Coordinate Geometry For Class 10 Maths Important Questions
Q 1 – If the centre of a circle is (3, 5) and end points of a diameter are (4, 7) and (2, y), then the value of y is
(a) 3
(b) – 3
(c) 7
(d) 4
(a) 3
Q 2 – Find the coordinates of the point equidistant from the points A (1, 2), B (3, – 4)and C (5, – 6).
(a) (2, 3)
(b) (11, 2)
(c) (0, 3)
(d) (1, 31
(b) (11,2)
Q 3 – If the point P (6, 2) divides the line segment joining A (6, 5) and B (4, y) in the ratio 3 : 1 then the value of y is
(a) 4
(b) 3
(c) 2
(d) 1
(d) 1
Q 4 – x – axis divides the line segment joining A (2 – 3) and B (5,6) in the ratio
(a) 2 : 3
(b) 3 : 5
(c) 1 : 2
(d) 2 : 1
(c) 1 : 2
Q 5 – The distance of the point (– 12, 5) from the origin is
(a) 12
(b) 5
(c) 13
(d) 169
(c) 13
Q 6 – Find the value of p for which the points (–5, 1), (1, p) and (4, –2) are collinear.
(a) –3
(b) –2
(c) 0
(d) –1
(d) –1
Q 7 – If the distance between the points A (2, –2) and B (–1, x) is equal to 5, then the value of x is:
(a) 2
(b) –2
(c) 1
(d) –1
(a) 2
Q 8 – The distance between the points P (0, 2) and Q (6, 0) is.
(a) 4√10
(b) 2√10
(c) √10
(d) 20
(b) 2√10
Q 9 – Assertion: The value of y is 6, for which the distance between the points P (2, –3) and Q(10, y) is 10.
Reason: Distance between two given points A and B is given, AB = 
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
Q 10 – If the point C (k, 4) divides the line segment joining two points A (2, 6) and B ( 5, 1) in ratio 2 : 3, the value of k is
. ⇒ If P(x, y) is the dividing point of the line joining AB then By Section Formula we have,
⇒ 
and 
where m and n is the ratio in which the point C divides the line AB
Finding the value of k:
⇒ m = 2 and n = 3
⇒ 
The value of k is 16/5.
Q 11 – What is the distance of point P ( 3, 4) from x -axis?
Q 12 – Find the distance of a point P (x , y) from the origin.
Distance between two points = √(( y2 – y1 )2 + ( x2 – x1 )2
Origin ( 0, 0 ) ( x1, y1 )
Point ( x, y ) ( x2, y2)
Distance = √(( y – 0 )2 + ( x – 0 )2
√(x) 2 + (y)2
Hence, the distance of a point P(x,y) from the origin is √(x)2 + (y)2.
Q 13 – Find the coordinates of a point A, where AB is diameter of a circle whose centre is (2, – 3) and B is the point (1, 4)
Given:-
A circle with centre (2,−3) and AB is the diameter of circle with B(1,4).
To find:- Coordinate of point A.
Let (x,y) be the coordinate of A.
Since AB is the diameter of the circle, the centre will be the mid-point of AB.
now, as centre is the mid-point of AB.
x-coordinate of centre
y-coordinate of centre
But given that centre of circle is . Therefore,
x+1/2 = ⇒x=3
y+4/2 ⇒ y=−10
Thus the coordinate of is .
Hence the correct answer is .
Q 14 – If the centre and radius of circle is (3, 4) and 7 units respectively, then what it the position of the point A(5, 8) with respect to circle?
First, we will find the distance between the given point (5,8) and the center of the circle (3,4).
Distance = √(5-3)²+(8-4)²
Distance = √(2)²+(4)²
Distance = √20
Distance = 2√5 units
Now, 7(radius)>2√5
So, the point (5,8) will lie inside the circle because the radius of the circle is greater than the distance between the point and the center.
Hence, point (5,8) lies inside the circle.
Q 15 – In Fig., find the area of triangle ABC (in sq. units)?
. The area of a triangle is given as:
Area =1/2 [x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
the given points are A(1,3), B(−1,0)and C(4,0)
by Substituting ,
Area=1/2[1(0−0)+(−1)(0−3)+4(3−0)]
=1/2 [3+12]=215=7.5
∴ Area of the given triangle ABC is 7.5 sq. units
Q 16 – If the point (0, 0) (1, 2) and (x, y) are collinear, then find x.
Q 17 – If the points A (0, 1) B (6, 3) and C (x, 5) are the vertices of a triangle, find the value of x such that area of ABC = 10
Q 18 – Locate a point Q on line segment AB such that BQ = 5/7 AB. What is the ratio of line segment in which AB is divided?
Here BQ = (5/7) AB
Q 19 – Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
A(1,5),B(2,3)and C(−2,−11)
Q 20 – Find the ratio in which the point P 
divides the line segment joining the point A 
and B (2, – 5) Mathematics.
Using section formula
The point which divide the line segment
joining the points A ( x1 , y1 ) , B ( x2 , y2 ) in the ratio k:1
is P ( kx2+ x1/k +1 , ky2 + y1 / k+ 1 )
______________________________________________
According to the given problem ,
A( x1 , y1 ) = ( 1/2 , 3 /2 )
B ( x2 , y2 ) = ( 2 , -5 )
P ( x , y ) = ( 3 / 4 , 5 / 12 )
Let the ratio = k : 1
x = 3/4 ( given
(kx2 + x1 ) / ( k+ 1 ) = x
( k× 2 + 1/2 ) / ( k + 1 ) = 3 / 4
2k + 1/2 = 3 /4 ( k + 1 )
4 ( 2k + 1 / 2) = 3 ( k + 1 )
8k + 2 = 3k + 3
8k – 3 k = 3 – 2
5k = 1
k = 1/5
Therefore ,
Required ratio = k : 1
= 1 /5 : 1
= 1 : 5
P divides the line segment joining the piints A and B
in the ratio 1 : 5
Q 21 – Find the point on the x-axis which is equidistant from (2 – 5) and (– 2, 9).
. Let point on axis be (x,0) to be equidistant from (2,−5) and (−2,9) ∴PA=PB Using the distance formula we get,
Q 22 – Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4)
Squaring both sides
⇒ (x−2)2 + 25 = (x+2)2 + 81
⇒ x2− 4x + 4−56 = x2 + 4x + 4
⇒ 8x = − 56
It gives x= −7
So, point equidistant from 2,−5) and −2,9) and lying on axis is −7,0).
Squaring both sides
⇒ (x−2)2 + 25 = (x+2)2 + 81
⇒ x2− 4x + 4−56 = x2 + 4x + 4
⇒ 8x = − 56
It gives x= −7
So, point equidistant from 2,−5) and −2,9) and lying on axis is −7,0).
Squaring both sides
⇒ (x−2)2 + 25 = (x+2)2 + 81
⇒ x2− 4x + 4−56 = x2 + 4x + 4
⇒ 8x = − 56
It gives x= −7
So, point equidistant from 2,−5) and −2,9) and lying on axis is −7,0).
Q 22 – Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4)
Q 23 – If (1,P/3) is the mid point of the line segment joining the points (2, 0) and (0,2/9) then show that the line 5x + 3y + 2 = 0 passes through the point (–1, 3p).
We need to show that line 5x+3y+2=0 passes through point (−1,3p)
Point =(−1,3p)=(−1,3×31)=(−1,1)
Putting (−1,1) in line
5x+3y+2=0
⇒5(−1)+3(1)+2=0
⇒−5+3+2=0
⇒−5+5=0
⇒0=0
Hence, passes through
Q 24 – Find a point on the y-axis equidistant from (–5, 2) and (9, –2).
Let us consider the point on the y-axis to be (0, y) as the x-coordinate is 0 on the y-axis.
So,the point (0, y) is equidistant from (- 5, 2) and (9, -2).
Since their distance is equal we can apply the distance formula.
Applying distance formula,
d = √[(x2 − x1)2+(y2 − y1)2]
= √[{0 – (- 5)}2 + (y – 2)2] = √[(9 – 0)2 + ( -2 – y)2]
Now simplifying step by step to find the value of y.
25 + y2 + 4 – 4y = 81 + 4 + y2 + 4y
29 – 4y = 85 + 4y
8y = -56
y = -56 / 8 = -7
The value of y is -7.
Q 25 – Points P (5, – 3) is one of the two points of trisection of the line segment joining points A (7, – 2) and B (1, – 5) near to A. find the coordinates of the other point of trisection.
Using the section formula, if a point (x,y) divides the line joining the point
(x1,y1) and (x2,y2) in the ratio m:n, then ‘
Also, we know that the points of trisection divide the line segment in ratio 2:1
and 1:2
Therefore, the points of trisection of line segment AB are given by
=(3,−4) and (5,−3)
Hence, the given statement is true.
Q 26 – Points P, Q, R and S divide a line segment joining A (2, 6) and B (7, – 4) in five equal parts. Find the coordinates of P and R.
P, Q, R, S divide the line segment AB on the ratio 1:4,2:3,3:2,4:1 respectively.
From above P divides AB in 1:4 ratio
coordinates of P using section formula,
here = 1 and = 4,x1 = 1,x2= 6,y1=2,y2=7
⇒ 
,
⇒ coordinates of P
Similarly,
Coordinates of Q,
Q divide AB in 2:3 ratio,
⇒ 3,4)= coordinates of Q
Similarly,
R divide AB in 3:2 ratio,
Coordinates of R,
⇒ 4,5) = coordinates of R
Q 27 – Find the area of the triangle ABC with A (1, – 4) and mid-points of sides through A being (2, – 1) and (0, – 1).
A(1,-4)
Considering,
AB (2,−1)
AC (0,−1)
Let B (x1,y1)
= (2,-1)
x1= -13 y1 = 2
B (3,2)
Let C(x2,y2)
= (0, -1)
x2 = −1,y2 =2
C (−1, 2)
Area = 1/2 [x1 (y2 – y3) + x2 (y3 – y1) + x3(y1 – y2)]
= 1/2 [3 (2 + 34) + (-1)( – 4 – 2) + 1(2 – 2)]
= [18 + 6 + 0]
= 24/2
Area of triangle 12 sq. units
Q 28 – If (a, b) is the mid-point of the segment joining the points A( 10, – 6) and B( k , 4) and a – 2b = 8, find the value of k and the distance AB.