# Class 10 Maths Worksheet for Coordinate Geometry

Given below are the Class 10 Maths Worksheet for Coordinate Geometry
(a) Multiple choice Questions

## Multiple choice Questions

Question 1
Find the centroid of the triangle XYZ whose vertices are X (3, - 5) Y (- 3, 4) and Z (9, - 2).
(a) (0, 0)
(b) (3, 1)
(c) (2, 3)
(d) (3,-1)
Solution
(d)
Centroid of the triangle is given by
$x=\frac{x_{1}+x_{2}+x_{3}}{3}=\frac{3-3+9}{3}=3$
$y=\frac{y_{1}+y_{2}+y_{3}}{3}=\frac{-5+4-2}{3}=-1$

Question 2
The area of the triangle ABC with coordinates as A (1, 2) B (2, 5) and C (- 2, - 5)
(a)-1
(b) .4
(c)2
(d) 1
Solution
(d)
$A=\frac{1}{2}[1(5+5)+2(-5-2)-2(2-5)]=1$

Question 3
Find the value of p for which these point are collinear  (7,-2) , (5,1) ,(3,p)?
(a) 2
(b) 4
(c) 3
(d) None of these
Solution
a
For these points to be collinear
A=0
Or
$\frac{1}{2}[7(1-p)+5(p+2)+3(-2-1)]=0$
7-7p+5p+10-9=0
p=2

Question 4
Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A (2, - 2) and B (3, 7).
(a) 2:9
(b) 1:9
(c)1:2
(d) 2:3
Solution
(a)
Let the ratio be m: n
Now
Coordinate of the intersection
$x=\frac{3m+3n}{m+n}$
$y=\frac{7m-2n}{m+n}$
Now these points should lie of the line, So
$2(\frac{3m+2n}{m+n})+(\frac{7m-n}{m+n})-4=0$
m:n=2:9

Question 5
If the mid-point of the line segment joining the points A (3, 4) and B (a, 4) is P (x, y) and x + y - 20 = 0,then find the value of a
(a) 0
(b) 1
(c) 40
(d)45
Solution (d)
id point (3+a)/2, 4
Now
(3+a)/2 -4 -20=0
3+a=48
A=45

Question 1
The coordinates of one end point of a diameter of a circle are (4, -1) and the co-ordinates of the centre of the circle are (1, -3). Find the co- ordinates of the other end of the diameter.
Answer :Co- ordinates are (-2, -5)
Question 2
If the mid- point of the line joining (3, 4) and (z, 7) is (x, y) and 2x + 2y + 1 = 0 find the value of z.
Question 3
Show that following points are vertices of a rectangle:
(a) (2 , -2) , ( 8, 4) , ( 5, 7 ) , (- 1, 1)
(b)(-4 , -1) , (-2 , 4) , ( 4, 0 ) , ( 2, 3 )
Question 4
If centre of circle passing through (a,-8), (b,-9) and (2,1) is (2,-4), find the value of a and b.
Question 5
Prove that (4, 3), (6, 4), (5, 6) and (3, 5) are the angular points of a square.

Question 1
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3)

Let the points be A(6, -6), B(3, -7) and C(3, 3)
Let the coordinates of the circumcentre of the triangle be O(x, y).
Hence
$OA= \sqrt {(x-6)^2 +(y+6)^2}$
$OB= \sqrt {(x-3)^2 +(y+7)^2}$
$OC= \sqrt {(x-3)^2 +(y+3)^2}$
Now
(x, y) will the equidistant from the vertices of the triangle as it is the circumcenter
OA=OB=OC
Taking OA=OB
$\sqrt {(x-6)^2 +(y+6)^2}=\sqrt {(x-3)^2 +(y+7)^2}$
Squaring both the side and cutting coming terms
3x + y = 7 --(A)
Taking OA=OC
$\sqrt {(x-6)^2 +(y+6)^2}=\sqrt {(x-3)^2 +(y+3)^2}$
Squaring both the side and cutting coming terms
x -3y=9 --(B)
Solving (A) and (B), we get
x=3 and y=-2
Therefore (3,-2) is the coordinate of circumcenter of the triangle

Question 2
Two opposite vertices of a square are (-1, 2) and (3, 2). Find thee co-ordinates of other two vertices.

Let ABCD is the square and A (-1,2) and C(3,2). Lets us assume the coordinate of B(x,y)
Now Sides are equal in square,Therefore
AB=BC
$\sqrt {(x+1)^2 +(y-2)^2}=\sqrt {(x-3)^2 +(y-2)^2}$
solving this
$x=1$ -(1)
Now
Diagonal= AC = $\sqrt {(-1-3)^2 +(2-2)^2}= 4$
Now $\text{Diagonal} = \text{side} \times \sqrt 2$
Therefore
$\text{side}= 2\sqrt {2}$
$\text{side}=AB= \sqrt {(x+1)^2 +(y-2)^2}$
or
$8 = (x+1)^2 +(y-2)^2$
Putting the value of x from (1), we get
$[(1+1)^2 +(y-2)^2]=8$
Solving it ,we get
$y^2 -4y=0$
Solving this quadratic equation ,we get
y=0 or 4
So coordinates of the other vertices are (1, 0) and (1, 4)

Question 3
Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1), taken in order, form a rhombus. Also, find its area.

The given points are P(3, 0), Q(4, 5), R(-1, 4) and S(-2, -1).We have
$PQ= \sqrt {(3-4)^2 +(0-5)^2}=\sqrt {26}$
$QR= \sqrt {(4+1)^2 +(5-4)^2} =\sqrt {26}$
$RS= \sqrt {(-1+2)^2 +(4+1)^2} =\sqrt {26}$
$SP= \sqrt {(-2-3)^2 +(-1-0)^2} =\sqrt {26}$
$PR= \sqrt {(3+1)^2 +(0-4)^2} = 4 \sqrt 2$
$QS= \sqrt {(4+2)^2 +(5+1)^2}=6 \sqrt 2$

Therefore Sides are equal
PQ=QR=RS=SP= $\sqrt {26}$
and Diagonal are not equal
$PR \ne QS$
Hence PQRS is a rhombus but not a square.
Now Area of rhombus PQRS is given by
$A= \frac {1}{2} D_1 \times D_2$
Where $D_1$ and $D_2$ are diagonals
So
$A=\frac {1}{2} 4 \sqrt 2 \times 6 \sqrt 2$
=24sq.units

Question 4
Find a point on y - axis which is equidistant from the points (5, -2) and (-3, 2).

Let the point on y axis is P(0,y) which is equidistant from the points A(5,-2) and B(-3,2). Then
PA=PB
$\sqrt {(0-5)^2 + (y+2)^2 } = \sqrt {(0+3)^2 + (y-2)^2 }$
Squaring both the sides
$(0-5)^2 + (y+2)^2=(0+3)^2 + (y-2)^2$
$25 + y^2 + 4 + 4y= 9 + y^2 +4 -4y$
$8y= 9-25$
y=-2

Question 5
Find a relation between a and b such that the point (a, b) is equidistant from the points (3, 6) and (-3, 4).

$\sqrt {(a-3)^2 + (b-6)^2 } = \sqrt {(a+3)^2 + (b-4)^2 }$
Squaring both the sides
$(a-3)^2 + (b-6)^2=(a+3)^2 + (b-4)^2$
$a^2 + 9 -6a+ b^2 +36 -12b= a^2 + 9 +6a+ b^2 + 16 -8b$
$12a + 4b=20$
3a + b = 5

Question 6
If the coordinates of the mid points of the sides of a triangle are (1, 2), (0, -1) and (2, -1). Find the coordinates of its vertices.

let ($x_1$,$y_1$) ,($x_2$,$y_2$) and ($x_3$,$y_3$) be the vertices,Then
($\frac {x_1+x_2}{2}$,$\frac {y_1+y_2}{2}$) = (1, 2)
($\frac {x_2+x_3}{2}$,$\frac {y_2+y_3}{2}$) = (0, -1)
($\frac {x_1+x_3}{2}$,$\frac {y_1+y_3}{2}$) = (2, -1)
So we get
$x_1+x_2=2$ -(a)
$x_2+x_3=0$ -(b)
$x_1+x_3=4$ -(c)
Subtracting (b) from (c)
$x_1 -x_2=4$
$2x_1 = 4$ or $x_1=3$
So $x_2= -1$ and $x_3=1$
Similary
$y_1+y_2=4$ -(p)
$y_2+y_3=-2$ -(q)
$y_1+y_3=-2$ -(r)
Solving same way, we get
y_1=2, y_2 =2, y_3=-4
So Vertices are
(1, -4), (3, 2), (-1, 2)

Question 7
Find the ratio in which the line segment joining (-2, -3) and (5, 6) is divided by (i) x - axis (ii) y -axis. Also, find the co- ordinates of the point of division in each case.

(i) 1 : 2; [1/3,0]
(ii) 2 : 5; [ 0,-3/7]

Question 8
Three vertices of a parallelogram are (a +b, a - b), (2a + b, 2a - b), (a - b, a + b).Find the fourth vertex.

(-b, b)

Question 9
If two vertices of a parallelogram are (3, 2), (-1, 0) and the diagonals cut at (2, -5), find the other vertices of the parallelogram.

(1, -12), (5, -10)

Question 10
Find the lengths of the medians of a ΔABC having vertices at A (0, -1), B (2, 1) and C (0, 3).

AD = 10 units, BE = 2units, CF = 10 units

Question 11
Find the lengths of the medians of a Δ A (5, 1), B (1, 5) and C (-3, -1).

AD = 37 units, BE = 5 units, CF = 2 13 units

Question 12
Find the co- ordinates of the points which divide the line segment joining the points (-4, 0) and (0, 6) in four equal parts.

Let the Points A and B and three points which divide them into four equal parts are P,Q and R
Now We can find coordinates using Mid point
Q divides A and B in 1:1
So Q is (-2, 3)
P divides A and Q in 1:1
So P is (-3, 1.5)
R divides Q and B in 1:1
So R is (-1, 4.5)

Question 13
If A (5, -1), B (-3, -2) and C (-1, 8) are the varieties of triangle ABC, find the length of median through A and the co- ordinates of the centroid.

Median =√65 units, co- ordinate = [1/3,5/3]

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### Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20