Given below are the Class 10 Maths Worksheet for Coordinate Geometry
(a) Multiple choice Questions
(b) Short Answer type
(c) Long answer 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
Short Answer type
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.
Answer: z =-15 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.
Long Answer Type
Question 1
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3) Solution
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. Solution
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. Solution
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). Solution
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). Solution
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. Solution
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)
$x_1 -x_2=4$
Adding the above with (a)
$2x_1 = 4$ 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. Solution
(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. Solution
(-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. Solution
(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). Solution
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). Solution
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. Solution
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. Solution
Median =√65 units, co- ordinate = [1/3,5/3]
link to this page by copying the following text Also Read