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Maths 3 Marks Question

P-2 Co-ordinate Geometry · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 78 questions.

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Question 13 Marks
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
According to the distance formula, the distance ' d ' between two points $( a , b )$ and $( c , d )$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
$\text { (i) } A(a, 0), B(0, a) $
$\text { i. } d=\sqrt{(a-0)^2+(0-a)^2} $
$=\sqrt{2 a^2}$
$=a \sqrt{2}$
$\text { (ii) } P(-6,-3), Q(-1,9) $
$d=\sqrt{(-6-(-1))^2+(-3-9)^2}$
$=\sqrt{(-5)^2+(-12)^2} $
$=\sqrt{25+144} $
$=\sqrt{169 }$
$=13$
$\text { (iii) } R(-3 a, a), S(a,-2 a) $
$d=\sqrt{(-3 a-a)^2+(a-(-2 a))^2} $
$=\sqrt{(-4 a)^2+(3 a)^2} $
$=\sqrt{16 a^2+9 a^2}$
$= \sqrt{{25}^2}$
$= 5a$
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Question 23 Marks
Find the point on X-axis which is equidistant from P(2,-5) and Q(-2,9).
Answer
According to the distance formula, the distance 'd' between two points $(a, b)$ and $(c, d)$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
As the point is on the $x=$ axis it is of the form $(x, 0)$
Distance from point $P=\sqrt{(2-x)^2+(-5-0)^2}=\sqrt{(2-x)^2+25}$
Distance from point $Q =\sqrt{(-2- x )^2+(9-0)^2}=\sqrt{(2+ x )^2+81}$
As the two points are equidistant from ( $x , 0$ )
$\sqrt{(2-x)^2+25}=\sqrt{(2+x)^2+81}$
Squaring both sides
$(2-x)^2+25=(2+x)^2+81$
Expanding and simplifying
$-4 x+25=4 x+81$
$8 x=-56$
$x=-7$
Hence the point is $(-7,0)$
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Question 33 Marks
Find the coordinates of the midpoint of the line segment joining P(0,6)and Q(12,20).
Answer
According to the mid point theorem the coordinates of the point P(x,y) dividing the line formed by A(a,b) and B(c,d) is given by:
$x =\frac{ a + c }{2} $
$y =\frac{ b + d }{2}$
In question $x =\frac{0+12}{2}=6$
$y=\frac{6+20}{2}=13$
Hence mid point is $(6,13)$
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Question 43 Marks
Determine whether the given points are collinear.
(1) A(0,2) , B(1,-0.5), C(2,-3)
(2) P(1, 2) , Q(2, 8/5) , R(3, 6/5)
(3) L(1,2) , M(5,3) , N(8,6)
Answer
If Three points $(a, b),(c, d),(e, f)$ are collinear then the area formed by the triangle by the three points is zero.
Area of a triangle $=\frac{1}{2}| a ( d - f )+ c ( f - b )+ e ( b - d )|$
1. area $=\frac{1}{2}(0(-0.5-(-3))+1(-3-2)+2(2-(-0.5)))=0$
Hence the points are collinear.
2. area $=\frac{1}{2}\left(1\left(\frac{8}{5}-\frac{6}{5}\right)+2\left(\frac{6}{5}-2\right)+3\left(2-\frac{8}{5}\right)\right)=0$
Hence the points are collinear
3. area $=\frac{1}{2}(1(3-6)+5(6-2)+8(2-3))=\frac{9}{2}$
Hence the points are not collinear
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Question 53 Marks
If A (1, -1),B (0, 4),C (-5, 3) are vertices of a triangle then find the slope of each side
Answer
Slope $m$ of a line passing through two points $A(a, b)$ and $B(c, d)$ is given by
$m =\frac{ d - b }{ c - a }$
Slope of $A B=$
$=\frac{4-(-1)}{0-1}=-5$
Slope of BC =
$\frac{3-4}{-5-0}=\frac{1}{5}$
Slope of $A C=$
$\frac{3-(-1)}{-5-1}=-\frac{2}{3}$
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Question 63 Marks
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, 5/ 2 ), N(4, -2)
Answer
Three points are said to be collinear if they all lie in a straight line.
If Three points $\left( x _1, y _1\right),\left( x _2, y _2\right),\left( x _3, y _3\right)$ are collinear then no triangle can be formed using three points and so the area formed by the triangle by the three points is zero.
Area of Triangle $=1 / 2\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$.
1. For triangle, $A(-1,-1), B(0,1), C(1,3)$
$\text { Area }=\frac{1}{2}(-1(1-3)+0(3-(-1))+1(-1-1))$
= 1/2 [ -1(-2) + 0(3+1) + 1(-1-1)]

= 1/2 [2 + 0 - 2]

= 1/2 [2-2]
= 0
Hence the points are collinear

2. For triangle, D(-2, -3), E(1, 0), F(2, 1)
Using 1,
Area = 1/2 [-2(0-1) + 1(1-(-3)) + 2(-3-0)]
= 1/2 [-2(-1) + 1(1+3) + 2(-3)]
= 1/2 [ 2 + 4 -6 ]
= 1/2 [ 6 - 6 ]
= 1/2 (0)
= 0
Hence the points are collinear.
3. For triangle, L(2, 5), M(3, 3), N(5, 1)
Using 1,
Area = 1/2 [2(3-1) + 3(1-5) + 5(5-3)]
= 1/2 [ 2(2) + 3(-4) + 5(2)]
= 1/2 [ 4 - 12 + 10 ]
= 1/2 [ 14 - 12 ]

= 1/2 (2)
= 1
Hence the points are not collinear.
4. For triangle, P(2, -5), Q(1, -3), R(-2, 3)
Using 1,
Area  $=\frac{1}{2}(2(-3-3)+1(3-(-5))+(-2)(-5-(-3))=0$
Area = 1/2 [ 2(-3-3) + 1(3-(-5)) + (-2)(-5-(-3))]
= 1/2 [ 2(-6) + 1(3+5) - 2 (-5+3)]
= 1/2 [ -12 + 8 -2(-2)]
= 1/2 [-12 + 8 +4]
= 1/2 [ -12 + 12]
= 1/2 (0)
= 0
Hence the points are collinear.
5. For triangle, R(1, -4), S(-2, 2), T(-3, 4)
Area $=\frac{1}{2}(1(2-4)+(-2)(4-(-4))+(-3)(-4-2))=0$
Hence the points are collinear.
6. For triangle, A(-4, 4), K(-2, 5/ 2 ), N(4, -2)
Area $=\frac{1}{2}\left((-4)\left(\frac{5}{2}-(-2)+(-2)(-2-4)+4\left(4-\frac{5}{2}\right)\right)=0\right.$
Hence the points are collinear.
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Question 73 Marks
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
The coordinates of the centroid (x,y) od a triangle formed by points (a,b), (c,d), (e,f) is given by
$x=\frac{a+c+e}{3} $
$ y=\frac{b+d+f}{3} $
$ \text { 1. } x=\frac{-7+2+8}{3}=1 $
$ y=\frac{6-2+5}{3}=3$
$ \text { 2. } x=\frac{3+4+11}{3}=6 $
$y=\frac{-5+3-4}{3}=-2 \\ \text { 3. } x=\frac{4+8+7}{3}=\frac{19}{3} $
$ y=\frac{7+4+11}{3}=\frac{22}{3}$
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Question 83 Marks
Find the ratio in which point T(-1, 6)divides the line segment joining the pointsP(-3, 10) and Q(6, -8).
Answer
A point $P(x, y)$ divides the line formed by points $(a, b)$ and $(c, d)$ in the ratio of $m: n$, then the coordinates of the point $P$ is given by
$x =\frac{ an + cm }{ m + n } \text { and } y =\frac{ bn + dm }{ m + n }$
In the given question,
Let the point T divide the line $PQ$ in the ratio m:n
Here $x=-1$ and $y=6$
$1=\frac{-3 n+6 m}{m+n}$
$6=\frac{10 n -8 m}{ m + n }$
Simplifying (1) we get,
$-m-n=-3 n+6 m$
$2 n=7 m$
Simplifying (2) we get,
$6 m+6 n=10 n-8 m$
$14 m=4 n$
From both we get $\frac{ m }{ n }=\frac{2}{7}$
Hence the point $T$ divides $P Q$ in the ratio $2: 7$
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Question 93 Marks
Find the coordinates of the centre of the circle passing through the points P(6,-6), Q(3,-7)and R(3,3).
Answer
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}............(1)$
Let the centre be A(x,y)
As it passes through the given points, distance between centre and the points is the radius.
$A P=\sqrt{(x-6)^2+(y+6)^2}$
$A Q=\sqrt{(x-3)^2+(y+7)^2}$
$A R=\sqrt{(x-3)^2+(y-3)^2}$
As AP = AQ
Squaring both sides
$(x-6)^2+(y+6)^2=(x-3)^2+(y+7)^2$
Simplifying
$12 y-12 x+72=14 y-6 x+58$
$2 y+6 x-14=0 \ldots \text { (a) }$
$A P=A R$
Squaring both sides and simplifying
$6 y-6 x+54=0$
Solving for $x$ and $y$ using (a) and (b)
We get $x=3 ; y=-2$
Hence centre is $(3,-2)$
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Question 103 Marks
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
A point P(x,y) divides the line formed by points (a,b) and (c,d) in the ratio of m:n, then the coordinates of the point P is given by
$x =\frac{ an + cm }{ m + n } \text { and } y =\frac{ bn + dm }{ m + n }$
Where m and n is defined as the ratio in which the line segments are divided

$\text { 1. } x=\frac{(-3)(1)+7(2)}{2+1}$
$x=\frac{11}{3}$
$y=\frac{(1) 7+(-4) 2}{2+1}$
$y=\frac{-1}{3}$
2. $x=\frac{(-2) 4+(4) 3}{4+3}$
$x=\frac{4}{7}$
$y=\frac{(-5) 4+(3) 3}{4+3}$
$y=\frac{-11}{7}$
$\text { 3. } x=\frac{2(2)+(-4) 1}{2+1}=0$
$y=\frac{(6) 2+1(1)}{2+1}$
$y=\frac{13}{3}$
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Question 113 Marks
The line segment AB is divided into five congruent parts at P, Q, R and S suchthat A-P-Q-R-S-B. If point Q(12, 14) and
S(4, 18) are given find thecoordinates of A, P, R,B.
Answer
$x=\frac{ an + cm }{ m + n } \text { and } y=\frac{ bn + dm }{ m + n }$
Coordinates of $R$ as $Q R: R S:: 1: 1$
$X=\frac{12+4}{2}=8$
$Y=\frac{14+18}{2}=16$
As RS:SB: :1:1
Coordinates of $B$
$4=\frac{8+x}{2}$
$x=0$
And
$18=\frac{16+y}{2}$
$Y=20$
As $PQ:QR::1:1$
Coordinates of $P$
$12=\frac{x+8}{2}$
$x=16$
and
$14=\frac{y+16}{2}$
$y=12$
$As AP:PQ::1:1$
Coodinates of $A$
$16=\frac{x+12}{2}$
$x=20$
and
$12=\frac{y+14}{2}$
$y=10$
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Question 123 Marks
If A (-14, -10), B(6, -2) is given, find the coordinates of the points whichdivide segment AB into four equal parts.
Answer
let the points dividing AB be C,D,E.
AC:CD:DE:EB∷1:1:1:1
A point P(x,y) divides the line formed by points (a,b) and (c,d) in the ratio of m:n, then the coordinates of the point P is given by
$x =\frac{ an + cm }{ m + n } \text { and } y =\frac{ bn + dm }{ m + n }$
$\text { For } C m:n :: 1: 3$
$x =\frac{(-14) 3+(6) 1}{1+3}=-9$
$y =\frac{(-10) 3+(-2) 1}{1+3}=-8$
For D m:n ::2:2
$x=\frac{(-14) 2+(6) 2}{2+2}=-4$
$y=\frac{(-10) 2+(-2) 2}{2+2}=-6$
For E m:n :: 3:1
$x=\frac{(-14) 1+(6) 3}{3+1}=1$
$y=\frac{(-10) 1+(-2) 3}{1+3}=-4$
Hence coordinates of $C=(-9,-8)$
$D=(-4,-6)$
$E=(1,-4)$
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Question 133 Marks
Find the co-ordinates of the points of trisection of the line segment AB with A(2, 7) and B(-4, -8).
Answer
let The points of trisection of a given line AB be P and Q
Then the ratio AP:PQ:QB = 1:1:1
Hence we get AP:PB = 1:2
And AQ:QB = 2:1
A point P(x,y) divides the line formed by points (a,b) and (c,d) in the ratio of m:n, then the coordinates of the point P is given by
$x=\frac{a n+c m}{m+n} \text { and } y=\frac{b n+d m}{m+n}$
To find point $P(x, y)$
$x=\frac{2(2)+(-4) 1}{2+1}$
$x=0$
$y=\frac{(7) 2+(-8) 1}{2+1}$
$y=2$
To find the point $Q\left(x^{\prime}, y^{\prime}\right)$
$x ^{\prime}=\frac{((2) 1+(-4) 2)}{2+1}$
$x ^{\prime}=-2$
$y ^{\prime}=\frac{(1) 7+(-8) 2}{2+1}$
$y ^{\prime}=-3$
Hence point $P=(0,2)$ and $Q=(-2,-3)$
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Question 143 Marks
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
A point $P(x, y)$ divides the line formed by points $(a, b)$ and $(c, d)$ in the ratio of $m: n$, then the coordinates of the point $P$ is given by
$x=\frac{a n+c m}{m+n} \text { and } y=\frac{b n+d m}{m+n}$
In the given question $x =\frac{(-1) 3+4(2)}{2+3}$
$x=\frac{8}{8}=1$
$y=\frac{7(3)+(-3)(2)}{2+3}$
$y=3$
Hence the coordinates of the point are $(1,3)$.
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Question 153 Marks
Find the type of the quadrilateral if points A(-4, -2), B(-3, -7) C(3, -2) andD(2, 3) are joined serially.
Answer
According to the distance formula, the distance ' $d$ ' between two points $(a, b)$ and $(c, d)$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
Slope of a line between two points $(a, b)$ and $(c, d)$ is
$m =\frac{d-b}{ c - a }$
$AB =\sqrt{(-4+3)^2+(-2+7)^2}=\sqrt{(26)}$
$BC =\sqrt{(-3-3)^2+(-7+2)^2}=\sqrt{(61)}$
$CD =\sqrt{(2-3)^2+(-2-3)^2}=\sqrt{(26)}$
$A D=\sqrt{(-4-2)^2+(-2-3)^2}=\sqrt{61}$
Slope $A B=\frac{-7+2}{-3+4}=-5$
Slope $B C=\frac{-2+7}{3+3}=\frac{5}{6}$
Slope $C D=\frac{3+2}{2-3}=-5$
Slope $A D=\frac{3+2}{2+4}=\frac{5}{6}$
As opposite sides are equal and parallel, it forms a parallelogram.
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Question 163 Marks
Find the coordinates of circumcentre and radius of circumcircle of triangle ABC ifA(7, 1), B(3, 5) and C(2, 0) are given.
Answer
Let the circumcentre be $(x, y)$
As the circumcentre is equidistant from all the 3 points, we get
$\sqrt{(3-x)^2+(5-y)^2}=\sqrt{(2-x)^2+y^2} \ldots \ldots i$
And
$\sqrt{(2-x)^2+y^2}=\sqrt{(7-x)^2+(y-1)^2}$
Squaring both sides of i and simplifying, we get
$-2 x-10 y=-30$
Squaring both sides of ii and simplifying, we get
$10 x+2 y=46$
Solving the above equations, we get
$x=\frac{25}{6}$
$y=\frac{13}{6}$
Radius of circumcircle is the distance between any point on the triangle and the circumcentre.
$\text { Radius }=\sqrt{\left(2-\frac{25}{6}\right)^2+\left(0-\frac{13}{6}\right)^2}=\frac{13}{6} \sqrt{2}$
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Question 173 Marks
Show that A(4, -1), B(6, 0), C(7, -2) and D(5, -3)are vertices of a square.
Answer
According to the distance formula, the distance 'd' between two points $(a, b)$ and $(c, d)$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
Slope of a line between two points $(a, b)$ and $(c, d)$ is
$m =\frac{ d - b }{ c - a }$
Note: If the Product of slopes of two lines $=-1$ then they are perpendicular to each other.
$A B=\sqrt{(6-4)^2+(-1+2)^2}=\sqrt{5}$
$B C=\sqrt{(6-7)^2+(-2-0)^{\wedge} 2}=\sqrt{5}$
$C D=\sqrt{(7-5)^2+(-2+3)^2}=\sqrt{(5)}$
$A D=\sqrt{(5-4)^2+(-1+3)^2}=\sqrt{5}$
Slope $A B=\frac{0-(-1)}{6-4}=\frac{1}{2}$
Slope $B C=\frac{-2-0}{7-6}=-2$
Slope $C D=\frac{-3+2}{5-7}=\frac{1}{2}$
Slope $A D=\frac{-3+1}{5-4}==2$
As all sides are equal and ajdacent sides are perndicular. Given points form a square.
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Question 183 Marks
Show that points A(-4, -7), B(-1, 2), C(8, 5) and D(5, -4) are vertices of a rhombus ABCD.
Answer
In a parallelogram, opposite sides are equal and parallel.
According to the distance formula, the distance 'd' between two points $(a, b)$ and $(c, d)$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
For the given points, length $P Q=\sqrt{(2-7)^2+(-2-3)^2}$
$P Q=\sqrt{50}$
Length $Q R=\sqrt{(11-7)^2+(3-(-1))^2}$
$Q R=\sqrt{16+16}=32$
Length $R S=\sqrt{(11-6)^2+(-1-(-6))^2}$
$R S=\sqrt{25+25}=\sqrt{50}$
Length $S P=\sqrt{(6-2)^2+(-6-(-2))^2}$
$S P=\sqrt{16+16}=\sqrt{32}$
As $P Q=R S$ and $Q R=S P$
Checking for slopes
Slope of a line between two points $(a, b)$ and $(c, d)$ is
$m =\frac{ d - b }{ c - a }$
Slope $P Q=\frac{7-2}{3-(-2)}=1$
Slope $QR =\frac{11-7}{-1-3}=-1$
Slope RS $=\frac{6-11}{-6-(-1)}=1$
Slope SP $=\frac{6-2}{-6-(-2)}=-1$
As $P Q=R S$ and their slope $=1$
And
$Q R=S P$ and their slope $=-1$
Hence the given points form a parallelogram.
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Question 193 Marks
Show that points P(2, -2), Q(7, 3), R(11, -1) and S (6, -6) are vertices of a parallelogram.
Answer
In a parallelogram, opposite sides are equal and parallel.
According to the distance formula, the distance 'd' between two points $(a, b)$ and $(c, d)$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
For the given points, length $P Q=\sqrt{(2-7)^2+(-2-3)^2}$
$P Q=\sqrt{50}$
Length $Q R=\sqrt{(11-7)^2+(3-(-1))^2}$
$Q R=\sqrt{16+16}=32$
Length $R S=\sqrt{(11-6)^2+(-1-(-6))^2}$
$R S=\sqrt{25+25}=\sqrt{50}$
Length $S P=\sqrt{(6-2)^2+(-6-(-2))^2}$
$S P=\sqrt{16+16}=\sqrt{32}$
As $P Q=R S$ and $Q R=S P$
Checking for slopes
Slope of a line between two points $(a, b)$ and $(c, d)$ is
$m =\frac{ d - b }{ c - a }$
Slope $P Q=\frac{7-2}{3-(-2)}=1$
Slope $QR =\frac{11-7}{-1-3}=-1$
Slope RS $=\frac{6-11}{-6-(-1)}=1$
Slope SP $=\frac{6-2}{-6-(-2)}=-1$
As $P Q=R S$ and their slope $=1$
And
$Q R=S P$ and their slope $=-1$
Hence the given points form a parallelogram.
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Question 203 Marks
Verify that points P(-2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.
Answer
In a right angles triangle ABC, right angled at B, according to the pythagoras theorem
$\mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2$
According to the distance formula, the distance 'd' between two points (a,b) and (c,d) is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}\dots(1)$
For the given points Distance between P and Q is
$P Q=\sqrt{(-2-2)^2+(2-2)^2}=\sqrt{16} $
$Q R=\sqrt{(2-2)^2+(7-2)^2}=\sqrt{25} $
$P R=\sqrt{(-2-2)^2+(2-7)^2}=\sqrt{16+25}=\sqrt{41}$
$\mathrm{PQ}^2=16$
$\mathrm{QR}^2=25$
$\mathrm{PR}^2=41$
As $P Q^2+Q^2=P R^2$
Hence the given points form a right angled triangle.
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Question 213 Marks
Find the point on the X-axis which is equidistant from A(-3, 4) and B(1, -4).
Answer
A point in the x = axis is of the form (a,0)
Distance d between two points(a,b) and (c,d)is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2}$
Distance between $(-3,4)$ and $(a, 0)=$
$D=\sqrt{(-3-a)^2+(0-4)^2} $
$D \sqrt{(3+a)^2+16}$
Distance between $(1,-4)$ and $(a, 0)$
$D=\sqrt{(1-a)^2+(0-(-4))^2} $
$D=\sqrt{(1-a)^2+16}$
As the two points are equidistant from the point (a.0)
$\sqrt{(1-a)^2+16}=\sqrt{(3+a)^2+16}$
Squaring both sides, we get
$(1-a)^2+16=(3+a)^2+16$
$1+a^2-2 a=9+a^2+6 a$
$8 a=-8$
$a=-1$
Hence the point is $(-1,0)$
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Question 223 Marks
Find the coordinates of the circumcentre of a triangle whose vertices are (-3,1),(0,-2) and (1,3)
Answer
The circumcentre is equidistant from all the points of the triangle. Let the coordinates of circumcentre be $(x, y)$
$\sqrt{(-3-x)^2+(1-y)^2}=\sqrt{x^2+(y+2)^2} \ldots i$
And
$\sqrt{x^2+(y+2)^2}=\sqrt{(1-x)^2+(3-y)^2} \ldots \text { ii }$
Squaring and simplifying $i$, we get
$2 x-2 y=-6$
Squaring and simplifying ii, we get
$2 x+10 y=6$
Solving the above equations, we get
$x=-\frac{1}{3}$
$y=\frac{2}{3}$
hence the coordinates of circumcircle is $\left(-\frac{1}{3}, \frac{2}{3}\right)$
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Question 233 Marks
If point $\mathrm{P}(-4,6)$ divides the line segment $\mathrm{AB}$ with $\mathrm{A}(-6,10)$ and $\mathrm{B}(\mathrm{r}, \mathrm{s})$ in the ratio $2: 1$, find the co-ordinates of $B$.
Answer
By section formula
$-4=\frac{2 \times r+1 \times(-6)}{2+1}$
$6=\frac{2 \times s+1 \times 10}{2+1}$
$\therefore-4=\frac{2 r-6}{3}$
$\therefore 6=\frac{2 s+10}{3}$
$\therefore-12=2 r-6$
$\therefore 18=2 \mathrm{~s}+10$
$\therefore 2 \mathrm{r}=-6$
$\therefore 2 \mathrm{~s}=8$
$\therefore \quad r=-3$
$\therefore \mathrm{s}=4$
$\therefore$ co-ordinates of point $\mathrm{B}$ are $(-3,4)$.
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Question 243 Marks
If point $T$ divides the segment $A B$ with $A(-7,4)$ and $B(-6,-5)$ in the ratio $7: 2$, find the co-ordinates of $T$.
Answer
Let the co-ordinates of $\mathrm{T}$ be $(x, y)$.
$\therefore$ by the section formula,
$
\begin{aligned}
x=\frac{m x_2+n x_1}{m+n} & =\frac{7 \times(-6)+2 \times(-7)}{7+2} \\
& =\frac{-42-14}{9}=\frac{-56}{9}
\end{aligned}
$
$
y=\frac{m y_2+n y_1}{m+n}=\frac{7 \times(-5)+2 \times(4)}{7+2}
$
$
=\frac{-35+8}{9}=\frac{-27}{9}=-3
$
$\therefore$ co-ordinates of point $\mathrm{T}$ are $\left(\frac{-56}{9},-3\right)$.

Image

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Question 253 Marks
$\mathrm{A}(-3,-4), \mathrm{B}(-5,0), \mathrm{C}(3,0)$ are the vertices of $\triangle \mathrm{ABC}$. Find the co-ordinates of the circumcentre of $\triangle \mathrm{ABC}$.
Answer
Let, $\mathrm{P}(a, b)$ be the circumcentre of $\triangle \mathrm{ABC}$.
$\therefore$ point $\mathrm{P}$ is equidistant from $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$.
$
\begin{aligned}
\therefore \mathrm{PA}^2=\mathrm{PB}^2=\mathrm{PC}^2 \ldots \ldots \ldots & \text { (I) } \quad \therefore \mathrm{PA}^2=\mathrm{PB}^2 \\
(a+3)^2+(\mathrm{b}+4)^2 & =(a+5)^2+(\mathrm{b}-0)^2 \\
\therefore a^2+6 a+9+\mathrm{b}^2+8 \mathrm{~b}+16 & =a^2+10 a+25+\mathrm{b}^2 \\
\therefore-4 a+8 \mathrm{~b} & =0 \\
\therefore a-2 \mathrm{~b} & =0 \ldots \ldots \ldots \text { (II) } \\
\text { Similarly } \mathrm{PA}^2 & =\mathrm{PC}^2 \ldots \ldots \ldots \text { (I) From } \\
\therefore(a+3)^2+(\mathrm{b}+4)^2 & =(a-3)^2+(\mathrm{b}-0)^2 \\
\therefore a^2+6 a+9+\mathrm{b}^2+8 \mathrm{~b}+16 & =a^2-6 a+9+\mathrm{b}^2 \\
\therefore 12 a+8 \mathrm{~b} & =-16 \\
\therefore 3 a+2 \mathrm{~b} & =-4 \ldots \ldots \ldots \text { (III) }
\end{aligned}
$
Solving (II) and (III) we get $a=-1, \mathrm{~b}=-\frac{1}{2}$
$\therefore$ co-ordinates of circumcentre are $\left(-1,-\frac{1}{2}\right)$.
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Question 263 Marks
Show that points $(1,7),(4,2),(-1,-1)$ and $(-4,4)$ are vertices of a square.
Answer
In a quadrilateral, if all sides are of equal length and both diagonals are of equal length, then it is a square.
$\therefore$ we will find lengths of sides and diagonals by using the distance formula.
Suppose, $A(1,7), B(4,2), C(-1,-1)$ and $D(-4,4)$ are the given points.
$ \mathrm{AB}=\sqrt{(1-4)^2+(7-2)^2}=\sqrt{9+25}=\sqrt{34}$
$\mathrm{BC}=\sqrt{(4+1)^2+(2+1)^2}=\sqrt{25+9}=\sqrt{34}$
$\mathrm{CD}=\sqrt{(-1+4)^2+(-1-4)^2}=\sqrt{9+25}=\sqrt{34}$
$\mathrm{DA}=\sqrt{(1+4)^2+(7-4)^2}=\sqrt{25+9}=\sqrt{34}$
$\mathrm{AC}=\sqrt{(1+1)^2+(7+1)^2}=\sqrt{4+64}=\sqrt{68}$
$\mathrm{BD}=\sqrt{(4+4)^2+(2-4)^2} \quad=\sqrt{64+4}=\sqrt{68}$
$\therefore \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \text { and } \mathrm{AC}=\mathrm{BD}$
$\therefore A B=B C=C D=D A \text { and } A C=B D$
So, the lengths of four sides are equal and two diagonals are equal.
$\therefore (1,7), (4,2), (-1,-1)$ and $(-4,4)$ are the vertices of a square.
Image
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Question 273 Marks
Verify, whether points $P(6,-6), Q(3,-7)$ and $R(3,3)$ are collinear.
Answer
$\mathrm{PQ}=\sqrt{(6-3)^2+(-6+7)^2}$ by distance formula
$
\begin{aligned}
& =\sqrt{(3)^2+(1)^2}=\sqrt{10} \\
\mathrm{QR} & =\sqrt{(3-3)^2+(-7-3)^2} \\
& =\sqrt{(0)^2+(-10)^2}=\sqrt{100} \\
\mathrm{PR} & =\sqrt{(3-6)^2+(3+6)^2} \\
& =\sqrt{(-3)^2+(9)^2}=\sqrt{90} .
\end{aligned}
$
From I, II and III out of $\sqrt{10}, \sqrt{100}$ and $\sqrt{90}, \sqrt{100}$ is the largest number.
Now we will verify whether $(\sqrt{100})$ and $(\sqrt{10}+\sqrt{90})$ are equal.
For this compare $(\sqrt{100})^2$ and $(\sqrt{10}+\sqrt{90})^2$.
You will see that $(\sqrt{10}+\sqrt{90})>(\sqrt{100}) \therefore \mathrm{PQ}+\mathrm{PR} \neq \mathrm{QR}$
$\therefore$ points $\mathrm{P}(6,-6), \mathrm{Q}(3,-7)$ and $\mathrm{R}(3,3)$ are not collinear.
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Question 303 Marks
Find the coordinates of centroid $G$ of $A B C$, if
(i) $A (8,9), B (4,5), C (6,2)$
(ii) $A (11,8), B (-6,5), C (1,-28)$
Answer
(i) $(6,5.33)$ (ii) $(2,-5)$
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Question 313 Marks
Find the coordinates of the points which divide the line segment joining the points $(-2,2)$ and $(6,-6)$ in four equal parts.
Answer
$(0,0)(2,-2)(4,-4)$
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Question 323 Marks
Find the ratio in which the line segment joining the points $(6,4)$ and $(1,-7)$ is divided by X -axis.
Answer
$4: 7$
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Question 333 Marks
In what ratio does the point $(1,3)$ divide line segment joining the points $(3,6)$ and $(-5,-6)$ ?
Answer
$1: 3$
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Question 343 Marks
Find the coordinates of the circumcentre of ABC, if A(2, 3), B(4, $-1$) and C(5, 2). Also, find circumradius.
Answer
$(3,1)$ circumradius $=\sqrt{5}$ units
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Question 353 Marks
Show that the points (2, 4), (2, 6) and $(2+\sqrt{3}, 5)$ are the vertices of an equilateral triangle.
Answer
self
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Question 373 Marks
Show that the points A(4, 7) B(8, 4) and C(7, 11) are the vertices of a right angled triangle.
Answer
self
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Question 383 Marks
Find the coordinates of the point on Y-axis which is equidistant from the points M(6, 5) and point N(-4, 3).
Answer
(0,9)
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Question 393 Marks
Find the relation between x and y, where point (x, y) is equidistant from (2, -4) and (-2, 6).
Answer
5y = x + 5
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Question 403 Marks
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)

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Question 433 Marks
Determine whether the given points are collinear.
(1) A(0,2) , B(1,-0.5), C(2,-3)
(2) P(1, 2) , Q(2, 8/5) , R(3, 6/5)
(3) L(1,2) , M(5,3) , N(8,6)

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Question 453 Marks
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, 5/ 2 ), N(4, -2)

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Question 463 Marks
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)

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Question 473 Marks
Find the ratio in which point T(-1, 6)divides the line segment joining the pointsP(-3, 10) and Q(6, -8).

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Question 483 Marks
Find the coordinates of the centre of the circle passing through the points P(6,-6), Q(3,-7)and R(3,3).

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Question 493 Marks
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

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Question 503 Marks
The line segment AB is divided into five congruent parts at P, Q, R and S suchthat A-P-Q-R-S-B. If point Q(12, 14) and
S(4, 18) are given find thecoordinates of A, P, R,B.

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