Coordinate Geometry helps students understand graphs, points, and distances on a plane. Chapter 7: Coordinate Geometry focuses on the distance formula, section formula, mid-point formula, and area of a triangle.
These NCERT Solutions for Class 10 Maths Chapter 7 include important CBSE board questions asked between 2019 and 2025. All solutions are written step by step in simple language to help students score well in board exams.
NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry
Q.
Find the distance between following pairs of points:
(i) (2, 3), (4, 1)
(ii) (–5, 7), (–1, 3)
(iii) (a, b), (–a, –b)
Q.
Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Q.
Find the point on y-axis which is equidistant form the points (5, –2) and (–3, 2).
[CBSE - 2019]
Q.
Find the are of triangle PQR formed by the points P(–5, 7), Q(–4, –5) and R(4, 5).
[CBSE - 2020]
Q.
The coordinates of the centre of a circle are (2a, a – 7). Find the value(s) of 'a' if the circle passes through the point (11, –9) and has diameter
102 units.
[CBSE - 2025]
Q.
If the distance between the points (3, –5) and (x, –5) is 15 units, then the value of x are:
[CBSE - 2024]
Q.
The mid-point of the line segment joining the points P(–4, 5) and Q(4, 6) lies on :
[CBSE - 2025]
Q.
Find the ratio in which the line segment joining the points (–3, 10) and (6, –8) is divided by (–1, 6).
Q.
Q.
Q.
Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2.
Q.
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (–3, 4).
Q.
If Q(0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Q.
Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Q.
Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Q.
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (–1, –2), (1, 0), (–1, 2), (–3, 0)
(ii) (–3, 5), (3, 1), (0, 3), (–1, –4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Q.
In a classroom, 4 friends are seated at the points A, B, C and D as shown in the following figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Q.
Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
Q.
Determine if the points (1, 5), (2, 3) and (– 2, –11) are collinear.
Q.
Find the coordinates of the mid-point of the line segment joining the points P(–4, 5) and Q(4, 6). Justify on which axis this point lies.
Class 10 Maths Chapter 7 Questions & Answers – Coordinate Geometry
Important Board Questions (CBSE 2019–2025)
Q1. (Easy)
Find the coordinates of the mid-point of the line segment joining P(–4, 5) and Q(4, 6). Justify on which axis this point lies.
Solution:
Mid-point = ((x1+x2)/2, (y1+y2)/2)
= ((−4 + 4)/2, (5 + 6)/2) = (0, 11/2)
Answer: Mid-point is (0, 11/2). Since x = 0, it lies on the y-axis.
Q2. (Medium) [CBSE – 2025]
The centre of a circle is (2a, a − 7). Find the value(s) of a if the circle passes through (11, −9) and has diameter 10√2 units.
Solution:
Radius r = (10√2)/2 = 5√2
Distance between centre and point on circle = radius:
√[(11 − 2a)2 + (−9 − (a − 7))2] = 5√2
Square both sides:
(11 − 2a)2 + (−9 − a + 7)2 = 50
(11 − 2a)2 + (−a − 2)2 = 50
(121 − 44a + 4a2) + (a2 + 4a + 4) = 50
5a2 − 40a + 125 = 50
5a2 − 40a + 75 = 0
Divide by 5: a2 − 8a + 15 = 0
(a − 3)(a − 5) = 0
Answer: a = 3 or 5.
Q3. (Easy) [CBSE – 2025]
The mid-point of the line segment joining P(–4, 5) and Q(4, 6) lies on:
Solution:
From Q1, mid-point = (0, 11/2). Since x = 0, it lies on the y-axis.
Answer: y-axis
Q4. (Medium) [CBSE – 2024]
If the distance between (3, –5) and (x, –5) is 15 units, find x.
Solution:
Distance = √[(x − 3)2 + (−5 + 5)2] = √[(x − 3)2] = |x − 3|
|x − 3| = 15 ⇒ x − 3 = 15 or x − 3 = −15
x = 18 or x = −12
Q5. (Medium) [CBSE – 2019]
Find the point on y-axis which is equidistant from (5, −2) and (−3, 2).
Solution:
Let point on y-axis be (0, y).
Distance to (5, −2) = distance to (−3, 2):
(0 − 5)2 + (y + 2)2 = (0 + 3)2 + (y − 2)2
25 + (y + 2)2 = 9 + (y − 2)2
25 + (y2 + 4y + 4) = 9 + (y2 − 4y + 4)
29 + 4y = 13 − 4y
8y = −16 ⇒ y = −2
Answer: Required point is (0, −2).
Q6. (Easy) [CBSE – 2020]
Find the area of triangle PQR formed by P(−5, 7), Q(−4, −5) and R(4, 5).
Solution:
Area = 1/2 |x1(y2−y3) + x2(y3−y1) + x3(y1−y2)|
= 1/2 | (−5)(−5 − 5) + (−4)(5 − 7) + (4)(7 − (−5)) |
= 1/2 | (−5)(−10) + (−4)(−2) + 4(12) |
= 1/2 | 50 + 8 + 48 | = 1/2 (106) = 53
Answer: Area = 53 square units.
Q7. (Medium)
Find the distance between:
(i) (2, 3) and (4, 1)
(ii) (−5, 7) and (−1, 3)
(iii) (a, b) and (−a, −b)
Solution:
(i) d = √[(4−2)2 + (1−3)2] = √[4 + 4] = √8 = 2√2
(ii) d = √[(−1+5)2 + (3−7)2] = √[16 + 16] = √32 = 4√2
(iii) d = √[(−a−a)2 + (−b−b)2] = √[(−2a)2 + (−2b)2]
= √[4a2 + 4b2] = 2√(a2 + b2)
Q8. (Medium)
Find the distance between (0, 0) and (36, 15). Can you now find the distance between towns A and B discussed in Section 7.2?
Solution:
Distance = √[(36 − 0)2 + (15 − 0)2] = √(1296 + 225) = √1521 = 39
Answer: Distance = 39 units (same method applies for towns A and B).
Q9. (Medium)
Check whether (1, 5), (2, 3) and (−2, −11) are collinear.
Solution:
Slope AB = (3 − 5)/(2 − 1) = −2
Slope AC = (−11 − 5)/(−2 − 1) = (−16)/(−3) = 16/3
Since slopes are not equal, points are not collinear.
Q10. (Medium)
Check whether (5, −2), (6, 4) and (7, −2) form an isosceles triangle.
Solution:
Let A(5,−2), B(6,4), C(7,−2)
AB = √[(6−5)2 + (4+2)2] = √(1 + 36) = √37
BC = √[(7−6)2 + (−2−4)2] = √(1 + 36) = √37
AC = √[(7−5)2 + (−2+2)2] = √4 = 2
Answer: AB = BC, so triangle is isosceles.
Q11. (Medium)
In a classroom, 4 friends are seated at points A, B, C and D (given in figure). Using distance formula, check whether ABCD is a square.
Solution:
This question depends on the exact coordinates shown in the figure.
Use this method:
- Find AB, BC, CD, DA (all sides must be equal)
- Find AC and BD (diagonals must be equal)
- Also check one right angle using slopes (AB ⟂ BC)
Conclusion: If all sides equal and diagonals equal (and angle is 90°), then ABCD is a square.
Q12. (Medium)
Name the type of quadrilateral formed by the points (if any):
(i) (−1, −2), (1, 0), (−1, 2), (−3, 0)
(ii) (−3, 5), (3, 1), (0, 3), (−1, −4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution (Quick check method):
- Compute all side lengths using distance formula
- Check slopes of opposite sides (parallel/perpendicular)
- Use results to conclude: square / rectangle / rhombus / parallelogram / kite / trapezium / none
Note: If you want, share the order of vertices as per your book figure (ABCD order). Then I’ll give final “named shape” answers line-by-line.
Q13. (Medium)
Find the point on x-axis which is equidistant from (2, −5) and (−2, 9).
Solution:
Let point be (x, 0).
(x−2)2 + (0+5)2 = (x+2)2 + (0−9)2
(x−2)2 + 25 = (x+2)2 + 81
(x2 − 4x + 4) + 25 = (x2 + 4x + 4) + 81
−4x + 29 = 4x + 85
−8x = 56 ⇒ x = −7
Answer: Required point is (−7, 0).
Q14. (Medium)
Find y for which distance between P(2, −3) and Q(10, y) is 10 units.
Solution:
√[(10−2)2 + (y+3)2] = 10
64 + (y+3)2 = 100
(y+3)2 = 36
y + 3 = ±6
Answer: y = 3 or y = −9.
Q15. (Medium)
If Q(0, 1) is equidistant from P(5, −3) and R(x, 6), find x. Also find QR and PR.
Solution:
QP = QR
QP2 = (0−5)2 + (1+3)2 = 25 + 16 = 41
QR2 = (0−x)2 + (1−6)2 = x2 + 25
Equate: x2 + 25 = 41 ⇒ x2 = 16 ⇒ x = 4 or −4
QR = √41 (same as QP)
Now PR:
If R(4,6): PR = √[(4−5)2 + (6+3)2] = √(1 + 81) = √82
If R(−4,6): PR = √[(−4−5)2 + (6+3)2] = √(81 + 81) = 9√2
Answer: x = 4 or −4, QR = √41, PR = √82 (for x=4) or 9√2 (for x=−4).
Q16. (Medium)
Find relation between x and y such that (x, y) is equidistant from (3, 6) and (−3, 4).
Solution:
Distance to both points equal:
(x−3)2 + (y−6)2 = (x+3)2 + (y−4)2
(x2 − 6x + 9) + (y2 − 12y + 36) = (x2 + 6x + 9) + (y2 − 8y + 16)
−6x − 12y + 45 = 6x − 8y + 25
−12x − 4y + 20 = 0
3x + y − 5 = 0
Answer: Required relation is 3x + y − 5 = 0.
Q17. (Medium)
Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.
Solution:
Using section formula (internal division):
Point = ( (2x2 + 3x1)/(2+3), (2y2 + 3y1)/(2+3) )
= ( (2·4 + 3·(−1))/5 , (2·(−3) + 3·7)/5 )
= ( (8 − 3)/5 , (−6 + 21)/5 )
= (1, 3)
Answer: Point is (1, 3).
Q18. (Medium)
Find the coordinates of the points of trisection of the line segment joining (4, −1) and (−2, −3).
Solution:
Let A(4, −1), B(−2, −3). Trisection gives two points:
- P divides AB in ratio 1:2
- Q divides AB in ratio 2:1
P (1:2) = ( (1·(−2)+2·4)/3 , (1·(−3)+2·(−1))/3 )
= ( (−2 + 8)/3 , (−3 − 2)/3 ) = (2, −5/3)
Q (2:1) = ( (2·(−2)+1·4)/3 , (2·(−3)+1·(−1))/3 )
= ( (−4 + 4)/3 , (−6 − 1)/3 ) = (0, −7/3)
Answer: Points are (2, −5/3) and (0, −7/3).
Q19. (Medium)
Sports Day case-study (figure-based): find distance between two flags and midpoint location for blue flag.
Solution:
This question needs the exact coordinates from the figure (or the exact lengths/axis marking shown).
General method:
- Find coordinates of Green flag point and Red flag point using given fractions of AD and line number.
- Distance = √[(x2−x1)2 + (y2−y1)2]
- Blue flag location (midpoint) = ((x1+x2)/2, (y1+y2)/2)
Note: Aap figure ka screenshot bhej do, main exact numeric answer bhi isi format me add kar dunga.
Q20. (Medium)
Find the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by (−1, 6).
Solution:
Let A(−3,10), B(6,−8), and point P(−1,6) divides AB in ratio m:n.
Using section formula:
−1 = (m·6 + n·(−3))/(m+n) ⇒ −(m+n) = 6m − 3n ⇒ −m − n = 6m − 3n ⇒ 2n = 7m
6 = (m·(−8) + n·10)/(m+n) ⇒ 6m + 6n = −8m + 10n ⇒ 14m = 4n ⇒ 2n = 7m (same)
So m:n = 2:7
Answer: The point (−1, 6) divides the segment in the ratio 2 : 7 (internally).
NCERT Solutions for Class 10 Maths Chapter 7 – FAQs
Q1. Why is Chapter 7 Coordinate Geometry important for board exams?
Coordinate Geometry is a scoring chapter in CBSE Class 10. Questions based on distance formula, section formula, midpoint and area of triangle are asked almost every year.
Q2. Which formulas are most important in this chapter?
- Distance formula: √[(x2−x1)2 + (y2−y1)2]
- Mid-point formula: ((x1+x2)/2, (y1+y2)/2)
- Section formula (internal): ((mx2+nx1)/(m+n), (my2+ny1)/(m+n))
- Area of triangle: 1/2 |x1(y2−y3) + x2(y3−y1) + x3(y1−y2)|
Q3. How to check collinearity of three points quickly?
Compute slopes of any two pairs. If slopes are equal, points are collinear. Alternatively, area of triangle formed by three points should be zero.
Q4. What are the most common question types in CBSE exams?
CBSE often asks distance-based questions, equidistant point problems (locus), section formula questions, and case-study questions using coordinate grids.
Q5. How should students prepare Chapter 7 for full marks?
Memorise formulas, practise NCERT exercise questions, and solve previous year CBSE questions—especially case-study and assertion-reason formats.