Mathematics Part ii Solutions Solutions for Class 10 Maths Chapter 5 - Co Ordinate Geometry

Share with your friends

Mathematics Part ii Solutions Solutions for Class 10 Maths Chapter 5 Co Ordinate Geometry are provided here with simple step-by-step explanations. These solutions for Co Ordinate Geometry are extremely popular among class 10 students for Maths Co Ordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 10 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 10 Maths are prepared by experts and are 100% accurate.

Page No 107:

Question 1:

Find the distance between each of the following pairs of points.
(1) A(2, 3), B(4, 1)

(2) P(–5, 7), Q(–1, 3)

(3) $R (0, - 3), S (0, \frac{5}{2})$

(4) L(5, –8), M(–7, –3)

(5) T(–3, 6), R(9, –10)

(6) $W (\frac{- 7}{2}, 4), X (11, 4)$

Answer:

(1) A(2, 3), B(4, 1)
$AB = \sqrt{{(2 - 4)}^{2} + {(3 - 1)}^{2}} = \sqrt{{(- 2)}^{2} + 2^{2}} = \sqrt{4 + 4} = 2 \sqrt{2}$

(2) P(–5, 7), Q(–1, 3)
$PQ = \sqrt{{(- 5 - (- 1))}^{2} + {(7 - 3)}^{2}} = \sqrt{{(- 4)}^{2} + 4^{2}} = \sqrt{16 + 16} = 4 \sqrt{2}$
(3) $R (0, - 3), S (0, \frac{5}{2})$
$RS = \sqrt{{(0 - 0)}^{2} + {(- 3 - \frac{5}{2})}^{2}} = \sqrt{{(\frac{- 11}{2})}^{2}} = \sqrt{\frac{121}{4}} = \frac{11}{2}$

(4) L(5, –8), M(–7, –3)
$LM = \sqrt{{(5 - (- 7))}^{2} + {(- 8 - (- 3))}^{2}} = \sqrt{12^{2} + {(- 5)}^{2}} = \sqrt{144 + 25} = \sqrt{169} = 13$

(5) T(–3, 6), R(9, –10)
$TR = \sqrt{{(- 3 - 9)}^{2} + {(6 - (- 10))}^{2}} = \sqrt{{(- 12)}^{2} + {(16)}^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20$

(6) $W (\frac{- 7}{2}, 4), X (11, 4)$
$WX = \sqrt{{(\frac{- 7}{2} - 11)}^{2} + {(4 - 4)}^{2}} = \sqrt{{(\frac{- 29}{2})}^{2}} = \sqrt{\frac{841}{4}} = \frac{29}{2}$

Page No 107:

Question 2:

Determine whether the points are collinear.
(1) A(1, –3), B(2, –5), C(–4, 7)
(2) L(–2, 3), M(1, –3), N(5, 4)
(3) R(0, 3), D(2, 1), S(3, –1)
(4) P(–2, 3), Q(1, 2), R(4, 1)

Answer:

(1) A(1, –3), B(2, –5), C(–4, 7)
$A B = \sqrt{{(1 - 2)}^{2} + {(- 3 - (- 5))}^{2}} = \sqrt{{(- 1)}^{2} + 2^{2}} = \sqrt{5} B C = \sqrt{{(2 - (- 4))}^{2} + {(- 5 - 7)}^{2}} = \sqrt{36 + 144} = \sqrt{180} = 6 \sqrt{5} A C = \sqrt{{(- 1 - (- 4))}^{2} + {(- 3 - 7)}^{2}} = \sqrt{25 + 100} = \sqrt{125} = 5 \sqrt{5}$
AB + AC = BC
$\sqrt{5} + 5 \sqrt{5} = 6 \sqrt{5}$
Thus, the given points lie on the same line. Hence, they are collinear.

(2) L(–2, 3), M(1, –3), N(5, 4)
$L M = \sqrt{{(- 2 - 1)}^{2} + {(3 - (- 3))}^{2}} = \sqrt{{(- 3)}^{2} + 6^{2}} = \sqrt{9 + 36} = \sqrt{45} = 3 \sqrt{5} M N = \sqrt{{(1 - 5)}^{2} + {(- 3 - 4)}^{2}} = \sqrt{16 + 49} = \sqrt{65} L N = \sqrt{{(- 2 - 5)}^{2} + {(3 - 4)}^{2}} = \sqrt{{(- 7)}^{2} + {(- 1)}^{2}} = \sqrt{49 + 1} = \sqrt{50} = 5 \sqrt{2}$
Sum of two sides is not equal to the third side.
Hence, the given points are not collinear.

(3) R(0, 3), D(2, 1), S(3, –1)
$R D = \sqrt{{(0 - 2)}^{2} + {(3 - 1)}^{2}} = \sqrt{{(- 2)}^{2} + 2^{2}} = \sqrt{4 + 4} = 2 \sqrt{2} D S = \sqrt{{(2 - 3)}^{2} + {(1 - (- 1))}^{2}} = \sqrt{{(- 1)}^{2} + 4} = \sqrt{1 + 4} = \sqrt{5} R S = \sqrt{{(0 - 3)}^{2} + {(3 - (- 1))}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
Sum of two sides is not equal to the third side.
Hence, the given points are not collinear.

(4) P(–2, 3), Q(1, 2), R(4, 1)
$P Q = \sqrt{{(- 2 - 1)}^{2} + {(3 - 2)}^{2}} = \sqrt{9 + 1} = \sqrt{10} Q R = \sqrt{{(1 - 4)}^{2} + {(2 - 1)}^{2}} = \sqrt{9 + 1} = \sqrt{10} P R = \sqrt{{(- 2 - 4)}^{2} + {(3 - 1)}^{2}} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}$
PQ + QR = PR
So, the given points lie on the same line. Hence, the given points are collinear.

Page No 107:

Question 3:

Find the point on the X–axis which is equidistant from A(–3, 4) and B(1, –4).

Answer:

Let the point on the x-axis be P(a, 0).
$PA = \sqrt{{(a - (- 3))}^{2} + {(0 - 4)}^{2}} = \sqrt{{(a + 3)}^{2} + 16} PB = \sqrt{{(a - 1)}^{2} + {(0 - (- 4))}^{2}} = \sqrt{{(a - 1)}^{2} + 16} {PA}^{2} = {PB}^{2} \Rightarrow {(a + 3)}^{2} + 16 = {(a - 1)}^{2} + 16 \Rightarrow a^{2} + 6 a + 9 = a^{2} - 2 a + 1 \Rightarrow 8 a = - 8 \Rightarrow a = - 1 (a, 0) = (- 1, 0)$

Page No 107:

Question 4:

Verify that points P(–2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.

Answer:

$PQ = \sqrt{{(- 2 - 2)}^{2} + {(2 - 2)}^{2}} = \sqrt{{(- 4)}^{2} + 0} = \sqrt{16} = 4 QR = \sqrt{{(2 - 2)}^{2} + {(2 - 7)}^{2}} = \sqrt{0 + {(- 5)}^{2}} = \sqrt{25} = 5 PR = \sqrt{{(- 2 - 2)}^{2} + {(2 - 7)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 5)}^{2}} = \sqrt{16 + 25} = \sqrt{41}$
${PQ}^{2} + {QR}^{2} = {PR}^{2} \Rightarrow 4^{2} + 5^{2} = 16 + 25 = 41 = PR$
Thus, the square of the third is equal to the sum of the squares of the other two sides.
Thus, they are the vertices of the right angled triangle.

Page No 108:

Question 5:

Show that points P(2, –2), Q(7, 3), R(11, –1) and S (6, –6) are vertices of a parallelogram.

Answer:

The given points are P(2, –2), Q(7, 3), R(11, –1) and S (6, –6).
$PQ = \sqrt{{(3 - (- 2))}^{2} + {(7 - 2)}^{2}} = \sqrt{25 + 25} = 5 \sqrt{2} QR = \sqrt{{(7 - 11)}^{2} + {(3 - (- 1))}^{2}} = \sqrt{16 + 16} = 4 \sqrt{2} RS = \sqrt{{(11 - 6)}^{2} + {(- 1 - (- 6))}^{2}} = \sqrt{25 + 25} = 5 \sqrt{2} PS = \sqrt{{(2 - 6)}^{2} + {(- 2 - (- 6))}^{2}} = \sqrt{16 + 16} = 4 \sqrt{2}$
So, PQ = RS and QR = PS
Thus, opposite sides are equal.
Hence, the given points form a parallelogram.

Page No 108:

Question 6:

Show that points A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4) are vertices of a rhombus ABCD.

Answer:

The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4).
$AB = \sqrt{{(- 4 - (- 1))}^{2} + {(- 7 - 2)}^{2}} = \sqrt{9 + 81} = \sqrt{90} = 3 \sqrt{10} BC = \sqrt{{(- 1 - 8)}^{2} + {(2 - 5)}^{2}} = \sqrt{81 + 9} = \sqrt{90} = 3 \sqrt{10} CD = \sqrt{{(8 - 5)}^{2} + {(5 - (- 4))}^{2}} = \sqrt{9 + 81} = \sqrt{90} = 3 \sqrt{10} DA = \sqrt{{(5 - (- 4))}^{2} + {(- 4 - (- 7))}^{2}} = \sqrt{81 + 9} = \sqrt{90} = 3 \sqrt{10}$
Thus, all the sides are equal.
$AC = \sqrt{{(- 4 - 8)}^{2} + {(- 7 - 5)}^{2}} = \sqrt{144 + 144} = 12 \sqrt{2} DB = \sqrt{{(8 - 5)}^{2} + {(2 - (- 4))}^{2}} = \sqrt{9 + 36} = 3 \sqrt{5}$
Thus, the diagonals are not equal.
So, the given vertices form a rhombus ABCD.

Page No 108:

Question 7:

Find x if distance between points L(x, 7) and M(1, 15) is 10.

Answer:

Distance between LM = 10
$LM = \sqrt{{(x - 1)}^{2} + {(7 - 15)}^{2}} = 10 \Rightarrow \sqrt{{(x - 1)}^{2} + {(- 8)}^{2}} = 10 \Rightarrow \sqrt{{(x - 1)}^{2} + 64} = 100$
Squaring both sides
${(x - 1)}^{2} + 64 = 100 \Rightarrow x^{2} - 2 x + 1 + 64 = 100 \Rightarrow x^{2} - 2 x - 35 = 0 \Rightarrow x^{2} - 7 x + 5 x - 35 = 0 \Rightarrow x (x - 7) + 5 (x - 7) = 0 \Rightarrow (x + 5) (x - 7) = 0 \Rightarrow x = - 5, 7$

Page No 108:

Question 8:

Show that the points A(1, 2), B(1, 6), $C (1 + 2 \sqrt{3}, 4)$ are vertices of an equilateral triangle.

Answer:

The given points are A(1, 2), B(1, 6), $C (1 + 2 \sqrt{3}, 4)$ .
$AB = \sqrt{{(1 - 1)}^{2} + {(2 - 6)}^{2}} = \sqrt{0 + {(- 4)}^{2}} = \sqrt{16} = 4 BC = \sqrt{{(1 - 1 - 2 \sqrt{3})}^{2} + {(6 - 4)}^{2}} = \sqrt{12 + 4} = \sqrt{16} = 4 AC = \sqrt{{(1 - 1 - 2 \sqrt{3})}^{2} + {(2 - 4)}^{2}} = \sqrt{12 + 4} = \sqrt{16} = 4$
AB = BC = AC = 4
Since, all the sides of the triangle formed by the points A, B and C are equal so, the tiangle formed is equilateral triangle.

Page No 115:

Question 1:

Find the coordinates of point P if P divides the line segment joining the points A(–1,7) and B(4,–3) in the ratio 2 : 3.

Answer:

Let the coordinates of point P be (x, y).
The ratio in which P divides A(–1,7) and B(4,–3) is 2 : 3.
Using the section formula we have
$x = \frac{4 \times 2 + 3 \times (- 1)}{2 + 3} = \frac{8 - 3}{5} = 1 y = \frac{2 \times (- 3) + 3 \times 7}{2 + 3} = \frac{- 6 + 21}{5} = 3$
Thus, the coordinates of point P are (1, 3).

Page No 115:

Question 2:

In each of the following examples find the co-ordinates of point A which divides segment PQ in the ratio a : b.
(1) P(–3, 7), Q(1, –4), a : b = 2 : 1
(2) P(–2, –5), Q(4, 3), a : b = 3 : 4
(3) P(2, 6), Q(–4, 1), a : b = 1 : 2

Answer:

Let the coordinates of point A be (x, y).
(1) P(–3, 7), Q(1, –4), a : b = 2 : 1
Using section formula
$x = \frac{2 \times 1 + 1 \times (- 3)}{2 + 1} = \frac{2 - 3}{3} = \frac{- 1}{3} y = \frac{2 \times (- 4) + 1 \times 7}{2 + 1} = \frac{- 8 + 7}{3} = \frac{- 1}{3} (x, y) = (\frac{- 1}{3}, \frac{- 1}{3})$

(2) P(–2, –5), Q(4, 3), a : b = 3 : 4
Using section formula
â€‹ $x = \frac{3 \times 4 + 4 \times (- 2)}{3 + 4} = \frac{12 - 8}{7} = \frac{4}{7} y = \frac{3 \times 3 + 4 \times (- 5)}{3 + 4} = \frac{9 - 20}{3} = \frac{- 11}{7} (x, y) = (\frac{4}{7}, \frac{- 11}{7})$

(3) P(2, 6), Q(–4, 1), a : b = 1 : 2
Using section formula
â€‹ $x = \frac{1 \times (- 4) + 2 \times 2}{1 + 2} = \frac{- 4 + 4}{3} = 0 y = \frac{1 \times 1 + 2 \times 6}{1 + 2} = \frac{1 + 12}{3} = \frac{13}{3} (x, y) = (0, \frac{13}{3})$

Page No 115:

Question 3:

Find the ratio in which point T(–1, 6)divides the line segment joining the points P(–3, 10) and Q(6, –8).

Answer:

Let the ratio be k : 1.
Using section formula we have
$- 1 = \frac{6 k - 3 \times 1}{k + 1} \Rightarrow - k - 1 = 6 k - 3 \Rightarrow - 1 + 3 = 6 k + k \Rightarrow 2 = 7 k \Rightarrow k = \frac{2}{7}$
Thus, the required ratio is 2 : 7.

Page No 115:

Question 4:

Point P is the centre of the circle and AB is a diameter . Find the coordinates of point B if coordinates of point A and P are (2, –3) and (–2, 0) respectively.

Answer:

Centre = P(–2, 0)
Diameter has A(2, –3) on one end and B(x, y) on the other.
The centre point P divides AB in two equal parts.
So, using the midpoint formula,
$- 2 = \frac{2 + x}{2} \Rightarrow - 4 = 2 + x \Rightarrow x = - 6 And 0 = \frac{- 3 + y}{2} \Rightarrow 0 = - 3 + y \Rightarrow y = 3 (x, y) = (- 6, 3)$

Page No 115:

Question 5:

Find the ratio in which point P(k, 7) divides the segment joining A(8, 9) and B(1, 2). Also find k.

Answer:

Let the ratio be x : 1.
Using the section formula,
$k = \frac{1 x + 1 \times 8}{x + 1} \Rightarrow k x + k = x + 8 . . . . . (1) Also, 7 = \frac{2 x + 9}{x + 1} \Rightarrow 7 x + 7 = 2 x + 9 \Rightarrow 5 x = 2 \Rightarrow x = \frac{2}{5}$
So, the required ratio is 2 : 5.
Putting this value of x in (1) we get
$k (\frac{2}{5} + 1) = \frac{2}{5} + 8 \Rightarrow \frac{7}{5} k = \frac{42}{5} \Rightarrow k = 6$

Page No 115:

Question 6:

Find the coordinates of midpoint of the segment joining the points (22, 20) and (0, 16).

Answer:

Let the given points be A (22, 20) and B(0, 16).
Let the coordinate of the mid point be (x, y).
Using the midpoint formula
$x = \frac{22 + 0}{2} \Rightarrow 2 x = 22 \Rightarrow x = 11 y = \frac{20 + 16}{2} \Rightarrow 2 y = 36 \Rightarrow y = 18$
(x, y) = (11, 18)

Page No 115:

Question 7:

Find the centroids of the triangles whose vertices are given below.
(1) (–7, 6), (2, –2), (8, 5)
(2) (3, –5), (4, 3), (11, –4)
(3) (4, 7), (8, 4), (7, 11)

Answer:

(1) (–7, 6), (2, –2), (8, 5)
The centroid of the triangle formed by the given coordinates is
$= (\frac{- 7 + 2 + 8}{3}, \frac{6 - 2 + 5}{3}) = (1, 3)$

(2) (3, –5), (4, 3), (11, –4)
The centroid of the triangle formed by the given coordinates is
$= (\frac{3 + 4 + 11}{3}, \frac{- 5 + 3 - 4}{3}) = (6, - 2)$

(3) (4, 7), (8, 4), (7, 11)
The centroid of the triangle formed by the given coordinates is

= (\frac{4 + 8 + 7}{3}, \frac{7 + 4 + 11}{3}) = (\frac{19}{3}, \frac{22}{3})

Page No 116:

Question 8:

In âˆ†ABC, G (–4, –7) is the centroid. If A (–14, –19) and B(3, 5) then find the co–ordinates of C.

Answer:

Let the coordinates of point C be (x, y).
Centroid = G (–4, –7)
$- 4 = (\frac{- 14 + 3 + x}{3}) \Rightarrow - 12 = x - 11 \Rightarrow x = - 1 A l s o, - 7 = (\frac{- 19 + 5 + y}{3}) \Rightarrow - 21 = - 14 + y \Rightarrow y = - 7$
Thus, point C will be $(- 1, - 7)$ .

Page No 116:

Question 9:

A(h, –6), B(2, 3) and C(–6, k) are the co–ordinates of vertices of a triangle whose centroid is G (1, 5). Find h and k.

Answer:

Centroid = G(1, 5)
The coordinates of the triangle are A(h, –6), B(2, 3) and C(–6, k).
$1 = (\frac{h + 2 - 6}{3}) \Rightarrow 3 = h - 4 \Rightarrow h = 7 A l s o, 5 = (\frac{- 6 + 3 + k}{3}) \Rightarrow 15 = k - 3 \Rightarrow k = 18$

Page No 116:

Question 10:

Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(–4, –8).

Answer:

Let P $(x_{1}, y_{1})$ and Q $(x_{2}, y_{2})$ divide the line AB into 3 equal parts.
AP = PQ = QB
$\frac{AP}{PB} = \frac{AP}{PQ + QB} = \frac{AP}{AP + AP} = \frac{AP}{2 AP} = \frac{1}{2}$
So, P divides AB in the ratio 1 : 2.
$x_{1} = \frac{1 \times (- 4) + 2 \times 2}{2 + 1} = 0 y_{1} = \frac{1 \times (- 8) + 2 \times 7}{2 + 1} = 2 (x_{1}, y_{1}) = (0, 2)$
Also, $\frac{AQ}{QB} = \frac{AP + PB}{QB} = \frac{QB + QB}{QB} = \frac{2}{1}$
$x_{2} = \frac{2 \times (- 4) + 1 \times 2}{2 + 1} = - 2 y_{2} = \frac{2 \times (- 8) + 1 \times 7}{2 + 1} = - 3 (x_{2}, y_{2}) = (- 2, - 3)$
Thus, the coordinates of the point of intersection are
P $(x_{1}, y_{1}) = (0, 2)$
Q $(x_{2}, y_{2})$ = $(- 2, - 3)$

Page No 116:

Question 11:

If A (–14, –10), B(6, –2) is given, find the coordinates of the points which divide segment AB into four equal parts.

Answer:

Let $P (x_{1}, y_{1}), Q (x_{2}, y_{2}), R (x_{3}, y_{3})$ be the points which divide the line segment AB into 4 equal parts.
AP = PQ = QR = RB
$\frac{AP}{PB} = \frac{AP}{PQ + QR + RB} = \frac{AP}{AP + AP + AP} = \frac{AP}{3 AP} = \frac{1}{3}$
$x_{1} = (\frac{6 \times 1 + 3 \times (- 14)}{1 + 3}) = \frac{6 - 42}{4} = - 9 y_{1} = (\frac{1 \times (- 2) + 3 \times (- 10)}{1 + 3}) = \frac{- 2 - 30}{4} = - 8 P (x_{1}, y_{1}) = (- 9, - 8)$
Now, $\frac{PQ}{QB} = \frac{PQ}{QR + RB} = \frac{PQ}{PQ + PQ} = \frac{1}{2}$
$x_{2} = (\frac{1 \times 6 + 2 \times (- 9)}{1 + 2}) = - 4 y_{2} = (\frac{1 \times (- 2) + 2 \times (- 8)}{1 + 2}) = - 6 Q (x_{2}, y_{2}) = (- 4, - 6)$
Now R divides QB into 2 equal parts so, using the midpoint formula we have
$x_{3} = \frac{- 4 + 6}{2} = 1 y_{3} = \frac{- 6 + (- 2)}{2} = - 4 R (x_{3}, y_{3}) = (1, - 4)$
Thus, $P (x_{1}, y_{1}), Q (x_{2}, y_{2}), R (x_{3}, y_{3})$ are $(- 9, - 8), (- 4, - 6) and (1, - 4)$ .

Page No 116:

Question 12:

If A (20, 10), B(0, 20) are given, find the coordinates of the points which divide segment AB into five congruent parts.

Answer:

Let the points $P (x_{1}, y_{1}), Q (x_{2}, y_{2}), R (x_{3}, y_{3}) and S (x_{4}, y_{4})$ be the points which divide the line segment AB into 5 equal parts.
$\frac{AP}{PB} = \frac{AP}{PQ + QR + RS} = \frac{AP}{4 AP} = \frac{1}{4}$
$x_{1} = (\frac{1 \times 0 + 4 \times 20}{1 + 4}) = 16 y_{1} = (\frac{1 \times 20 + 4 \times 10}{1 + 4}) = 12 P (x_{1}, y_{1}) = (16, 12)$
$\frac{PQ}{QB} = \frac{PQ}{QR + RS + SB} = \frac{PQ}{PQ + PQ + PQ} = \frac{PQ}{3 PQ} = \frac{1}{3}$
$x_{2} = (\frac{1 \times 0 + 3 \times 16}{1 + 3}) = 12 y_{2} = (\frac{1 \times 20 + 3 \times 12}{1 + 3}) = 14 Q (x_{2}, y_{2}) = (12, 14)$
$\frac{QR}{RB} = \frac{QR}{RS + SB} = \frac{QR}{QR + QR} = \frac{QR}{2 QR} = \frac{1}{2}$
$x_{3} = (\frac{1 \times 0 + 2 \times 12}{1 + 2}) = 8 y_{3} = (\frac{1 \times 20 + 2 \times 14}{1 + 3}) = 16 R (x_{3}, y_{3}) = (8, 16)$
S is the midpoint of RB so, using the midpoint formula
$x_{4} = \frac{8 + 0}{2} = 4 y_{4} = \frac{16 + 20}{2} = 18 S (x_{4}, y_{4}) = (4, 18)$
So, the points
$P (x_{1}, y_{1}) = (16, 12) Q (x_{2}, y_{2}) = (12, 14) R (x_{3}, y_{3}) = (8, 16) S (x_{4}, y_{4}) = (4, 18)$

Page No 121:

Question 1:

Angles made by the line with the positive direction of X–axis are given. Find the slope of these lines.
(1) 45° (2) 60° (3) 90°

Answer:

(1) 45°
$m = \tan 45 ° = 1$
Thus, slope = 1

(2) 60°
$m = \tan 60 ° = \sqrt{3}$
Thus, slope = $\sqrt{3}$

(3) 90°
$m = \tan 90 ° = undefined$
Thus, the slope cannot be defined.

Page No 121:

Question 2:

Find the slopes of the lines passing through the given points.
(1) A (2, 3) , B (4, 7)
(2) P (–3, 1) , Q (5, –2)
(3) C (5, –2) , âˆ† (7, 3)
(4) L (–2, –3) , M (–6, –8)
(5) E(–4, –2) , F (6, 3)
(6) T (0, –3) , S (0, 4)

Answer:

(1) A (2, 3) , B (4, 7)
Slope = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{7 - 3}{4 - 2} = \frac{4}{2} = 2$

(2) P (–3, 1) , Q (5, –2)
Slope = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 2 - 1}{5 - (- 3)} = \frac{- 3}{8}$

(3) C (5, –2) , âˆ† (7, 3)
Slope = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3 - (- 2)}{7 - 5} = \frac{5}{2}$

(4) L (–2, –3) , M (–6, –8)
Slope = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{- 8 - (- 3)}{- 6 - (- 2)} = \frac{- 5}{- 4} = \frac{5}{4}$

(5) E(–4, –2) , F (6, 3)
Slope = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{3 - (- 2)}{6 - (- 4)} = \frac{5}{10} = \frac{1}{2}$

(6) T (0, –3) , S (0, 4)
Slope = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{4 - (- 3)}{0 - 0} = \frac{7}{0} = slope not defined$

Page No 121:

Question 3:

Determine whether the following points are collinear.
(1) A(–1, –1), B(0, 1), C(1, 3)

(2) D(–2, –3), E(1, 0), F(2, 1)

(3) L(2, 5), M(3, 3), N(5, 1)

(4) P(2, –5), Q(1, –3), R(–2, 3)

(5) R(1, –4), S(–2, 2), T(–3, 4)

(6) A(–4, 4), $K (- 2, \frac{5}{2}),$ N (4, –2)

Answer:

(1) A(–1, –1), B(0, 1), C(1, 3)
Slope of AB = $\frac{1 - (- 1)}{0 - (- 1)} = \frac{2}{1} = 2$
Slope of BC = $\frac{3 - 1}{1 - 0} = \frac{2}{1} = 2$
Slope of AB = Slope of BC = 2
Thus, the given points are collinear.

(2) D(–2, –3), E(1, 0), F(2, 1)
$Slope of DE = \frac{0 - (- 3)}{1 - (- 2)} = \frac{3}{3} = 1 Slope of EF = \frac{1 - 0}{2 - 1} = \frac{1}{1} = 1$
Slope of DE = Slope of EF = 1
So, the given points are collinear.

(3) L(2, 5), M(3, 3), N(5, 1)
$Slope of LM = \frac{3 - 5}{3 - 2} = \frac{- 2}{1} = - 2 Slope of MN = \frac{1 - 3}{5 - 3} = \frac{- 2}{2} = - 1$
Slope of LM not equal to slope of MN. Thus, the given points are not collinear.

(4) P(2, –5), Q(1, –3), R(–2, 3)
$Slope of PQ = \frac{- 3 - (- 5)}{1 - 2} = \frac{2}{- 1} = - 2 Slope of QR = \frac{3 - (- 3)}{- 2 - 1} = \frac{6}{- 3} = - 2$
Slope of PQ = Slope of QR
So, the given points are collinear.

(5) R(1, –4), S(–2, 2), T(–3, 4)
$Slope of RS = \frac{2 - (- 4)}{- 2 - 1} = \frac{6}{- 3} = - 2 Slope of ST = \frac{4 - 2}{- 3 - (- 2)} = \frac{2}{- 1} = - 2$
Slope of RS = Slope of ST
So, the given points are collinear.

(6) A(–4, 4), $K (- 2, \frac{5}{2}),$ N (4, –2)
$Slope of AK = \frac{\frac{5}{2} - 4}{- 2 - (- 4)} = \frac{\frac{- 3}{2}}{2} = \frac{- 3}{4} Slope of KN = \frac{- 2 - \frac{5}{2}}{4 - (- 2)} = \frac{- 3}{4}$
Slope of AK=Slope of KN
Thus, the given points are collinear.

Page No 121:

Question 4:

If A (1, –1), B (0, 4), C (–5, 3) are vertices of a triangle then find the slope of each side.

Answer:

A (1, –1), B (0, 4), C (–5, 3) form a triangle.
Slope of AB = $\frac{4 - (- 1)}{0 - 1} = \frac{5}{- 1} = - 5$
Slope of BC = $\frac{3 - 4}{- 5 - 0} = \frac{- 1}{- 5} = \frac{1}{5}$
Slope of AC = $\frac{3 - (- 1)}{- 5 - 1} = \frac{4}{- 6} = \frac{- 2}{3}$

Page No 121:

Question 5:

Show that A(–4, –7), B (–1, 2), C (8, 5) and D (5, –4) are the vertices of a parallelogram.

Answer:

The given points are A(–4, –7), B(–1, 2), C(8, 5) and D(5, –4).
Slope of AB = $\frac{2 - (- 7)}{- 1 - (- 4)} = \frac{9}{3} = 3$
Slope of BC = $\frac{5 - 2}{8 - (- 1)} = \frac{3}{9} = \frac{1}{3}$
Slope of CD = $\frac{- 4 - 5}{5 - 8} = \frac{- 9}{- 3} = 3$
Slope of AD = $\frac{- 4 - (- 7)}{5 - (- 4)} = \frac{3}{9} = \frac{1}{3}$
Slope of AB = Slope of CD
Slope of BC = Slope of AD
So, AB || CD and BC || AD
Hence, ABCD is a parallelogram.

Page No 122:

Question 6:

Find k, if R(1, –1), S (–2, k) and slope of line RS is –2.

Answer:

Slope of line RS is –2
Slope of RS will be
$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{k - (- 1)}{- 2 - 1} = \frac{k + 1}{- 3} = - 2 \Rightarrow k + 1 = 6 \Rightarrow k = 5$

Page No 122:

Question 7:

Find k, if B(k, –5), C (1, 2) and slope of the line is 7.

Answer:

Slope of the line BC is 7.
Slope of BC =
$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 - (- 5)}{1 - k} = 7 \Rightarrow \frac{7}{1 - k} = 7 \Rightarrow 1 - k = 1 \Rightarrow k = 0$

Page No 122:

Question 8:

Find k, if PQ || RS and P(2, 4), Q (3, 6), R(3, 1), S(5, k).

Answer:

Given P(2, 4), Q (3, 6), R(3, 1), S(5, k)
PQ || RS so, slope of PQ = slope of RS
$\frac{6 - 4}{3 - 2} = \frac{k - 1}{5 - 3} \Rightarrow \frac{2}{1} = \frac{k - 1}{2} \Rightarrow k - 1 = 4 \Rightarrow k = 5$

Page No 122:

Question 1:

Fill in the blanks using correct alternatives.
(1) Seg AB is parallel to Y-axis and coordinates of point A are (1,3) then co–ordinates of point B can be ........ .
(A) (3,1)
(B) (5,3)
(C) (3,0)
(D) (1,–3)

(2) Out of the following, point ........ lies to the right of the origin on X– axis.
(A) (–2,0)
(B) (0,2)
(C) (2,3)
(D) (2,0)

(3) Distance of point (–3,4) from the origin is ...... .
(A) 7
(B) 1
(C) 5
(D) –5

(4) A line makes an angle of 30° with the positive direction of X– axis. So the slope of the line is .......... .

(A) $\frac{1}{2}$

(B) $\frac{\sqrt{3}}{2}$

(C) $\frac{1}{\sqrt{3}}$

(D) $\sqrt{3}$

Answer:

(1) Slope of y-axis will be not defined because the denominator will be 0.
Since AB is parallel to y-axis, so the slope will be same.
Let the coordinate of point B be (x, y).
Slope of AB = $\frac{y - 3}{x - 1}$ = slope of y-axis
Slope of y-axis is not defined as the denominator is 0 so, the denominator of $\frac{y - 3}{x - 1}$ will also be 0.
So,
$x - 1 = 0 \Rightarrow x = 1$
with $x = 1$ there is only 1 option that (D).
Hence, the correct answer is option D.

(2) A point lying on the right of the origin on X – axis will have positive x-coordinate and non-zero y-coordinate.
So, among the given points (2, 3) lies on the right of the origin on X – axis.
Hence, the correct answer is option (C).

(3) Let the given point be P(–3,4).
Distance of P(–3,4) from the origin O(0, 0) is
$\sqrt{{(0 - (- 3))}^{2} + {(0 - 4)}^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$
Hence, the correct answer is option (C).

(4) Slope of line will be $\tan 30 ° = \frac{1}{\sqrt{3}}$
Hence, the correct answer is option (C).

Page No 122:

Question 2:

Determine whether the given points are collinear.
(1) A(0,2), B(1,–0.5), C(2,–3)

(2) $P (1, 2), Q (2, \frac{8}{5}), R (3, \frac{6}{5})$

(3) L(1,2), M(5,3) , N(8,6)

Answer:

(1) A(0,2), B(1,–0.5), C(2,–3)
Slope of AB = $\frac{- 0.5 - 2}{1 - 0} = \frac{- 2.5}{1} = - 2.5$
Slope of BC = $\frac{- 3 - (- 0.5)}{2 - 1} = \frac{- 2.5}{1} = - 2.5$
So, the slope of AB = slope of BC.
Point B lies on both the lines.
Hence, the given points are collinear.

(2) $P (1, 2), Q (2, \frac{8}{5}), R (3, \frac{6}{5})$
Slope of PQ = $\frac{\frac{8}{5} - 2}{2 - 1} = \frac{\frac{- 2}{5}}{1} = \frac{- 2}{5}$
Slope of QR = $\frac{\frac{6}{5} - \frac{8}{5}}{3 - 2} = \frac{\frac{- 2}{5}}{1} = \frac{- 2}{5}$
So, the slope of PQ = slope of QR.
Point Q lies on both the lines.
Hence, the given points are collinear.

(3) L(1,2), M(5,3) , N(8,6)
Slope of LM $= \frac{3 - 2}{5 - 1} = \frac{1}{4}$
Slope of MN = $\frac{6 - 3}{8 - 5} = \frac{3}{3} = 1$
Thus, the slope of LM not equal to slope MN.
So, the given points are not collinear.

Page No 122:

Question 3:

Find the coordinates of the midpoint of the line segment joining P(0,6) and Q(12,20).

Answer:

The given points are P(0,6) and Q(12,20).
Midpoint of PQ = $= (\frac{0 + 12}{2}, \frac{6 + 20}{2}) = (6, 13)$ .

Page No 122:

Question 4:

Find the ratio in which the line segment joining the points A(3,8) and B(–9, 3) is divided by the Y– axis.

Answer:

Let the ratio in which the line AB is divided by the y-axis be k : 1.
The point on the y - axis be (0, y).
$(0, y) = (\frac{k \times (- 9) + 1 \times 3}{k + 1}, \frac{k \times 3 + 1 \times 8}{k + 1}) \Rightarrow (0, y) = (\frac{- 9 k + 3}{k + 1}, \frac{3 k + 8}{k + 1}) \Rightarrow 0 = \frac{- 9 k + 3}{k + 1} \Rightarrow - 9 k + 3 = 0 \Rightarrow k = \frac{1}{3}$
Thus, the ratio is 1 : 3.

Page No 122:

Question 5:

Find the point on X–axis which is equidistant from P(2,–5) and Q(–2,9).

Answer:

Let the point on x-axis equidistant from P(2,–5) and Q(–2,9) be $A (x, 0)$ .
$AP = \sqrt{{(x - 2)}^{2} + {(0 - (- 5))}^{2}} = \sqrt{{(x - 2)}^{2} + 25} QA = \sqrt{{(x - (- 2))}^{2} + {(0 - 9)}^{2}} = \sqrt{{(x + 2)}^{2} + 81} AP = QA \Rightarrow \sqrt{{(x - 2)}^{2} + 25} = \sqrt{{(x + 2)}^{2} + 81}$
Squaring both sides
${(x - 2)}^{2} + 25 = {(x + 2)}^{2} + 81 \Rightarrow x^{2} + 4 - 4 x + 25 = x^{2} + 4 + 4 x + 81 \Rightarrow - 8 x = 56 \Rightarrow x = - 7$
Thus, the required point is $(- 7, 0)$ .

Page No 122:

Question 6:

Find the distances between the following points.
(i) A(a, 0), B(0, a) (ii) P(–6, –3), Q(–1, 9) (iii) R(–3a, a), S(a, –2a)

Answer:

(i) A(a, 0), B(0, a)
$AB = \sqrt{{(0 - a)}^{2} + {(a - 0)}^{2}} = \sqrt{a^{2} + a^{2}} = \sqrt{2 a^{2}} = a \sqrt{2}$

(ii) P(–6, –3), Q(–1, 9)
$PQ = \sqrt{{(- 6 - (- 1))}^{2} + {(- 3 - 9)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$

(iii) R(–3a, a), S(a, –2a)
$RS = \sqrt{{(- 3 a - a)}^{2} + {(a - (- 2 a))}^{2}} = \sqrt{16 a^{2} + 9 a^{2}} = \sqrt{25 a^{2}} = 5 a$

Page No 122:

Question 7:

Find the coordinates of the circumcentre of a triangle whose vertices are (–3,1), (0,–2) and (1,3)

Answer:

Let the given vertices A(–3, 1), B(0, –2) and C(1, 3).
Circumcircle passes through the vertices of a triangle.
Let the circumcentre be P(a, b)
The distance of point P will be equal from A, B and C.
PA = PB = PC
$\Rightarrow$ ${PA}^{2} = {PB}^{2} = {PC}^{2}$
${PB}^{2} = {PC}^{2} \Rightarrow {(1 - a)}^{2} + {(3 - b)}^{2} = {(0 - a)}^{2} + {(- 2 - b)}^{2} \Rightarrow 1 + a^{2} - 2 a + 9 + b^{2} - 6 b = a^{2} + 4 + b^{2} - 2 \times b \times (- 2) \Rightarrow 10 + a^{2} - 2 a + 9 + b^{2} - 6 b = a^{2} + 4 + b^{2} + 4 b \Rightarrow a + 5 b = 3 . . . . . (1) {PB}^{2} = {PA}^{2} \Rightarrow {(a - 0)}^{2} + {(b + 2)}^{2} = {(a + 3)}^{2} + {(b - 1)}^{2} \Rightarrow a^{2} + 9 + 6 a + b^{2} + 1 - 2 b = a^{2} + b^{2} + 4 + 4 b \Rightarrow a - b = 1 . . . . . (2)$
solving (1) and (2) we get
$a = \frac{- 1}{3}, b = \frac{2}{3}$
Thus, the circumcentre of the triangle is $P (\frac{- 1}{3}, \frac{2}{3})$

Page No 123:

Question 8:

In the following examples, can the segment joining the given points form a triangle ? If triangle is formed, state the type of the triangle considering sides of the triangle.
(1) L(6,4) , M(–5,–3) , N(–6,8)

(2) P(–2,–6) , Q(–4,–2), R(–5,0)

(3) $A (\sqrt{2}, \sqrt{2}), B (- \sqrt{2}, - \sqrt{2}), C (- \sqrt{6}, \sqrt{6})$

Answer:

(1) L(6,4) , M(–5,–3) , N(–6,8)
$LM = \sqrt{{(6 + 5)}^{2} + {(4 + 3)}^{2}} = \sqrt{121 + 49} = \sqrt{170} = 2 \sqrt{35} MN = \sqrt{{(- 5 + 6)}^{2} + {(- 3 - 8)}^{2}} = \sqrt{1 + 121} = \sqrt{122} = 2 \sqrt{61} LN = \sqrt{{(6 + 6)}^{2} + {(4 - 8)}^{2}} = \sqrt{144 + 16} = \sqrt{160} = 4 \sqrt{10}$
LM, MN and LN form a scalene triangle.

(2) P(–2,–6) , Q(–4,–2), R(–5,0)
$PQ = \sqrt{{(- 2 + 4)}^{2} + {(- 6 + 2)}^{2}} = \sqrt{4 + 16} = \sqrt{20} = 2 \sqrt{5} QR = \sqrt{{(- 4 + 5)}^{2} + {(- 2 + 0)}^{2}} = \sqrt{1 + 4} = \sqrt{5} PR = \sqrt{{(- 2 + 5)}^{2} + {(- 6 + 0)}^{2}} = \sqrt{9 + 36} = \sqrt{45} = 3 \sqrt{5} PQ + QR = PR$
Since the sum of the two sides is not greater than the third side so, the given vertices do not form a triangle.

(3) $A (\sqrt{2}, \sqrt{2}), B (- \sqrt{2}, - \sqrt{2}), C (- \sqrt{6}, \sqrt{6})$
$AB = \sqrt{{(\sqrt{2} + \sqrt{2})}^{2} + {(\sqrt{2} + \sqrt{2})}^{2}} = \sqrt{8 + 8} = \sqrt{16} = 4 BC = \sqrt{{(- \sqrt{2} + \sqrt{6})}^{2} + {(- \sqrt{2} - \sqrt{6})}^{2}} = \sqrt{16} = 4 AC = \sqrt{{(\sqrt{2} + \sqrt{6})}^{2} + {(\sqrt{2} - \sqrt{6})}^{2}} = \sqrt{16} = 4 AB = BC = AC$
So, these vertices form an equilateral triangle.

Page No 123:

Question 9:

Find k if the line passing through points P(–12, –3) and Q(4, k) has slope $\frac{1}{2}$ .

Answer:

Slope = $\frac{1}{2}$
Given points are P(–12, –3) and Q(4, k)
Slope of PQ =
$\frac{k + 3}{4 + 12} = \frac{1}{2} \Rightarrow 2 k + 6 = 16 \Rightarrow 2 k = 10 \Rightarrow k = 5$

Page No 123:

Question 10:

Show that the line joining the points A(4, 8) and B(5, 5) is parallel to the line joining the points C(2, 4) and D(1, 7).

Answer:

Slope of the line joining the points A(4, 8) and B(5, 5) will be
$= \frac{5 - 8}{5 - 4} = \frac{- 3}{1} = - 3$
Slope of the line joining the points C(2, 4) and D(1, 7) will be
$= \frac{7 - 4}{1 - 2} = \frac{3}{- 1} = - 3$
Since the slope of AB = slope of CD
so, the given lines are parallel.

Page No 123:

Question 11:

Show that points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram

Answer:

The given points are P(1, –2), Q(5, 2), R(3, –1), S(–1, –5).
$Slope of PQ = \frac{2 + 2}{5 - 1} = \frac{4}{4} = 1 Slope of QR = \frac{- 1 - 2}{3 - 5} = \frac{- 3}{- 2} = \frac{3}{2} Slope of RS = \frac{- 5 + 1}{- 1 - 3} = \frac{- 4}{- 4} = 1 Slope of PS = \frac{- 5 + 2}{- 1 - 1} = \frac{- 3}{- 2} = \frac{3}{2}$
Slope of PQ = Slope of RS and Slope of QR = Slope of PS
So, PQ || RS and QR || PS
Thus, PQRS is a parallelogram.

Page No 123:

Question 12:

Show that the â–¢PQRS formed by P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5) is a rectangle

Answer:

The given points are P(2, 1), Q(–1, 3), R(–5, –3) and S(–2, –5).
PQ = $\sqrt{{(- 1 - 2)}^{2} + {(3 - 1)}^{2}} = \sqrt{9 + 4} = \sqrt{13}$
QR = $\sqrt{{(- 5 + 1)}^{2} + {(- 3 - 3)}^{2}} = \sqrt{16 + 36} = \sqrt{52} = 2 \sqrt{13}$
RS = $\sqrt{{(- 2 + 5)}^{2} + {(- 5 + 3)}^{2}} = \sqrt{9 + 4} = \sqrt{13}$
PS = $\sqrt{{(- 5 - 1)}^{2} + {(- 2 - 2)}^{2}} = \sqrt{36 + 16} = \sqrt{52} = 2 \sqrt{13}$
PQ = RS and QR = PS
Opposite sides are equal.
$PR = \sqrt{{(- 3 - 1)}^{2} + {(- 5 - 2)}^{2}} = \sqrt{16 + 49} = \sqrt{65} QS = \sqrt{{(- 2 + 1)}^{2} + {(- 5 - 3)}^{2}} = \sqrt{1 + 64} = \sqrt{65}$
Diagonals are equal.
Thus, the given vertices form a rectangle.

Page No 123:

Question 13:

Find the lengths of the medians of a triangle whose vertices are A(–1, 1), B(5, –3) and C(3, 5).

Answer:

Let the medians meet the lines BC, AC and AB at points be $P (x_{1}, y_{1})$ , $Q (x_{2}, y_{2})$ and $R (x_{3}, y_{3})$ respectively.
P is thus the mid point of line BC
$P (x_{1}, y_{1}) = (\frac{5 + 3}{2}, \frac{5 - 3}{2}) = (4, 2) AP = \sqrt{{(4 + 1)}^{2} + {(1 - 1)}^{2}} = \sqrt{25} = 5$
Q is the mid point of line AC.
$Q (x_{2}, y_{2}) = (\frac{- 1 + 3}{2}, \frac{1 + 5}{2}) = (1, 3) BQ = \sqrt{{(5 - 1)}^{2} + {(- 3 - 3)}^{2}} = \sqrt{16 + 36} = \sqrt{52} = 2 \sqrt{13}$
R is the mid point of AB.
$R (x_{3}, y_{3}) = (\frac{- 1 + 5}{2}, \frac{1 - 3}{2}) = (2, - 1) RC = \sqrt{{(3 - 2)}^{2} + {(5 + 1)}^{2}} = \sqrt{1 + 36} = \sqrt{37}$

Page No 123:

Question 14:

Find the coordinates of centroid of the triangles if points D(–7, 6), E(8, 5) and F(2, –2) are the mid points of the sides of that triangle.

Answer:

The given points are D(–7, 6), E(8, 5) and F(2, –2).
Centroid is the meeting point of the medians.
Let the centroid be O(a, b)
$(a, b) = (\frac{2 + 8 - 7}{3}, \frac{- 2 + 5 + 6}{3}) = (1, 3)$

Page No 123:

Question 15:

Show that A(4, –1), B(6, 0), C(7, –2) and D(5, –3) are vertices of a square.

Answer:

The given points are A(4, –1), B(6, 0), C(7, –2) and D(5, –3).
AB = $\sqrt{{(6 - 4)}^{2} + {(0 + 1)}^{2}} = \sqrt{4 + 1} = \sqrt{5}$
BC = $\sqrt{{(6 - 7)}^{2} + {(0 + 2)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
CD = $\sqrt{{(7 - 5)}^{2} + {(- 2 + 3)}^{2}} = \sqrt{4 + 1} = \sqrt{5}$
AD = $\sqrt{{(5 - 4)}^{2} + {(- 3 + 1)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$
AB = BC = CD = DA
Slope of AB = $\frac{0 + 1}{6 - 4} = \frac{1}{2}$
Slope of BC = $\frac{- 2 - 0}{7 - 6} = - 2$
Slope of CD = $\frac{- 3 + 2}{5 - 7} = \frac{1}{2}$
Slope of AD = $\frac{- 3 + 1}{5 - 4} = - 2$
Thus, AB perpendicular to BC and AD. Also, CD perpendicular to AD and AB.
So, all the sides are equal to each other and are perpendicular.
Thus, they form a square.

Page No 123:

Question 16:

Find the coordinates of circumcentre and radius of circumcircle of âˆ†ABC if A(7, 1), B(3, 5) and C(2, 0) are given.

Answer:

Let the circumcentre be $P (a, b)$ .
The given points are A(7, 1), B(3, 5) and C(2, 0).
The circumcircle passes through the points A, B and C and the thus,
PA = PB = PC
$\Rightarrow {PA}^{2} = {PB}^{2} = {PC}^{2}$
$P A^{2} = P B^{2} \Rightarrow {(3 - a)}^{2} + {(5 - b)}^{2} = {(7 - a)}^{2} + {(1 - b)}^{2} \Rightarrow 9 + a^{2} - 6 a + 25 + b^{2} - 10 b = 49 + a^{2} - 14 a + 1 + b^{2} - 2 b \Rightarrow a - b = 2 . . . . . (1) P A^{2} = P C^{2} \Rightarrow {(7 - a)}^{2} + {(1 - b)}^{2} = {(2 - a)}^{2} + {(0 - b)}^{2} \Rightarrow 49 + a^{2} - 14 a + 1 + b^{2} - 2 b = 4 + a^{2} - 4 a + b^{2} \Rightarrow 5 a + b = 23 . . . . . (2) (1) + (2) a = \frac{25}{6}, b = \frac{13}{6}$
Radius = PC =
$= \sqrt{{(\frac{25}{6} - 2)}^{2} + {(\frac{13}{6} - 0)}^{2}} = \sqrt{{(\frac{13}{6})}^{2} + {(\frac{13}{6})}^{2}} = \frac{13}{6} \sqrt{2}$

Page No 123:

Question 17:

Given A(4, –3), B(8, 5). Find the coordinates of the point that divides segment AB in the ratio 3 : 1.

Answer:

Let he coordinate of the point which divide the line AB in the ratio 3 : 1 be P(a, b)
$a = \frac{3 \times 8 + 1 \times 4}{3 + 1} = \frac{24 + 4}{4} = 7 b = \frac{3 \times 5 + 1 \times (- 3)}{3 + 1} = \frac{15 - 3}{4} = 3$
P(a, b) = (7, 3)

Page No 123:

Question 18:

Find the type of the quadrilateral if points A(–4, –2), B(–3, –7) C(3, –2) and D(2, 3) are joined serially.

Answer:

The given points are A(–4, –2), B(–3, –7) C(3, –2) and D(2, 3).
If they are joined serially so,
Slope of AB = $\frac{- 7 + 2}{- 3 + 4} = - 5$
Slope of BC = $\frac{- 2 + 7}{3 + 3} = \frac{5}{6}$
Slope of CD = $\frac{3 + 2}{2 - 3} = - 5$
Slope of AD = $\frac{3 + 2}{2 + 4} = \frac{5}{6}$
Opposite sides are parallel.
AC = $\sqrt{{(3 + 4)}^{2} + {(- 2 + 2)}^{2}} = \sqrt{49} = 7$
BD = $\sqrt{{(3 + 7)}^{2} + {(2 + 3)}^{2}} = \sqrt{125} = 5 \sqrt{5}$
Diagonals are not equal.
Hence, the given points form a parallelogram.

Page No 123:

Question 19:

The line segment AB is divided into five congruent parts at P, Q, R and S such that A–P–Q–R–S–B. If point Q(12, 14) and S(4, 18) are given find the coordinates of A, P, R, B.

Answer:

Let the coordinates be $A (x_{1}, y_{1})$ $P (x_{2}, y_{2})$ , $R (x_{3}, y_{3})$ and B $(x_{4}, y_{4})$ .
QR = RS
$R (x_{3}, y_{3}) = (\frac{12 + 4}{2}, \frac{14 + 18}{2}) = (8, 16)$
RS = SB
$B (x_{4}, y_{4}) = \frac{8 + x_{4}}{2} = 4, \frac{16 + y_{4}}{2} = 18 \Rightarrow 8 + x_{4} = 8, y_{4} = 36 - 16 \Rightarrow x_{4} = 0, y_{4} = 20 B (x_{4}, y_{4}) = (0, 20)$
$\frac{AQ}{QB} = \frac{AP + PQ}{QR + RS + SB} = \frac{AP + AP}{AP + AP + AP} = \frac{2}{3}$
$\frac{2 \times 0 + 3 \times x_{1}}{2 + 3} = 12, \frac{2 \times 20 + 3 \times y_{1}}{2 + 3} = 14 \Rightarrow 3 x_{1} = 60, 40 + 3 y_{1} = 14 \Rightarrow x_{1} = 20, y_{1} = 10$
$A (x_{1}, y_{1})$ = (20, 10)
AP = PQ
$\frac{20 + 12}{2} = x_{2}, \frac{10 + 14}{2} = y_{2} \Rightarrow \frac{32}{2} = x_{2}, \frac{24}{2} = y_{2} \Rightarrow x_{2} = 16, y_{2} = 12$

Page No 123:

Question 20:

Find the coordinates of the centre of the circle passing through the points P(6, –6), Q(3, –7) and R (3, 3).

Answer:

Let the centre be P(a, b)
RO = PO = QO
${RO}^{2} = {PO}^{2}$
$\Rightarrow {(3 - a)}^{2} + {(3 - b)}^{2} = {(6 - a)}^{2} + {(- 6 - b)}^{2} \Rightarrow 9 + a^{2} - 6 a + 9 + b^{2} - 6 b = 36 + a^{2} - 12 a + 36 + b^{2} + 12 b \Rightarrow 6 a - 18 b = 54 \Rightarrow a - 3 b = 9 . . . . . (1)$
${PO}^{2} = {OQ}^{2}$
${(6 - a)}^{2} + {(- 6 - b)}^{2} = {(3 - a)}^{2} + {(- 7 - b)}^{2} \Rightarrow 36 + a^{2} - 12 a + 36 + b^{2} + 12 b = 9 + a^{2} - 6 a + 49 + b^{2} + 14 b \Rightarrow 3 a + b = 7 . . . . . (2)$
Multiplying (2) with 3 we get
$a - 3 b = 9 . . . . . (3) 9 a + 3 b = 21 . . . . . (4) Solving (3) and (4) a = 3, b = - 2$

Page No 123:

Question 21:

Find the possible pairs of coordinates of the fourth vertex D of the parallelogram, if three of its vertices are A(5, 6), B(1, –2) and C(3, –2).

Answer:

ABCD is a parallelogram. So, AD = CB and CD = AB.
$A D^{2} = B C^{2} \Rightarrow {(a - 5)}^{2} + {(b - 6)}^{2} = {(3 - 1)}^{2} + {(- 2 + 2)}^{2} \Rightarrow a^{2} + 25 - 10 a + b^{2} + 36 - 12 b = 4 \Rightarrow a^{2} + b^{2} - 10 a - 12 b + 57 = 0 . . . . . (1) C D^{2} = A B^{2} \Rightarrow {(a - 3)}^{2} + {(b + 2)}^{2} = {(5 - 1)}^{2} + {(6 + 2)}^{2} \Rightarrow a^{2} + b^{2} - 6 a + 4 b + 13 = 80 \Rightarrow a^{2} + b^{2} - 6 a + 4 b - 67 = 0 . . . . . (2)$
Point D lies on the line passing through the point A. So, the ordinate of the point D will also be same as that of point A which is 6.
So, b = 6
Putting the value of b in (1) we get
$a^{2} + {(6)}^{2} - 10 a - 12 \times 6 + 57 = 0 \Rightarrow a^{2} + 36 - 10 a - 72 + 57 = 0 \Rightarrow a^{2} - 10 a + 21 = 0 \Rightarrow a^{2} - 7 a - 3 a + 21 = 0 \Rightarrow a (a - 7) - 3 (a - 7) = 0 \Rightarrow (a - 3) (a - 7) = 0 \Rightarrow a = 3, 7$
Thus, the possible values of point D are (3, 6) and (7, 6).

Page No 123:

Question 22:

Find the slope of the diagonals of a quadrilateral with vertices A(1, 7), B(6, 3), C(0, –3) and D(–3, 3).

Answer:

The quadrilateral is formed with the coordinates A(1, 7), B(6, 3), C(0, –3) and D(–3, 3).
Slope of BD = $\frac{3 - 3}{6 + 3} = 0$
Slope of AC = $\frac{7 + 3}{1 - 0} = \frac{10}{1} = 10$

View NCERT Solutions for all chapters of Class 10