + 16 + 21] = (\frac{1}{2} \times 48) = 24 sq units$

Page No 668:

Question 5:

Find the length of the line segment AB whose end points are A(2, −5) and B(−4, 7).

Answer:

The given points are $A (2, - 5) and B (- 4, 7)$ .
Then, $(x_{1} = 2, y_{1} = - 5) and (x_{2} = - 4, y_{2} = 7)$
Therefore,
$A B = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} = \sqrt{{(- 4 - 2)}^{2} + {(7 + 5)}^{2}} = \sqrt{{(- 6)}^{2} + {(12)}^{2}} = \sqrt{36 + 144} = \sqrt{180} = 6 \sqrt{5} units$

Page No 668:

Question 6:

Find the point on x-axis that is equidistant from the points A(−3, 2) and B(5, −2).

Answer:

Let the point P(x,0) be equidistant from the points $A (- 3, 2) and B (5, - 2)$ .
Then,
$P A = P B \Rightarrow P A^{2} = P B^{2} \Rightarrow {(x + 3)}^{2} + {(0 - 2)}^{2} = {(x - 5)}^{2} + {(0 + 2)}^{2} \Rightarrow x^{2} + 6 x + 13 = x^{2} - 10 x + 29 \Rightarrow 16 x = 16 \Rightarrow x = 1$
Therefore,
$P (x, 0) \equiv P (1, 0)$

Page No 668:

Question 7:

Find the coordinates of point P that divides the join of A(4, −3) and B(9, 7) in the ratio 3 : 2.

Answer:

The given points are $A (4, - 3) and B (9, 7)$ .
Then, $(x_{1} = 4, y_{1} = - 3) and (x_{2} = 9, y_{2} = 7)$
Also, m = 3 and n = 2
Let the required point be P(x, y).
By section formula, we have:
$x = \frac{(3 \times 9 + 2 \times 4)}{(3 + 2)}, y = \frac{(3 \times 7 + 2 \times (- 3))}{(3 + 2)} \Rightarrow x = \frac{35}{5}, y = \frac{15}{5} \Rightarrow x = 7, y = 3$
Hence, the coordinates of P are (7, 3).

Page No 668:

Question 8:

If A(−2, 4), B(0, 0) and C(4, 2) are the vertices of a âˆ†ABC, find the length of the median through A.

Answer:

Let D be the midpoint of BC.
So, the coordinates of D are
$D (\frac{0 + 4}{2}, \frac{0 + 2}{2}) [B (0, 0) and C (4, 2) i . e, (x_{1} = 0, y_{1} = 0) and (x_{2} = 4, y_{2} = 2)] i . e . D (2, 1)$
Length of the median:

$A D = \sqrt{{(2 + 2)}^{2} + {(1 - 4)}^{2}} = \sqrt{{(4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units$

Page No 668:

Question 9:

If the points A(2, 3), B(4, k) and C(6, −3) are collinear, find the value of k.

Answer:

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

Page No 668:

Question 10:

If the points A(4, 3) and B(x, 5) lie on the circle with centre O(2, 3), find the value of x.

Answer:

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

Page No 668:

Question 11:

In what ratio does the line x − y − 2 = 0 divide the line segment joining the points A(3, −1) and B(8, 9)?

Answer:

Let the line x − y − 2=0 divide the line segment joining A(3, −1) and B(8, 9) in the ratio k : 1 at the point P.
Then, by the section formula, 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:
$\frac{8 k + 3}{k + 1} - \frac{9 k - 1}{k + 1} - 2 = 0 \Rightarrow (8 k + 3) - (9 k - 1) - (2 k + 2) = 0 \Rightarrow 3 k = 2 \Rightarrow k = \frac{2}{3}$
So, the required ratio is $\frac{2}{3} : 1$ ,which is same as 2 : 3.

Page No 668:

Question 12:

Find the area of âˆ†ABC with vertices A(1, −3), B(4, −3) and C(−9, 7).

Answer:

The given points are $A (1, - 3), B (4, - 3) and C (- 9, 7)$ .
Here, $(x_{1} = 1, y_{1} = - 3), (x_{2} = 4, y_{2} = - 3) and (x_{3} = - 9, y_{3} = 7)$
Therefore,
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \frac{1}{2} |1 (- 3 - 7) + 4 (7 + 3) - 9 (- 3 + 3)| = \frac{1}{2} |- 10 + 40 + 0| = (\frac{1}{2} \times 30) = 15 sq units$

Page No 668:

Question 13:

Show that A(−3, −1), B(−4, −1), C(3, 3) and D(4, 3) are the vertices of a rhombus.

Answer:

Disclaimer:- It should be parallelogram in place of rhombus in the question.
The given vertices are A(−3, −1), B(−4, −1), C(3, 3) and Dâ€‹(4, 3).
Then,
$A B = \sqrt{{(- 4 + 3)}^{2} + {(- 1 + 1)}^{2}} = \sqrt{{(- 1)}^{2} + {(0)}^{2}} = \sqrt{1 + 0} = 1 unit B C = \sqrt{{(3 + 4)}^{2} + {(3 + 1)}^{2}} = \sqrt{{(7)}^{2} + {(4)}^{2}} = \sqrt{49 + 16} = \sqrt{65} units C D = \sqrt{{(4 - 3)}^{2} + {(3 - 3)}^{2}} = \sqrt{{(1)}^{2} + {(0)}^{2}} = \sqrt{1 + 0} = 1 unit D A = \sqrt{{(- 3 - 4)}^{2} + {(- 1 - 3)}^{2}} = \sqrt{{(- 7)}^{2} + {(- 4)}^{2}} = \sqrt{49 + 16} = \sqrt{65} units$
Here, AB = CD
and BC = DA
Therefore, ABCD is a parallelogram.

Page No 668:

Question 14:

Find the points on y-axis, which are at a distance of 13 units each from the point (−5, 7).

Answer:

Let the point on y-axis be B(0, b) and the given point be A(−5, 7).The distance between A and B is 13 units.
Now,
$A B = 13 \Rightarrow \sqrt{{(0 + 5)}^{2} + {(b - 7)}^{2}} = 13 \Rightarrow 25 + b^{2} + 49 - 14 b = 169 \Rightarrow b^{2} - 14 b - 95 = 0 \Rightarrow (b - 19) (b + 5) = 0 \Rightarrow b = 19 or - 5$
Hence, the points are (0, 19) and (0, −5).

Page No 668:

Question 15:

If three consecutive vertices of a parallelogram ABCD are A(1, −2), B(3, 6) and C(5, 10), find its fourth vertex D.

Answer:

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

$\frac{3 + a}{2} = 3 and \frac{6 + b}{2} = 4 \Rightarrow 3 + a = 6 and 6 + b = 8$

$\Rightarrow a = 3 and b = 2$

Hence, the fourth vertex is D(3, 2).

Page No 668:

Question 16:

If the points A(x, y), B(−5, 7) and C(−4, 5) are collinear, show that 2x + y + 3 = 0.

Answer:

The given points are $A (x, y), B (- 5, 7) and C (- 4, 5)$ .
Here, $(x_{1} = x, y_{1} = y), (x_{2} = - 5, y_{2} = 7) and (x_{3} = - 4, y_{3} = 5)$
A,B and Câ€‹ are collinear. Then,
$x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) = 0 \Rightarrow x (7 - 5) - 5 (5 - y) - 4 (y - 7) = 0 \Rightarrow 7 x - 5 x - 25 + 5 y - 4 y + 28 = 0 \Rightarrow 2 x + y + 3 = 0$

Page No 668:

Question 17:

Find the area of quadrilateral ABCD whose vertices are A(−4, −2), B(−3, −5), C(3, −2) and D(2, 3).

Answer:

Join A and C . Then,

Area of quadilateral ABCD $= (area of ∆ A B C) + (area of ∆ A C D)$
$Area of ∆ A B C = \frac{1}{2} |\{- 4 (- 5 + 2) - 3 (- 2 + 2) + 3 (- 2 + 5)\}| = \frac{1}{2} |- 4 (- 3) - 3 (0) + 3 (3)| = \frac{1}{2} |12 - 0 + 9| = \frac{21}{2} sq units$

$Area of ∆ A C D = \frac{1}{2} |\{- 4 (- 2 - 3) + 3 (3 + 2) + 2 (- 2 + 2)\}| = \frac{1}{2} |- 4 (- 5) + 3 (5) + 2 (0)| = \frac{1}{2} |20 + 15 + 0| = \frac{35}{2} sq units$
Therefore, area of quadilateral ABCD $= (\frac{21}{2} + \frac{35}{2})$
$= 28 sq units$

Page No 668:

Question 18:

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 2), B(4, 4) and C(2, 6).

Answer:

Let A(2, 2), B(4, 4) and C(2, 6) be the vertices of a triangle ABC.
Let D,E and F be the midpoints of AB, BC and CA, respectively.
Then, the coordinates of D, E and Fâ€‹ are:
$D (\frac{2 + 4}{2}, \frac{2 + 4}{2}), E (\frac{4 + 2}{2}, \frac{4 + 6}{2}) and F (\frac{2 + 2}{2}, \frac{2 + 6}{2})$
$i.e. D (3, 3), E (3, 5) and F (2, 4) Then, (x_{1} = 3, y_{1} = 3), (x_{2} = 3, y_{2} = 5) and (x_{3} = 2, y_{3} = 4)$
$Area of ∆ D E F = \frac{1}{2} |x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})|$

$= \frac{1}{2} |3 (5 - 4) + 3 (4 - 3) + 2 (3 - 5)| = \frac{1}{2} |3 (1) + 3 (1) + 2 (- 2)| = \frac{1}{2} |3 + 3 - 4| = (\frac{1}{2} \times 2) = 1 sq unit$

Page No 668:

Question 19:

Prove that the points A(−3, 0), B(1, −3) and C(4, 1) are the vertices of an isosceles right-angled triangle. Find the area of this triangle.

Answer:

Let $A (- 3, 0), B (1, - 3) and C (4, 1)$ be the given vertices. Then,
$A B = \sqrt{{(1 - (- 3))}^{2} + {(- 3 - 0)}^{2}} = \sqrt{{(4)}^{2} + {(- 3)}^{2}} = \sqrt{16 + 9} = \sqrt{25} = 5 units B C = \sqrt{{(4 - 1)}^{2} + {(1 - (- 3))}^{2}} = \sqrt{{(3)}^{2} + {(4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 units A C = \sqrt{{(4 - (- 3))}^{2} + {(1 - 0)}^{2}} = \sqrt{{(7)}^{2} + {(1)}^{2}} = \sqrt{49 + 1} = \sqrt{50} = 5 \sqrt{2} units$
Thus, AB = BC = 5 units.
Therefore, $∆ A B C$ is an isosceles triangle.
Also,
$(A B^{2} + B C^{2}) = (5^{2} + 5^{2})$
= 50 units.
$and A C^{2} = {(5 \sqrt{2})}^{2}$
= 50 units
Thus, $A B^{2} + B C^{2} = A {C^{2}}^{}$
This shows that $∆ A B C$ is right-angled at B.
So, âˆ†ABC is an isosceles right-angled triangle.
In $∆ A B C$ , we have:
base BC = 5 units and height AB= 5 units
$Area of ∆ A B C = (\frac{1}{2} \times base \times height)$
$= (\frac{1}{2} \times 5 \times 5) = 12.5 sq units$

Page No 668:

Question 20:

If the point P(−1, 2) divides the line segment joining A(2, 5) and B in the ratio 3 : 4, find the coordinates of B.

Answer:

Let the coordinates of B be $(x_{1}, y_{1})$ .
Then, P divides the line joining A(2, 5) and B $(x_{1}, y_{1})$ in the ratio 3 : 4.
So, by section formula, the coordinates of P are
$P (\frac{3 x_{1} + 8}{3 + 4}, \frac{3 y_{1} + 20}{3 + 4}) i . e . P (\frac{3 x_{1} + 8}{7}, \frac{3 y_{1} + 20}{7})$

But, the coordinates of P are (−1, 2).

$\frac{3 x_{1} + 8}{7} = - 1 and \frac{3 y_{1} + 20}{7} = 2$

$\Rightarrow 3 x$ ₁+8=−7 and 3y₁+20=14

$\Rightarrow x_{1} = - 15 and 3 y_{1} = - 6 \Rightarrow x_{1} = - 5 and y_{1} = - 2$

Hence, the coordinates of B are (−5, −2).

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