CBSE Class 10 Maths Chapter-7 Coordinate Geometry – Free PDF Download
Free PDF download of Important Questions with Answers for CBSE Class 10 Maths Chapter 7 – Coordinate Geometry prepared by expert Maths teachers from latest edition of CBSE(NCERT) books only by CoolGyan to score more marks in CBSE board examination.
CBSE Class 10 Maths Chapter-7 Coordinate Geometry Important Questions
CBSE Class 10 Maths Important Questions Chapter 7 – Coordinate Geometry
2 Mark Questions
1. Find the distance between the point 
Ans. 
2. Determine if the points 
collinear.
Ans. Let 
Hence, A, B and C are not collinear.
3. Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order form a rhombus.
Ans. Let 
Since AB = BC = CD = DA
Hence, ABCD is a rhombus.
4. Show that (4, 4), (3, 5), (-1, 1) are vertices of a right-angled triangle.
Ans. Let A (4, 4), B (3, 5) and C (-1, 1)
AB2 = (3 – 4)2 + (5 – 4)2 = 2
AC2 = (-1 – 4)2 + (5 – 4)2 = 34
BC2 = (-1 – 3)2 + (1 – 5)2=32
Since AC2 = AB2 +BC2
Hence, ABC is a right-angled triangle.
5. Find the coordinates of the points which divide the line segment joining the points (-2, 0) and (0, 8) in four equal parts.
Ans. Q is the mid-point of AB
Coordinate of 
= (-1, 4)
Coordinate of 
Coordinate of 
6. Find the area of the rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
Ans. Let A (3, 0), B (4, 5), C (-1, 4) and D (-2, -1)
Area of rhombus = 
7. If the coordinates A and B are (-2, -2) and (2, -4) respectively. Find the coordinates of P such that AP = 
and P lies on the line segment AB.
Ans. Coordinate of P are
8. In what ratio is the line segment joining the points (-2, 3) and (3, 7) divided by the y-axis?
Ans. Let A (-2, -3) and B (3, 7)
P (0, y) and ratio be K:1
Coordinate of P are 
9. For what value of P are the points (2, 1) (p, -1) and (-1, 3) collinear?
Ans. For collinear
10. Find the third vertex of a 
, if two of its vertices are at (1, 2) and (3, 5) and the centroid is at the origin.
Ans. Let third vertex of the be 
11. In a seating arrangement of desks in a classroom, three students are seated at 
and 
respectively. Are they seated in line?
Ans. 
Hence, they seated in a line.
12. Show that 
are the vertices of an equilateral triangle.
Ans. Let 
Since AB = BC = CA, then 
is equilateral triangle.
13. Find the distance between the points (0, 0) and (36, 15).Also, find the distance between towns A and B if town B is located at 36 km east and15 km north of town A.
Ans. Applying Distance Formula to find distance between points (0, 0) and (36, 15), we get
d = 
Town B is located at 36 km east and 15 km north of town A. So, the location of town A and B can be shown as:
Clearly, the coordinates of point A are (0, 0) and coordinates of point B are (36, 15).
To find the distance between them, we use Distance formula:
d = 
14. Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
Ans. Let A = (1, 5), B = (2, 3) and C = (–2, –11)
Using Distance Formula to find distance AB, BC and CA.
AB = 
BC = 
CA = 
Since AB + AC ≠ BC, BC + AC 
AB and AC BC.
Therefore, the points A, B and C are not collinear.
15. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
Ans. Let A = (5, –2), B = (6, 4) and C = (7, –2)
Using Distance Formula to find distances AB, BC and CA.
AB = 
BC = 
CA = 
Since AB = BC.
Therefore, A, B and C are vertices of an isosceles triangle.
16. Find the values of y for which the distance between the points P (2, –3) and Q (10, y) is 10 units.
Ans. Using Distance formula, we have
⇒ 
⇒ 
Squaring both sides, we get
100=73+y2+6y
⇒ y2+6y – 27=0
Solving this Quadratic equation by factorization, we can write
⇒ y2+9y−3y – 27=0
⇒ y(y+9) – 3(y+9)=0
⇒ (y+9)(y−3)=0
⇒ y=3,−9
17. Find a relation between x and y such that the point (x, y)is equidistant from the point (3, 6) and (–3, 4).
Ans. It is given that (x,y) is equidistant from (3, 6) and (–3, 4).
Using Distance formula, we can write
⇒ 
Squaring both sides, we get
⇒ x2+9−6x+y2+36−12y
=x2+9+6x+y2+16−8y
⇒ −6x−12y+45
=6x−8y+25
⇒ 12x+4y=20
⇒ 3x+y=5
18. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2:3.
Ans. Let x1=−1, x2=4, y1=7 and y2=−3, m1=2 and m2=3
Using Section Formula to find coordinates of point which divides join of (–1, 7) and (4, –3) in the ratio 2:3, we get
Therefore, the coordinates of point are (1,3) which divides join of (–1, 7) and (4, –3) in the ratio 2:3.
19. Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).
Ans. Let (–1, 6) divides line segment joining the points (–3, 10) and (6, –8) in k:1.
Using Section formula, we get
⇒ −k – 1=(−3+6k)
⇒ −7k=−2
⇒ k=
Therefore, the ratio is :1 which is equivalent to 2:7.
Therefore, (–1, 6) divides line segment joining the points (–3, 10) and (6, –8) in 2:7.
20. Find the ratio in which the line segment joining A(1, –5) and B(–4, 5) is divided by the x–axis. Also find the coordinates of the point of division.
Ans. Let the coordinates of point of division be (x, 0) and suppose it divides line segment joining A(1, –5) and B(–4, 5) in k:1.
According to Section formula, we get
… (1)
⇒ 5=5k
⇒ k=1
Putting value of k in (1), we get
Therefore, point (
, 0) on x–axis divides line segment joining A(1, –5) and B(–4, 5) in 1:1.
21. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Ans. Let A = (1,2), B = (4,y), C= (x, 6) and D= (3, 5)
We know that diagonals of parallelogram bisect each other. It means that coordinates of midpoint of diagonal AC would be same as coordinates of midpoint of diagonal BD. … (1)
Using Section formula, the coordinates of midpoint of AC are:
Using Section formula, the coordinates of midpoint of BD are:
According to condition (1), we have
⇒ (1+x)=7
⇒ x=6
Again, according to condition (1), we also have
⇒ 8=5+y
⇒ y=3
Therefore, x=6 and y=3
22. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4).
Ans. We want to find coordinates of point A. AB is the diameter and coordinates of center are (2, –3) and, coordinates of point B are (1, 4).
Let coordinates of point A are (x, y). Using section formula, we get
⇒ 4=x+1
⇒ x=3
Using section formula, we get
⇒ −6=4+y
⇒ y=−10
Therefore, Coordinates of point A are (3, –10).
23. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (–1, 4) and (–2, –1) taken in order. {Hint: Area of a rhombus = ½ (product of its diagonals)}
Ans. Let A = (3, 0), B = (4, 5), C =(–1, 4) and D = (–2, –1)
Using Distance Formula to find length of diagonal AC, we get
Using Distance Formula to find length of diagonal BD, we get
Area of rhombus = ½ (product of its diagonals)
=
= 24 sq. units
24. Find the area of the triangle whose vertices are:
(i) (2, 3), (–1, 0), (2, –4)
(ii) (–5, –1), (3, –5), (5, 2)
Ans. (i) (2, 3), (–1, 0), (2, –4)
Area of Triangle = 
= ½ [2 {0−(−4)} – 1(−4−3)+2(3−0)]
= ½ [2(0+4) – 1(−7)+2(3)] = ½ (8+7+6)=
sq. units
(ii) (–5, –1), (3, –5), (5, 2)
Area of Triangle =
=½[−5(−5−2)+3 {2−(−1)} +5 {−1−(−5)}]
= ½ [−5(−7)+3(3)+5(4)]
= ½ (35+9+20)
= ½ (64)
=32 sq. units