Chapter 7 · Written-answer practice

Coordinate Geometry 2-Mark Questions Class 10: CBSE PYQs with Answers

Practise 63 selected Coordinate Geometry questions worth 2 marks each, from CBSE Class 10 Basic and Standard board papers, and official sample papers. Use distance, midpoint and section formulae to find coordinates and prove geometric results. Try each question before opening its solution.

63 numbered questions · 102 board-paper appearances + 5 official-sample appearances
84 worked solutions, including both alternatives in 21 OR pairs
By Siddharth Agrawal, Lead Maths Tutor, TutorVertex · Solutions prepared with Team TutorVertex
Updated 4 November 2026

Key results for Coordinate Geometry 2-mark questions

For A(x₁, y₁) and B(x₂, y₂):

Use	Result
Distance	AB = √((x₂ − x₁)² + (y₂ − y₁)²).
Midpoint	M = ( $\frac{(x₁ + x₂)/(}{2)}$ , $\frac{(y₁ + y₂)/(}{2)}$ ).
Internal division	For AP : PB = m : n, P = ( $\frac{(mx₂ + nx₁)/(}{m + n)}$ , $\frac{(my₂ + ny₁)/(}{m + n)}$ ).
Axis intersections	Set y = 0 for the x-axis, or x = 0 for the y-axis.
Equidistant points	PA = PB implies PA² = PB².
Quadrilateral check	Use side lengths and the relevant diagonal or right-angle properties.

Which formula should I use?

Given	Starting point
A length or radius	Use the distance formula, keeping each negative coordinate in brackets.
A median or diameter	Find the midpoint first, then use distance if a length is required.
A division ratio or trisection point	Keep the endpoint order fixed and use the section formula.
Equal distances	Equate squared distances, then simplify.
A shape or collinearity proof	Compare side lengths; use the appropriate geometric converse.

TopicLevelSource

Showing all 63 questions

Questions on this page — Show list

Distance formula, medians and radii

Question 1 · Standard
Question 2 · Standard
Question 3 · Standard
Question 4 · Basic
Question 5 · Standard
Question 6 · Standard
Question 7 · Standard
Question 8 · Basic
Question 9 · Standard
Question 10 · Standard
Question 11 · Basic
Question 12 · Basic
Question 13 · Standard

Section formula, trisection and axis intersections

Question 14 · Basic
Question 15 · Standard
Question 16 · Standard
Question 17 · Standard
Question 18 · Basic
Question 19 · Basic
Question 20 · Basic
Question 21 · Basic
Question 22 · Standard
Question 23 · Basic
Question 24 · Basic
Question 25 · Basic
Question 26 · Standard
Question 27 · Standard
Question 28 · Basic
Question 29 · Basic
Question 30 · Basic
Question 31 · Basic
Question 32 · Standard
Question 33 · Standard
Question 34 · Basic
Question 35 · Basic
Question 36 · Standard
Question 37 · Standard
Question 38 · Standard

Equidistant points and coordinate relations

Question 39 · Basic
Question 40 · Basic
Question 41 · Standard
Question 42 · Basic
Question 43 · Basic
Question 44 · Standard
Question 45 · Basic
Question 46 · Standard
Question 47 · Standard
Question 48 · Basic
Question 49 · Basic

Triangle types, collinearity and quadrilateral proofs

Question 50 · Standard
Question 51 · Basic
Question 52 · Standard
Question 53 · Standard
Question 54 · Standard
Question 55 · Basic
Question 56 · Standard
Question 57 · Standard
Question 58 · Basic
Question 59 · Basic
Question 60 · Standard
Question 61 · Standard
Question 62 · Standard
Question 63 · Standard

Distance formula, medians and radii

Standard2 marks

Q1. The vertices of a △ABC are A(−2, 4), B(4, 3) and C(1, −6). Find length of the median BD.

Appeared in: 2024 · Standard · Main

Show answer and solution

D is the midpoint of AC, so D = ( $\frac{(−2 + 1)/(}{2)}$ , $\frac{(4 − 6)/(}{2)}$ ) = ( $\frac{(−1)/(}{2)}$ , −1).

BD = √((4 + $\frac{(1)/(}{2)}$ )² + (3 + 1)²).

BD = √( $\frac{(81)/(}{4)}$ + 16) = √( $\frac{(145)/(}{4)}$ ).

BD = $\frac{(√145)/(}{2)}$ units

Standard2 marks

Q2. 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 10√2 units.

Appeared in: 2025 · Standard · Main

Show answer and solution

Radius = $\frac{(10√2)/(}{2)}$ = 5√2.

By the distance formula, (2a − 11)² + (a + 2)² = 50.

5a² − 40a + 75 = 0.

a² − 8a + 15 = 0, so (a − 3)(a − 5) = 0.

a = 3 or 5

Standard2 marks

Q3. Find the length of the median through the vertex B of △ABC with vertices A(9, −2), B(−3, 7) and C(−1, 10).

Appeared in: 2025 · Standard · Main

Show answer and solution

Let M be the midpoint of AC.

M = ( $\frac{(9 − 1)/(}{2)}$ , $\frac{(−2 + 10)/(}{2)}$ ) = (4, 4).

BM = √((4 + 3)² + (4 − 7)²) = √(49 + 9).

Median BM = √58 units

Basic2 marks

Q4. (a) Find the values of a for which the distance between the points P(1, 2) and Q(−2, a) is 5 units.

Show answer and solution

PQ² = (−2 − 1)² + (a − 2)² = 25.

(a − 2)² = 16, so a − 2 = ±4.

a = 6 or −2

(b) Find the coordinates of the point which divides the join of (4, −3) and (8, −5) in the ratio 3 : 1.

Appeared in: 2025 · Basic · Supplementary

Show answer and solution

By the section formula, P = ( $\frac{(3(8) + 1(4))/(}{4)}$ , $\frac{(3(−5) + 1(−3))/(}{4)}$ ).

P = ( $\frac{(28)/(}{4)}$ , $\frac{(−18)/(}{4)}$ ).

P = (7, $\frac{(−9)/(}{2)}$ )

Standard2 marks

Q5. The coordinates of the centre of a circle are (x − 7, 2x). Find the value(s) of x, if the circle passes through the point (−9, 11) and has radius 5√2 units.

Appeared in: 2026 · Standard · Main

Show answer and solution

By the distance formula, (x − 7 + 9)² + (2x − 11)² = (5√2)².

(x + 2)² + (2x − 11)² = 50.

5x² − 40x + 75 = 0.

(x − 3)(x − 5) = 0.

x = 3 or 5

Standard2 marks

Q6. (a) Vertices of a right triangle ABC with ∠B = 90° are A(3, 4), B(1, 1) and C(−8, 7). Find the value of tan A.

Show answer and solution

AB = √((3 − 1)² + (4 − 1)²) = √13.

BC = √((−8 − 1)² + (7 − 1)²) = √117 = 3√13.

tan A = $\frac{(BC)/(}{AB)}$ = $\frac{(3√13)/(}{√13)}$ .

tan A = 3

(b) Using distance formula, prove that the points A(2, 3), B(−7, 0) and C(−1, 2) are collinear.

Appeared in: 2026 · Standard · Main

The full printed OR pair is retained; alternative (a) also uses trigonometry.

Show answer and solution

AC = √((−1 − 2)² + (2 − 3)²) = √10.

CB = √((−7 + 1)² + (0 − 2)²) = 2√10.

AB = √((−7 − 2)² + (0 − 3)²) = 3√10.

AC + CB = AB, so C lies on AB.

A, B and C are collinear.

Standard2 marks

Q7. Find the point(s) on the x-axis which is at a distance of √41 units from the point (8, −5).

Appeared in: 2024–25 · Standard · Official sample

Show answer and solution

Let P = (x, 0). By the distance formula, (x − 8)² + (0 + 5)² = 41.

(x − 8)² = 16, so x − 8 = ±4.

(4, 0) and (12, 0)

Basic2 marks

Q8. Find the radius of the circle whose end points of a diameter are (0, 0) and (6, 8).

Appeared in: 2025–26 · Basic · Official sample

Accessible alternative to the tangent-and-coordinate diagram question in the same sample.

Show answer and solution

Diameter = √((6 − 0)² + (8 − 0)²) = 10.

Radius = $\frac{(diameter)/(}{2)}$ = $\frac{(10)/(}{2)}$ .

Radius = 5 units

Standard2 marks

Q9. Find the points on the x-axis, each of which is at a distance of 10 units from the point A(11, −8).

Appeared in: 2023 · Standard · Main

Show answer and solution

Let P = (x, 0).

(x − 11)² + (0 + 8)² = 100.

(x − 11)² = 36, so x − 11 = ±6.

(5, 0) and (17, 0)

Standard2 marks

Q10. If the distance between the points (4, p) and (1, 0) is 5, what is the value of p?

Appeared in: 2026 · Standard · Main

Show answer and solution

By the distance formula, (4 − 1)² + (p − 0)² = 5².

9 + p² = 25, hence p² = 16.

p = ±4

Basic2 marks

Q11. A(−2, 3) and B(4, 1) are end points of the diameter of a semi-circle. The semi-circle intersects y-axis at point P. Find the co-ordinates of the point P.

Appeared in: 2026 · Basic · Main

Show answer and solution

The centre is the midpoint O = ( $\frac{(−2 + 4)/(}{2)}$ , $\frac{(3 + 1)/(}{2)}$ ) = (1, 2).

Radius² = OB² = (4 − 1)² + (1 − 2)² = 10.

Let P = (0, y). Then OP² = 1 + (y − 2)² = 10.

(y − 2)² = 9, so y = 5 or −1.

The printed question does not specify which half-circle; the marking scheme lists both possible intersections.

P = (0, 5) or (0, −1)

Basic2 marks

Q12. In the given figure, PQ is a tangent to a circle with centre O(−5, 3). If coordinates of P and Q are (3, 1) and (0, 6) respectively, then using distance formula, show that PQ ⟂ OQ.

Circle with centre O(−5, 3), tangent PQ, contact point Q(0, 6), and external point P(3, 1). — CBSE question diagram.

Appeared in: 2026 · Basic · Main

Show answer and solution

PQ² = (3 − 0)² + (1 − 6)² = 34.

OQ² = (−5 − 0)² + (3 − 6)² = 34.

OP² = (3 + 5)² + (1 − 3)² = 68.

PQ² + OQ² = OP², so ∠PQO = 90° by the converse of Pythagoras theorem.

PQ ⟂ OQ

Standard2 marks

Q13. If AOBC is rectangle whose vertices are A(0, 3), O(0, 0), B(5, 0) and C(5, 3), then find the length of its diagonal.

Appeared in: 2026 · Standard · Second board examination

Show answer and solution

A diagonal joins A to B.

AB = √((5 − 0)² + (0 − 3)²) = √(25 + 9).

Diagonal = √34 units

Section formula, trisection and axis intersections

Basic2 marks

Q14. (a) Find the coordinates of the point which divides the line segment joining the points (7, −1) and (−3, 4) internally in the ratio 2 : 3.

Show answer and solution

For AP : PB = m : n, P = ( $\frac{(mx₂ + nx₁)/(}{m + n)}$ , $\frac{(my₂ + ny₁)/(}{m + n)}$ ).

P = ( $\frac{(2(−3) + 3(7))/(}{5)}$ , $\frac{(2(4) + 3(−1))/(}{5)}$ ).

P = ( $\frac{(15)/(}{5)}$ , $\frac{(5)/(}{5)}$ ).

P = (3, 1)

(b) Find the value(s) of y for which the distance between the points A(3, −1) and B(11, y) is 10 units.

Appeared in: 2023 · Basic · Main

Show answer and solution

AB² = (11 − 3)² + (y + 1)².

100 = 64 + (y + 1)².

y + 1 = ±6.

y = 5 or −7

Standard2 marks

Q15. (a) The line segment joining the points A(4, −5) and B(4, 5) is divided by the point P such that AP : AB = 2 : 5. Find the coordinates of P.

Show answer and solution

AP : PB = 2 : (5 − 2) = 2 : 3.

By the section formula, P = ( $\frac{(2(4) + 3(4))/(}{5)}$ , $\frac{(2(5) + 3(−5))/(}{5)}$ ).

P = ( $\frac{(20)/(}{5)}$ , $\frac{(−5)/(}{5)}$ ).

P = (4, −1)

(b) Point P(x, y) is equidistant from points A(5, 1) and B(1, 5). Prove that x = y.

Appeared in: 2023 · Standard · Main

Show answer and solution

PA = PB, so (x − 5)² + (y − 1)² = (x − 1)² + (y − 5)².

−10x − 2y + 26 = −2x − 10y + 26.

8y = 8x.

x = y

Standard2 marks

Q16. Find the ratio in which y-axis divides the line segment joining the points (5, −6) and (−1, −4).

Appeared in: 2023 · Standard · Main

Show answer and solution

Let AP : PB = m : n. On the y-axis, the x-coordinate is 0.

By the section formula, $\frac{(−m + 5n)/(}{m + n)}$ = 0.

m = 5n.

AP : PB = 5 : 1

Standard2 marks

Q17. Find the ratio in which line y = x divides the line segment joining the points (6, −3) and (1, 6).

Appeared in: 2023 · Standard · Main

Show answer and solution

Let AP : PB = m : n.

By the section formula, P = ( $\frac{(m + 6n)/(}{m + n)}$ , $\frac{(6m − 3n)/(}{m + n)}$ ).

Since y = x, m + 6n = 6m − 3n.

5m = 9n.

AP : PB = 9 : 5

Basic2 marks

Q18. (a) Find the coordinates of the point which divides the join of A(−1, 7) and B(4, −3) in the ratio 2 : 3.

Show answer and solution

By the section formula, P = ( $\frac{(2(4) + 3(−1))/(}{5)}$ , $\frac{(2(−3) + 3(7))/(}{5)}$ ).

P = ( $\frac{(5)/(}{5)}$ , $\frac{(15)/(}{5)}$ ).

P = (1, 3)

(b) If the points A(2, 3), B(−5, 6), C(6, 7) and D(p, 4) are the vertices of a parallelogram ABCD, find the value of p.

Appeared in: 2023 · Basic · Main

Show answer and solution

The diagonals of a parallelogram bisect each other.

Midpoint of AC = ( $\frac{(2 + 6)/(}{2)}$ , $\frac{(3 + 7)/(}{2)}$ ) = (4, 5).

Midpoint of BD = ( $\frac{(p − 5)/(}{2)}$ , $\frac{(6 + 4)/(}{2)}$ ).

$\frac{(p − 5)/(}{2)}$ = 4.

p = 13

Basic2 marks

Q19. Find the ratio in which a line segment joining points (1, −4) and (6, 5) is divided by the x-axis.

Appeared in: 2023 · Basic · Supplementary

Show answer and solution

Let AP : PB = m : n. On the x-axis, y = 0.

By the section formula, $\frac{(5m − 4n)/(}{m + n)}$ = 0.

5m = 4n.

AP : PB = 4 : 5

Basic2 marks

Q20. Find the ratio in which the point P(m, 6) divides the line segment joining A(−4, 3) and B(2, 8). Also, find the value of m.

Appeared in: 2023 · Basic · Supplementary

Show answer and solution

Let AP : PB = k : 1.

6 = $\frac{(8k + 3)/(}{k + 1)}$ , so 6k + 6 = 8k + 3.

k = $\frac{(3)/(}{2)}$ , hence AP : PB = 3 : 2.

m = $\frac{(3(2) + 2(−4))/(}{5)}$ .

AP : PB = 3 : 2; m = $\frac{(−2)/(}{5)}$

Basic2 marks

Q21. Find the ratio in which a line segment joining the points (−1, 1) and (5, 4) is divided by the y-axis.

Appeared in: 2023 · Basic · Supplementary

Show answer and solution

Let AP : PB = m : n. On the y-axis, x = 0.

$\frac{(5m − n)/(}{m + n)}$ = 0.

n = 5m.

AP : PB = 1 : 5

Standard2 marks

Q22. Find the ratio in which the point (−1, k) divides the line segment joining the points (−3, 10) and (6, −8). Hence, find the value of k.

Appeared in: 2023 · Standard · Supplementary

Show answer and solution

Let AP : PB = m : n.

−1 = $\frac{(6m − 3n)/(}{m + n)}$ , so 7m = 2n.

AP : PB = 2 : 7.

k = $\frac{(2(−8) + 7(10))/(}{9)}$ = 6.

Ratio = 2 : 7; k = 6

Basic2 marks

Q23. Find the ratio in which the Y-axis divides the line segment joining the points A(5, −6) and B(−1, −4). Also, find the point of intersection.

Appeared in: 2024 · Basic · Main

Show answer and solution

Let AP : PB = m : n. On the y-axis, x = 0.

$\frac{(−m + 5n)/(}{m + n)}$ = 0, so m : n = 5 : 1.

y = $\frac{(5(−4) + 1(−6))/(}{6)}$ = $\frac{(−13)/(}{3)}$ .

Ratio = 5 : 1; intersection = (0, $\frac{(−13)/(}{3)}$ )

Basic2 marks

Q24. Point P(x, y) divides the line segment joining the points A(−1, 3) and B(9, 8) such that AP : PB = k : 1. If the co-ordinates of P are such that x = y, then find the value of k.

Appeared in: 2024 · Basic · Main

Show answer and solution

By the section formula, x = $\frac{(9k − 1)/(}{k + 1)}$ and y = $\frac{(8k + 3)/(}{k + 1)}$ .

Since x = y, 9k − 1 = 8k + 3.

k = 4

Basic2 marks

Q25. (a) Find the ratio in which the point (3, y), divides the line segment joining the points (−2, −5) and (6, 3). Also, find the value of y.

Show answer and solution

Let AP : PB = m : n.

3 = $\frac{(6m − 2n)/(}{m + n)}$ , so 3m = 5n.

AP : PB = 5 : 3.

y = $\frac{(5(3) + 3(−5))/(}{8)}$ = 0.

Ratio = 5 : 3; y = 0

(b) Find a point on y-axis which is equidistant from the points A(6, 5) and B(−4, 3).

Appeared in: 2024 · Basic · Main

Show answer and solution

Let P = (0, y). Since PA = PB, 36 + (y − 5)² = 16 + (y − 3)².

61 − 10y = 25 − 6y.

4y = 36.

P = (0, 9)

Standard2 marks

Q26. (a) In what ratio is the line segment joining the points (3, −5) and (−1, 6) divided by the line y = x?

Show answer and solution

Let AP : PB = m : n.

P = ( $\frac{(−m + 3n)/(}{m + n)}$ , $\frac{(6m − 5n)/(}{m + n)}$ ).

Since x = y, −m + 3n = 6m − 5n.

7m = 8n.

AP : PB = 8 : 7

(b) A(3, 0), B(6, 4) and C(−1, 3) are vertices of a triangle ABC. Find length of its median BE.

Appeared in: 2024 · Standard · Main

Show answer and solution

E is the midpoint of AC, so E = ( $\frac{(3 − 1)/(}{2)}$ , $\frac{(0 + 3)/(}{2)}$ ) = (1, $\frac{(3)/(}{2)}$ ).

BE = √((6 − 1)² + (4 − $\frac{(3)/(}{2)}$ )²).

BE = √(25 + $\frac{(25)/(}{4)}$ ) = √( $\frac{(125)/(}{4)}$ ).

BE = $\frac{(5√5)/(}{2)}$ units

Standard2 marks

Q27. (a) Find the ratio in which the point P(−4, 6) divides the line segment joining the points A(−6, 10) and B(3, −8).

Show answer and solution

Let AP : PB = m : n.

−4 = $\frac{(3m − 6n)/(}{m + n)}$ , so 7m = 2n.

The y-coordinate check gives $\frac{(2(−8) + 7(10))/(}{9)}$ = 6.

AP : PB = 2 : 7

(b) Prove that the points (3, 0), (6, 4) and (−1, 3) are the vertices of an isosceles triangle.

Appeared in: 2024 · Standard · Main

Show answer and solution

Let the points be A, B and C.

AB² = (6 − 3)² + (4 − 0)² = 25.

AC² = (−1 − 3)² + (3 − 0)² = 25.

AB = AC = 5, hence the triangle is isosceles.

The triangle is isosceles.

Basic2 marks

Q28. Find the coordinates of the points of trisection of line segment joining the points (−4, 1) and (6, 5).

Appeared in: 2025 · Basic · Main

Show answer and solution

The trisection points divide AB in the ratios 1 : 2 and 2 : 1.

P = ( $\frac{(6 + 2(−4))/(}{3)}$ , $\frac{(5 + 2(1))/(}{3)}$ ) = ( $\frac{(−2)/(}{3)}$ , $\frac{(7)/(}{3)}$ ).

Q = ( $\frac{(2(6) − 4)/(}{3)}$ , $\frac{(2(5) + 1)/(}{3)}$ ) = ( $\frac{(8)/(}{3)}$ , $\frac{(11)/(}{3)}$ ).

( $\frac{(−2)/(}{3)}$ , $\frac{(7)/(}{3)}$ ) and ( $\frac{(8)/(}{3)}$ , $\frac{(11)/(}{3)}$ )

Basic2 marks

Q29. Point P(x, 0) divides the line segment joining the points (2, 8) and (−3, −5) in a certain ratio. Find the ratio and hence find the value of x.

Appeared in: 2025 · Basic · Main

Show answer and solution

Let AP : PB = m : n.

0 = $\frac{(−5m + 8n)/(}{m + n)}$ , so m : n = 8 : 5.

x = $\frac{(8(−3) + 5(2))/(}{13)}$ = $\frac{(−14)/(}{13)}$ .

Ratio = 8 : 5; x = $\frac{(−14)/(}{13)}$

Basic2 marks

Q30. Find the ratio in which point P(−1, m) divides the line segment joining the points A(2, 5) and B(−5, −2). Hence, find the value of m.

Appeared in: 2025 · Basic · Main

Show answer and solution

Let AP : PB = k : 1.

−1 = $\frac{(−5k + 2)/(}{k + 1)}$ , so 4k = 3.

AP : PB = 3 : 4.

m = $\frac{(3(−2) + 4(5))/(}{7)}$ = 2.

Ratio = 3 : 4; m = 2

Basic2 marks

Q31. Find the ratio in which the segment joining the points (2, −5) and (5, 3) is divided by x-axis. Also, find coordinates of the point on x-axis.

Appeared in: 2025 · Basic · Main

Show answer and solution

Let AP : PB = m : n.

0 = $\frac{(3m − 5n)/(}{m + n)}$ , so m : n = 5 : 3.

x = $\frac{(5(5) + 3(2))/(}{8)}$ = $\frac{(31)/(}{8)}$ .

Ratio = 5 : 3; point = ( $\frac{(31)/(}{8)}$ , 0)

Standard2 marks

Q32. Find the coordinates of the point C which lies on the line AB produced such that AC = 2BC, where coordinates of points A and B are (−1, 7) and (4, −3) respectively.

Appeared in: 2025 · Standard · Main

Show answer and solution

Since C lies beyond B, AC = AB + BC.

Given AC = 2BC, so AB = BC; B is the midpoint of AC.

4 = $\frac{(−1 + x)/(}{2)}$ and −3 = $\frac{(7 + y)/(}{2)}$ .

x = 9 and y = −13.

C = (9, −13)

Standard2 marks

Q33. The coordinates of the end points of the line segment AB are A(−2, −2) and B(2, −4). P is the point on AB such that BP = $\frac{(4)/(}{7)}$ AB. Find the coordinates of point P.

Appeared in: 2025 · Standard · Main

Show answer and solution

AP = AB − BP = ( $\frac{(3)/(}{7)}$ )AB, so AP : PB = 3 : 4.

P = ( $\frac{(3(2) + 4(−2))/(}{7)}$ , $\frac{(3(−4) + 4(−2))/(}{7)}$ ).

P = ( $\frac{(−2)/(}{7)}$ , $\frac{(−20)/(}{7)}$ ).

P = ( $\frac{(−2)/(}{7)}$ , $\frac{(−20)/(}{7)}$ )

Basic2 marks

Q34. (a) Find the coordinates of the point which divides the join of (−1, 7) and (4, −3) in the ratio 2 : 3.

Same division subquestion as Q18(a).

Show answer and solution

By the section formula, P = ( $\frac{(2(4) + 3(−1))/(}{5)}$ , $\frac{(2(−3) + 3(7))/(}{5)}$ ).

P = (1, 3).

P = (1, 3)

(b) Using distance formula, prove that the points A(3, 1), B(6, 4) and C(8, 6) are collinear.

Appeared in: 2025 · Basic · Supplementary

Show answer and solution

AB = √((6 − 3)² + (4 − 1)²) = 3√2.

BC = √((8 − 6)² + (6 − 4)²) = 2√2.

AC = √((8 − 3)² + (6 − 1)²) = 5√2.

AB + BC = AC, so B lies on AC.

A, B and C are collinear.

Basic2 marks

Q35. The vertices of a △ABC are A(−1, 3), B(2, −3) and C(4, 5). Find the coordinates of a point P on median AD such that AP : PD = 2 : 3.

Appeared in: 2026 · Basic · Main

Show answer and solution

D is the midpoint of BC, so D = ( $\frac{(2 + 4)/(}{2)}$ , $\frac{(−3 + 5)/(}{2)}$ ) = (3, 1).

By the section formula, P = ( $\frac{(2(3) + 3(−1))/(}{5)}$ , $\frac{(2(1) + 3(3))/(}{5)}$ ).

P = ( $\frac{(3)/(}{5)}$ , $\frac{(11)/(}{5)}$ ).

P = ( $\frac{(3)/(}{5)}$ , $\frac{(11)/(}{5)}$ )

Standard2 marks

Q36. Find the coordinates of the point R which lies on the line PQ produced such that PR = 2QR, where coordinates of points P and Q are (7, −1) and (−3, 4) respectively.

Appeared in: 2026 · Standard · Second board examination

Show answer and solution

R lies beyond Q, so PR = PQ + QR.

Given PR = 2QR, hence PQ = QR and Q is the midpoint of PR.

−3 = $\frac{(7 + x)/(}{2)}$ and 4 = $\frac{(−1 + y)/(}{2)}$ .

x = −13 and y = 9.

R = (−13, 9)

Standard2 marks

Q37. Find the coordinates of the point which divides the line segment joining (−1, 3) and (4, −7) internally in the ratio 3 : 4.

Appeared in: 2026 · Standard · Second board examination

Show answer and solution

By the section formula, P = ( $\frac{(3(4) + 4(−1))/(}{7)}$ , $\frac{(3(−7) + 4(3))/(}{7)}$ ).

P = ( $\frac{(8)/(}{7)}$ , $\frac{(−9)/(}{7)}$ ).

P = ( $\frac{(8)/(}{7)}$ , $\frac{(−9)/(}{7)}$ )

Standard2 marks

Q38. Find the co-ordinates of point A, where AB is diameter of the circle whose centre is (2, −3) and B is (1, 4).

Appeared in: 2026 · Standard · Second board examination

Show answer and solution

Let A = (x, y). The centre is the midpoint of AB.

$\frac{(x + 1)/(}{2)}$ = 2 and $\frac{(y + 4)/(}{2)}$ = −3.

x = 3 and y = −10.

A = (3, −10)

Equidistant points and coordinate relations

Basic2 marks

Q39. (a) Find the value(s) of x so that PQ = QR, where the coordinates of P, Q and R are (6, −1), (1, 3) and (x, 8) respectively.

Show answer and solution

By the distance formula, PQ² = (6 − 1)² + (−1 − 3)² = 41.

QR² = (x − 1)² + (8 − 3)² = (x − 1)² + 25.

PQ = QR gives (x − 1)² = 16.

x − 1 = ±4.

x = 5 or −3

(b) The vertices of a triangle are (−2, 0), (2, 3) and (1, −3). Is the triangle equilateral, isosceles or scalene?

Appeared in: 2023 · Basic · Main

Show answer and solution

Let the points be A, B and C respectively.

AB² = (2 + 2)² + (3 − 0)² = 25.

BC² = (1 − 2)² + (−3 − 3)² = 37.

CA² = (−2 − 1)² + (0 + 3)² = 18.

All three side lengths are different.

Scalene triangle

Basic2 marks · Includes accessible-paper appearance

Q40. (a) Find the positive value of y for which the distance between the points A(3, −1) and B(11, y) is 10 units.

Show answer and solution

By the distance formula, (11 − 3)² + (y + 1)² = 10².

(y + 1)² = 36, so y = 5 or −7.

The required value is positive.

y = 5

(b) Find the ratio in which y-axis divides the line segment joining the points A(5, −6) and B(−1, −4).

Same requested ratio as Q16.

Appeared in: 2023 · Basic · Supplementary

Show answer and solution

Let AP : PB = m : n. On the y-axis, x = 0.

$\frac{(−m + 5n)/(}{m + n)}$ = 0.

m = 5n.

AP : PB = 5 : 1

Standard2 marks · Includes accessible-paper appearance

Q41. (a) If the point P(3, −3) is equidistant from the points A(4, 9) and B(−9, k), find the value(s) of k.

Show answer and solution

PA² = (3 − 4)² + (−3 − 9)² = 145.

PB² = (3 + 9)² + (−3 − k)² = 144 + (k + 3)².

PA = PB gives (k + 3)² = 1.

k + 3 = ±1.

k = −2 or −4

(b) Find the ratio in which the y-axis divides the line segment joining the points (5, −6) and (−1, −4). Also, find the point of intersection.

Same ratio-and-intersection subquestion as Q23.

Appeared in: 2023 · Standard · Supplementary

Show answer and solution

Let AP : PB = m : n. The x-coordinate is 0.

$\frac{(−m + 5n)/(}{m + n)}$ = 0, so AP : PB = 5 : 1.

y = $\frac{(5(−4) + 1(−6))/(}{6)}$ = $\frac{(−13)/(}{3)}$ .

Source note: the 2023 Standard $\frac{(30)/(}{C)}$ /B scheme, PDF p. 9, prints $\frac{(−13)/(}{2)}$ after $\frac{(−26)/(}{6)}$ . The displayed fraction simplifies to $\frac{(−13)/(}{3)}$ .

Ratio = 5 : 1; intersection = (0, $\frac{(−13)/(}{3)}$ )

Basic2 marks

Q42. (a) If Q(0, 2) is equidistant from P(5, −3) and R(x, 7), find the value(s) of x.

Show answer and solution

QP² = (0 − 5)² + (2 + 3)² = 50.

QR² = (0 − x)² + (2 − 7)² = x² + 25.

QP = QR gives x² + 25 = 50.

x = ±5

(b) If A(1, 1) and B(7, 9) are the end points of a diameter of a circle, then find the co-ordinates of the centre of the circle.

Appeared in: 2024 · Basic · Main

Show answer and solution

The centre is the midpoint of the diameter.

O = ( $\frac{(1 + 7)/(}{2)}$ , $\frac{(1 + 9)/(}{2)}$ ).

Centre = (4, 5)

Basic2 marks

Q43. Find a point which is equidistant from the points A(−1, 5) and B(2, 1). How many such points are there?

Appeared in: 2024 · Basic · Main

Show answer and solution

For P(x, y), PA² = PB² gives (x + 1)² + (y − 5)² = (x − 2)² + (y − 1)², hence 6x − 8y + 21 = 0.

One such point is the midpoint M = ( $\frac{(−1 + 2)/(}{2)}$ , $\frac{(5 + 1)/(}{2)}$ ) = ( $\frac{(1)/(}{2)}$ , 3).

Every point on the perpendicular bisector of AB is equidistant from A and B.

One point is ( $\frac{(1)/(}{2)}$ , 3); infinitely many such points exist.

Standard2 marks

Q44. (a) Find a relation between x and y such that the point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).

Show answer and solution

PA = PB gives (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)².

−14x − 2y + 50 = −6x − 10y + 34.

8x − 8y = 16.

x − y = 2

(b) Points A(−1, y) and B(5, 7) lie on a circle with centre O(2, −3y) such that AB is a diameter of the circle. Find the value of y. Also, find the radius of the circle.

Appeared in: 2024 · Standard · Main

Show answer and solution

The centre is the midpoint of AB, so −3y = $\frac{(y + 7)/(}{2)}$ .

−6y = y + 7, hence y = −1.

O = (2, 3) and A = (−1, −1).

Radius OA = √((2 + 1)² + (3 + 1)²) = 5.

y = −1; radius = 5 units

Basic2 marks

Q45. Establish a relation between x and y such that point (x, y) is equidistant from points (−2, 5) and (3, 9).

Appeared in: 2025 · Basic · Main

Show answer and solution

Equal squared distances give (x + 2)² + (y − 5)² = (x − 3)² + (y − 9)².

4x − 10y + 29 = −6x − 18y + 90.

10x + 8y = 61.

10x + 8y = 61

Standard2 marks

Q46. Prove that abscissa of a point P which is equidistant from points with coordinates A(7, 1) and B(3, 5) is 2 more than its ordinate.

Same coordinate relation as Q44(a); expressed using abscissa and ordinate.

Appeared in: 2025 · Standard · Main

Show answer and solution

Let P = (x, y). Since PA = PB, (x − 7)² + (y − 1)² = (x − 3)² + (y − 5)².

−14x − 2y + 50 = −6x − 10y + 34.

8x − 8y = 16, hence x = y + 2.

The abscissa is 2 more than the ordinate.

Standard2 marks

Q47. Find the coordinates of a point P on x-axis equidistant from the points A(−1, 0) and B(5, 0), using distance formula.

Appeared in: 2026 · Standard · Second board examination

Show answer and solution

Let P = (x, 0). Since PA = PB, (x + 1)² = (x − 5)².

2x + 1 = −10x + 25.

12x = 24.

P = (2, 0)

Basic2 marks

Q48. (a) If A(a, b) is a point which is equidistant from points B(2, 4) and C(−2, −3), then find the relation between a and b.

Show answer and solution

AB² = AC² gives (a − 2)² + (b − 4)² = (a + 2)² + (b + 3)².

−4a − 8b + 20 = 4a + 6b + 13.

8a + 14b = 7

(b) Find the ratio in which the line x + 3y = 12 divides the line segment joining (1, 3) and (2, 5).

Appeared in: 2026 · Basic · Second board examination

Show answer and solution

Let AP : PB = m : n.

P = ( $\frac{(2m + n)/(}{m + n)}$ , $\frac{(5m + 3n)/(}{m + n)}$ ).

Since x + 3y = 12, 2m + n + 15m + 9n = 12m + 12n.

5m = 2n.

AP : PB = 2 : 5

Basic2 marks

Q49. (a) P(x, y) is a point equidistant from the points A(4, 3) and B(3, 4). Prove that x − y = 0.

Show answer and solution

PA = PB, so (x − 4)² + (y − 3)² = (x − 3)² + (y − 4)².

−8x − 6y + 25 = −6x − 8y + 25.

2x = 2y.

x − y = 0

(b) In the given figure, △ABC is an equilateral triangle. Coordinates of vertices A and B are (0, 3) and (0, −3) respectively. Find the coordinates of points C.

Equilateral triangle ABC with A(0, 3), B(0, −3), and C on the positive x-axis. — CBSE question diagram.

Appeared in: 2024–25 · Basic · Official sample

Show answer and solution

The midpoint of AB is M(0, 0), and AB = 6.

In an equilateral triangle, CM is perpendicular to AB and AM = 3.

CM² = CA² − AM² = 36 − 9 = 27.

In the given figure, C is on the positive x-axis, so its x-coordinate is 3√3.

C = (3√3, 0)

Triangle types, collinearity and quadrilateral proofs

Standard2 marks

Q50. Show that the points (−2, 3), (8, 3) and (6, 7) are the vertices of a right-angled triangle.

Appeared in: 2023 · Standard · Main

Show answer and solution

Let the points be A, B and C respectively.

AB² = (8 + 2)² + (3 − 3)² = 100.

BC² = (6 − 8)² + (7 − 3)² = 20.

CA² = (−2 − 6)² + (3 − 7)² = 80.

BC² + CA² = AB², so ∠C = 90° by the converse of Pythagoras theorem.

The triangle is right-angled at (6, 7).

Basic2 marks

Q51. (a) Show that A(1, 2), B(5, 4), C(3, 8) and D(−1, 6) are vertices of a parallelogram ABCD.

Show answer and solution

Midpoint of AC = ( $\frac{(1 + 3)/(}{2)}$ , $\frac{(2 + 8)/(}{2)}$ ) = (2, 5).

Midpoint of BD = ( $\frac{(5 − 1)/(}{2)}$ , $\frac{(4 + 6)/(}{2)}$ ) = (2, 5).

The diagonals bisect each other, so ABCD is a parallelogram.

ABCD is a parallelogram.

(b) Show that the points A(3, 0), B(6, 4) and C(−1, 3) are vertices of a right-angled triangle.

Same vertices as Q27(b); this version asks for a right-angle proof.

Appeared in: 2023 · Basic · Main

Show answer and solution

AB² = (6 − 3)² + (4 − 0)² = 25.

AC² = (−1 − 3)² + (3 − 0)² = 25.

BC² = (−1 − 6)² + (3 − 4)² = 50.

AB² + AC² = BC², so ∠A = 90° by the converse of Pythagoras theorem.

△ABC is right-angled at A.

Standard2 marks

Q52. (a) Show that the points (−3, −3), (3, 3) and (−3√3, 3√3) are the vertices of an equilateral triangle.

Show answer and solution

Let the points be A, B and C.

AB² = (3 + 3)² + (3 + 3)² = 72.

AC² = (3 − 3√3)² + (3 + 3√3)² = 72.

BC² = (−3√3 − 3)² + (3√3 − 3)² = 72.

AB = BC = CA = 6√2.

The triangle is equilateral.

(b) Prove that A(4, 3), B(6, 4), C(5, 6), D(3, 5) are the vertices of a square ABCD.

Appeared in: 2023 · Standard · Supplementary

Show answer and solution

By the distance formula, AB² = BC² = CD² = DA² = 5.

AC² = (5 − 4)² + (6 − 3)² = 10.

BD² = (3 − 6)² + (5 − 4)² = 10.

The four sides are equal and the diagonals are equal; hence ABCD is a square.

ABCD is a square.

Standard2 marks

Q53. (a) If A(−2, −1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, then find the values of a and b.

Show answer and solution

Diagonals AC and BD have the same midpoint.

$\frac{(−2 + 4)/(}{2)}$ = $\frac{(a + 1)/(}{2)}$ , so a = 1.

$\frac{(−1 + b)/(}{2)}$ = $\frac{(0 + 2)/(}{2)}$ , so b = 3.

a = 1; b = 3

(b) The three vertices of a parallelogram ABCD, taken in order, are A(−1, 0), B(3, 1) and C(2, 2). Find the coordinates of the fourth vertex D.

Appeared in: 2023 · Standard · Supplementary

Show answer and solution

Let D = (x, y). The diagonals bisect each other.

$\frac{(−1 + 2)/(}{2)}$ = $\frac{(3 + x)/(}{2)}$ , so x = −2.

$\frac{(0 + 2)/(}{2)}$ = $\frac{(1 + y)/(}{2)}$ , so y = 1.

D = (−2, 1)

Standard2 marks

Q54. Find the type of triangle ABC formed whose vertices are A(1, 0), B(−5, 0) and C(−2, 5).

Appeared in: 2024 · Standard · Main

Show answer and solution

AB² = (−5 − 1)² = 36.

BC² = (−2 + 5)² + 5² = 34.

CA² = (1 + 2)² + (0 − 5)² = 34.

BC = CA = √34.

△ABC is isosceles.

Basic2 marks

Q55. Using distance formula, prove that the points (1, 5), (2, 3) and (3, 1) are collinear.

Appeared in: 2025 · Basic · Main

Show answer and solution

Let the points be A, B and C.

AB = √((2 − 1)² + (3 − 5)²) = √5.

BC = √((3 − 2)² + (1 − 3)²) = √5.

AC = √((3 − 1)² + (1 − 5)²) = 2√5.

AB + BC = AC; hence B lies on AC.

The three points are collinear.

Standard2 marks · Includes accessible-paper appearance

Q56. Show that the points (a, a), (−a, −a) and (−√3a, √3a) are the vertices of an equilateral triangle.

Appeared in: 2025 · Standard · Supplementary

Show answer and solution

Let the points be A, B and C.

AB² = (−2a)² + (−2a)² = 8a².

AC² = (−√3a − a)² + (√3a − a)² = 8a².

BC² = (−√3a + a)² + (√3a + a)² = 8a².

For a ≠ 0, all three sides have the same positive length. If a = 0, the points coincide.

Equilateral for a ≠ 0.

Standard2 marks

Q57. Show that the points A(−5, 6), B(3, 0) and C(9, 8) are the vertices of an isosceles triangle.

Appeared in: 2024–25 · Standard · Official sample

Show answer and solution

AB² = (3 + 5)² + (0 − 6)² = 100.

BC² = (9 − 3)² + (8 − 0)² = 100.

AC² = (9 + 5)² + (8 − 6)² = 200.

AB = BC = 10.

△ABC is isosceles.

Basic2 marks

Q58. (a) Points A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.

Show answer and solution

The diagonals of a parallelogram bisect each other.

Midpoint of AC = ( $\frac{(3 + a)/(}{2)}$ , $\frac{(1 + b)/(}{2)}$ ).

Midpoint of BD = ( $\frac{(5 + 4)/(}{2)}$ , $\frac{(1 + 3)/(}{2)}$ ) = ( $\frac{(9)/(}{2)}$ , 2).

3 + a = 9 and 1 + b = 4.

a = 6; b = 3

(b) Find a linear relation between x and y such that P(x, y) is equidistant from the points A(1, 4) and B(−1, 2).

Appeared in: 2026–27 · Basic · Official sample

Show answer and solution

PA = PB gives (x − 1)² + (y − 4)² = (x + 1)² + (y − 2)².

−2x − 8y + 17 = 2x − 4y + 5.

4x + 4y = 12.

x + y = 3

Basic2 marks

Q59. Using distance formula, show that the points (−1, 3), (6, 2) and (3, −1) are vertices of a right-angled triangle.

Appeared in: 2025 · Basic · Main

Show answer and solution

Let the points be A, B and C.

AB² = (6 + 1)² + (2 − 3)² = 50.

BC² = (3 − 6)² + (−1 − 2)² = 18.

AC² = (3 + 1)² + (−1 − 3)² = 32.

BC² + AC² = AB², so ∠C = 90° by the converse of Pythagoras theorem.

The triangle is right-angled at (3, −1).

Standard2 marks

Q60. Do the points P(1, 0), Q(−5, 0) and R(−2, 5) form a triangle? If so, name the type of triangle formed.

Same coordinates as Q54; this version also checks whether a triangle is formed.

Appeared in: 2026 · Standard · Main

Show answer and solution

PQ = √((−5 − 1)² + (0 − 0)²) = 6.

PR = √((−2 − 1)² + 5²) = √34.

QR = √((−2 + 5)² + 5²) = √34.

The triangle inequalities hold, including 2√34 > 6.

PR = QR, so the triangle is isosceles.

Yes; an isosceles triangle.

Standard2 marks

Q61. If the points A(4, 5), B(m, 6), C(4, 3) and D(1, n) taken in this order are the vertices of a parallelogram ABCD, then find the values of m and n.

Appeared in: 2026 · Standard · Main

Show answer and solution

The diagonals bisect each other.

Midpoint of AC = ( $\frac{(4 + 4)/(}{2)}$ , $\frac{(5 + 3)/(}{2)}$ ) = (4, 4).

Midpoint of BD = ( $\frac{(m + 1)/(}{2)}$ , $\frac{(6 + n)/(}{2)}$ ).

m + 1 = 8 and 6 + n = 8.

m = 7; n = 2

Standard2 marks

Q62. (a) Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are (9, −2) and (1, 6) respectively. (i) Find the co-ordinates of point P. (ii) Find the length of the side of the square.

Square ABCD with diagonals AC and BD meeting at P. The opposite vertices are B(9, −2) and D(1, 6). — CBSE question diagram.

Show answer and solution

(i) The diagonals bisect each other, so P is the midpoint of BD.

P = ( $\frac{(9 + 1)/(}{2)}$ , $\frac{(−2 + 6)/(}{2)}$ ) = (5, 2).

(ii) BD² = (9 − 1)² + (−2 − 6)² = 128.

For a square, BD² = 2AB², hence AB² = 64.

(i) P = (5, 2); (ii) side = 8 units

(b) Find the coordinates of a point on the line x + y = 5 which is equidistant from (6, 4) and (5, 2).

Appeared in: 2026 · Standard · Main

Show answer and solution

Let P = (x, y). Equal squared distances give (x − 6)² + (y − 4)² = (x − 5)² + (y − 2)².

2x + 4y = 23, so x + 2y = $\frac{(23)/(}{2)}$ .

Subtract x + y = 5 to get y = $\frac{(13)/(}{2)}$ .

x = 5 − $\frac{(13)/(}{2)}$ = $\frac{(−3)/(}{2)}$ .

P = ( $\frac{(−3)/(}{2)}$ , $\frac{(13)/(}{2)}$ )

Standard2 marks

Q63. If points A(1, 2), B(4, y), C(x, 6) and D(3, 5) are the vertices of a parallelogram ABCD, then find values of x and y.

Appeared in: 2026 · Standard · Second board examination

Show answer and solution

The diagonals AC and BD have the same midpoint.

$\frac{(1 + x)/(}{2)}$ = $\frac{(4 + 3)/(}{2)}$ , so x = 6.

$\frac{(2 + 6)/(}{2)}$ = $\frac{(y + 5)/(}{2)}$ , so y = 3.

x = 6; y = 3

Coordinate Geometry: frequently asked questions

What is the distance formula for two points?

For A(x₁, y₁) and B(x₂, y₂), AB = √((x₂ − x₁)² + (y₂ − y₁)²). Substitute each coordinate with its sign. The squared terms are added, so the distance is non-negative.

How do I use the section formula without reversing the ratio?

For AP : PB = m : n, multiply B’s coordinates by m and A’s coordinates by n, then divide their sums by m + n. Keep the names A and B fixed. For a midpoint, m = n = 1.

How do I find where an axis divides a line segment?

Let the division ratio be m : n and write the section-formula coordinates. Set y = 0 for the x-axis or x = 0 for the y-axis. Solve for m : n, then substitute it to find the other coordinate if asked.

What is the difference between AP : AB and AP : PB?

AB is the whole segment, while PB is its remaining part. If AP : AB = 2 : 5, then PB corresponds to 5 − 2 = 3 parts, so AP : PB = 2 : 3. Convert to the divided-segment ratio before using the section formula.

How do I find the length of a median?

A median joins a vertex to the midpoint of the opposite side. First use the midpoint formula to find that point. Then apply the distance formula between the vertex and the midpoint. Do not use the opposite side’s full endpoint coordinates as the median’s endpoint.

How can I prove three points are collinear using distance?

Find the three pairwise distances and identify the largest. If the sum of the two smaller distances equals the largest, the points are collinear and the middle point lies between the other two. Use exact values rather than rounded decimals for the equality.

Are equal sides enough to prove that a quadrilateral is a square?

No. Four equal sides establish a rhombus, which need not be a square. Also establish a right angle or equal diagonals. For a rectangle, show the relevant parallelogram property and a right angle or equal diagonals.

Can an equidistant-point question have more than one answer?

Yes. Points equidistant from two distinct points lie on their perpendicular bisector, so infinitely many exist unless another condition is given. A condition such as lying on an axis can restrict the answer. Keep both values when both satisfy the original question.

How do I prove a parallelogram using coordinates?

Find the midpoint of each diagonal. If the two midpoints coincide, the diagonals bisect each other, so the quadrilateral is a parallelogram. Keep the vertices in the stated order: in ABCD, compare the midpoints of AC and BD.

Do I answer both OR alternatives?

Answer one alternative where an OR choice is printed. Both alternatives are solved here for practice. A full OR pair is one numbered question, while each alternative has its own worked solution and answer-summary entry.

Answer summary

Open the answer summary

Question	Answer
Q1	BD = $\frac{(√145)/(}{2)}$ units
Q2	a = 3 or 5
Q3	Median BM = √58 units
Q4(a)	a = 6 or −2
Q4(b) OR	P = (7, $\frac{(−9)/(}{2)}$ )
Q5	x = 3 or 5
Q6(a)	tan A = 3
Q6(b) OR	A, B and C are collinear.
Q7	(4, 0) and (12, 0)
Q8	Radius = 5 units
Q9	(5, 0) and (17, 0)
Q10	p = ±4
Q11	P = (0, 5) or (0, −1)
Q12	PQ ⟂ OQ
Q13	Diagonal = √34 units
Q14(a)	P = (3, 1)
Q14(b) OR	y = 5 or −7
Q15(a)	P = (4, −1)
Q15(b) OR	x = y
Q16	AP : PB = 5 : 1
Q17	AP : PB = 9 : 5
Q18(a)	P = (1, 3)
Q18(b) OR	p = 13
Q19	AP : PB = 4 : 5
Q20	AP : PB = 3 : 2; m = $\frac{(−2)/(}{5)}$
Q21	AP : PB = 1 : 5
Q22	Ratio = 2 : 7; k = 6
Q23	Ratio = 5 : 1; intersection = (0, $\frac{(−13)/(}{3)}$ )
Q24	k = 4
Q25(a)	Ratio = 5 : 3; y = 0
Q25(b) OR	P = (0, 9)
Q26(a)	AP : PB = 8 : 7
Q26(b) OR	BE = $\frac{(5√5)/(}{2)}$ units
Q27(a)	AP : PB = 2 : 7
Q27(b) OR	The triangle is isosceles.
Q28	( $\frac{(−2)/(}{3)}$ , $\frac{(7)/(}{3)}$ ) and ( $\frac{(8)/(}{3)}$ , $\frac{(11)/(}{3)}$ )
Q29	Ratio = 8 : 5; x = $\frac{(−14)/(}{13)}$
Q30	Ratio = 3 : 4; m = 2
Q31	Ratio = 5 : 3; point = ( $\frac{(31)/(}{8)}$ , 0)
Q32	C = (9, −13)
Q33	P = ( $\frac{(−2)/(}{7)}$ , $\frac{(−20)/(}{7)}$ )
Q34(a)	P = (1, 3)
Q34(b) OR	A, B and C are collinear.
Q35	P = ( $\frac{(3)/(}{5)}$ , $\frac{(11)/(}{5)}$ )
Q36	R = (−13, 9)
Q37	P = ( $\frac{(8)/(}{7)}$ , $\frac{(−9)/(}{7)}$ )
Q38	A = (3, −10)
Q39(a)	x = 5 or −3
Q39(b) OR	Scalene triangle
Q40(a)	y = 5
Q40(b) OR	AP : PB = 5 : 1
Q41(a)	k = −2 or −4
Q41(b) OR	Ratio = 5 : 1; intersection = (0, $\frac{(−13)/(}{3)}$ )
Q42(a)	x = ±5
Q42(b) OR	Centre = (4, 5)
Q43	One point is ( $\frac{(1)/(}{2)}$ , 3); infinitely many such points exist.
Q44(a)	x − y = 2
Q44(b) OR	y = −1; radius = 5 units
Q45	10x + 8y = 61
Q46	The abscissa is 2 more than the ordinate.
Q47	P = (2, 0)
Q48(a)	8a + 14b = 7
Q48(b) OR	AP : PB = 2 : 5
Q49(a)	x − y = 0
Q49(b) OR	C = (3√3, 0)
Q50	The triangle is right-angled at (6, 7).
Q51(a)	ABCD is a parallelogram.
Q51(b) OR	△ABC is right-angled at A.
Q52(a)	The triangle is equilateral.
Q52(b) OR	ABCD is a square.
Q53(a)	a = 1; b = 3
Q53(b) OR	D = (−2, 1)
Q54	△ABC is isosceles.
Q55	The three points are collinear.
Q56	Equilateral for a ≠ 0.
Q57	△ABC is isosceles.
Q58(a)	a = 6; b = 3
Q58(b) OR	x + y = 3
Q59	The triangle is right-angled at (3, −1).
Q60	Yes; an isosceles triangle.
Q61	m = 7; n = 2
Q62(a)	(i) P = (5, 2); (ii) side = 8 units
Q62(b) OR	P = ( $\frac{(−3)/(}{2)}$ , $\frac{(13)/(}{2)}$ )
Q63	x = 6; y = 3

Both OR alternatives are listed for practice. The answer summary follows your selected filters.

How to write a 2-mark Coordinate Geometry solution

Identify what is required and follow any specified method.
Identify the endpoints, the required point and any division ratio.
Show one logical operation per line, keeping signs and brackets.
Check the required geometric condition and state the coordinates, ratio or length, including units when given.

For the geometric facts used in these proofs, revise Triangles 2-mark questions and Circles MCQs with answers.

About these questions and solutions

Exact repeats of full question groups are merged; original OR pairs are retained, so a subquestion can recur in another pair.
Solutions prepared by Team TutorVertex in NCERT style, with source references supplied for checking against the official marking schemes. How we check solutions.

Official sources — question papers and marking schemes

PDF pages are given for the displayed version. Repeat appearances are identified by paper code and question number. Archive member names identify the exact files inside official ZIPs. Question wording follows the first listed appearance; spacing and punctuation are normalised.

Question 1 sources

2024 · Standard · Main · 30/2/2 · Q24Official question paper · PDF p. 15
MATHEMATICS_STANDARD/30-2-2(Mathematics Standard).pdfOfficial marking scheme · PDF p. 10
Mathematics_Standard/MS 041_30-2-2 Mathematics 2023-24.pdf

Question 2 sources

2025 · Standard · Main · 30/1/1 · Q23Official question paper · PDF p. 15
041_Mathematics_Standard/30-1-1_Mathematics Standard.pdfOfficial marking scheme · PDF p. 8
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-1-1_2024-25.pdf
2025 · Standard · Main · 30/1/2 · Q25Official question paper · PDF p. — see Q25
041_Mathematics_Standard/30-1-2_Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q25
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-1-2_2024-25.pdf

Question 3 sources

2025 · Standard · Main · 30/1/3 · Q25Official question paper · PDF p. 17
041_Mathematics_Standard/30-1-3_Mathematics Standard.pdfOfficial marking scheme · PDF p. 9
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-1-3_2024-25.pdf

Question 4 sources

2025 · Basic · Supplementary · 430/S/2 · Q21Official question paper · PDF p. 13
Maths_Basic/430-S-2Math Basic.pdfOfficial marking scheme · PDF p. 8
MATHEMATICS_BASIC/MS_X_Mathematics(Basic)_241_430_S_2.pdf

Question 5 sources

2026 · Standard · Main · 30/1/2 · Q22Official question paper · PDF p. 9
Mathematics_Standard/30-1-2P-Maths (Std)_X.pdfOfficial marking scheme · PDF p. 8
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-1-2_2025-26 (1).pdf
2026 · Standard · Main · 30/1/1 · Q23Official question paper · PDF p. — see Q23
Mathematics_Standard/30-1-1P-Maths (Std)_X.pdfOfficial marking scheme · PDF p. — see Q23
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-1-1_2025-26 (1).pdf

Question 6 sources

2026 · Standard · Main · 30/5/1 · Q22Official question paper · PDF p. 13
Mathematics_Standard/1172-1_30-5-1 (Mathematics Standard).pdfOfficial marking scheme · PDF pp. 8, 9
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-5-1_2025-26 (1).pdf
2026 · Standard · Main · 30/5/2 · Q25Official question paper · PDF p. — see Q25
Mathematics_Standard/1172-2_30-5-2 (Mathematics Standard).pdfOfficial marking scheme · PDF p. — see Q25
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-5-2_2025-26 (1).pdf

Question 7 sources

2024–25 · Standard · Official sample · Standard sample · Q24Official question paper · PDF p. 4
2024-Standard-SQP.pdfOfficial marking scheme · PDF p. 2
2024-Standard-MS.pdf

Question 8 sources

2025–26 · Basic · Official sample · Basic sample · Accessible alternative · Q22Official question paper · PDF p. 6
2025-Basic-SQP.pdfOfficial marking scheme · PDF p. 3
2025-Basic-MS.pdf

Question 9 sources

2023 · Standard · Main · 30/6/2 · Q24Official question paper · PDF p. 13
MATHEMATICS_STANDARD/30_6_2_Maths Standard.pdfOfficial marking scheme · PDF p. 9
Maths_Standard/MS 041_30-6-2 Mathematics 2022-23.pdf

Question 10 sources

2026 · Standard · Main · 30/2/3 · Q21Official question paper · PDF p. 17
Mathematics_Standard/30_2_3_Mathematics Std._cl.10.pdfOfficial marking scheme · PDF p. 8
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-2-3_2025-26 (1).pdf

Question 11 sources

2026 · Basic · Main · 430/4/1 · Q23Official question paper · PDF p. 13
Mathematics_Standard/1158-1_430-4-1 Mathematics (Basic).pdfOfficial marking scheme · PDF p. 8
241_MATHEMATICS BASIC\X_241_MS_430_4_1_unsigned.pdf
2026 · Basic · Main · 430/4/2 · Q24Official question paper · PDF p. — see Q24
Mathematics_Standard/1158-2_430-4-2 Mathematics (Basic).pdfOfficial marking scheme · PDF p. — see Q24
241_MATHEMATICS BASIC\X_241_MS_430_4_2_unsigned.pdf

Question 12 sources

2026 · Basic · Main · 430/5/3 · Q25Official question paper · PDF p. 15
Mathematics_Standard/1159-3_430-5-3 Mathematics (Basic).pdfOfficial marking scheme · PDF p. 8
241_MATHEMATICS BASIC\X_241_MS_ 430_5_3_unsigned.pdf

Question 13 sources

2026 · Standard · Second board examination · 30/8/3 · Q23Official question paper · PDF p. 17
Mathmatics_Std/30-8-3_Maths Std_E2_10_2026.pdfOfficial marking scheme · PDF p. 9
Mathematics_Standard/MS_X_041_Mathematics_30-8-3_2025-26 F (1).pdf

Question 14 sources

2023 · Basic · Main · 430/1/2 · Q23Official question paper · PDF p. 13
MATHEMATICS_BASIC/430_1_2_Maths Basic.pdfOfficial marking scheme · PDF p. 8
Math_Basic/430_1_2.pdf
2023 · Basic · Main · 430/1/3 · Q23Official question paper · PDF p. — see Q23
MATHEMATICS_BASIC/430_1_3_Maths Basic.pdfOfficial marking scheme · PDF p. — see Q23
Math_Basic/430_1_3.pdf
2023 · Basic · Main · 430/1/1 · Q21Official question paper · PDF p. — see Q21
MATHEMATICS_BASIC/430_1_1_Maths Basic.pdfOfficial marking scheme · PDF p. — see Q21
Math_Basic/430_1_1.pdf

Question 15 sources

2023 · Standard · Main · 30/5/1 · Q21Official question paper · PDF p. 11
MATHEMATICS_STANDARD/30_5_1_Maths Standard.pdfOfficial marking scheme · PDF p. 9
Maths_Standard/MS 041_30-5-1 Mathematics 2022-23.pdf
2023 · Standard · Main · 30/5/2 · Q24Official question paper · PDF p. — see Q24
MATHEMATICS_STANDARD/30_5_2_Maths Standard.pdfOfficial marking scheme · PDF p. — see Q24
Maths_Standard/MS 041_30-5-2 Mathematics 2022-23.pdf
2023 · Standard · Main · 30/5/3 · Q25Official question paper · PDF p. — see Q25
MATHEMATICS_STANDARD/30_5_3_Maths Standard.pdfOfficial marking scheme · PDF p. — see Q25
Maths_Standard/MS 041_30-5-3 Mathematics 2022-23.pdf

Question 16 sources

2023 · Standard · Main · 30/5/1 · Q24Official question paper · PDF p. 13
MATHEMATICS_STANDARD/30_5_1_Maths Standard.pdfOfficial marking scheme · PDF p. 11
Maths_Standard/MS 041_30-5-1 Mathematics 2022-23.pdf
2023 · Standard · Main · 30/5/3 · Q23Official question paper · PDF p. — see Q23
MATHEMATICS_STANDARD/30_5_3_Maths Standard.pdfOfficial marking scheme · PDF p. — see Q23
Maths_Standard/MS 041_30-5-3 Mathematics 2022-23.pdf
2023 · Standard · Main · 30/6/3 · Q21Official question paper · PDF p. — see Q21
MATHEMATICS_STANDARD/30_6_3_Maths Standard.pdfOfficial marking scheme · PDF p. — see Q21
Maths_Standard/MS 041_30-6-3 Mathematics 2022-23.pdf

Question 17 sources

2023 · Standard · Main · 30/5/2 · Q22Official question paper · PDF p. 11
MATHEMATICS_STANDARD/30_5_2_Maths Standard.pdfOfficial marking scheme · PDF pp. 10, 11
Maths_Standard/MS 041_30-5-2 Mathematics 2022-23.pdf

Question 18 sources

2023 · Basic · Main · 430/4/1 · Q22Official question paper · PDF p. 6
MATHEMATICS_BASIC/430_4_1 Maths Basic.pdfOfficial marking scheme · PDF p. 6
Math_Basic/430_4_1.pdf
2023 · Basic · Main · 430/4/2 · Q24Official question paper · PDF p. — see Q24
MATHEMATICS_BASIC/430_4_2 Maths Basic.pdfOfficial marking scheme · PDF p. — see Q24
Math_Basic/430_4_2.pdf
2023 · Basic · Main · 430/4/3 · Q24Official question paper · PDF p. — see Q24
MATHEMATICS_BASIC/430_4_3 Maths Basic.pdfOfficial marking scheme · PDF p. — see Q24
Math_Basic/430_4_3.pdf

Question 19 sources

2023 · Basic · Supplementary · 430/C/1 · Q21Official question paper · PDF p. 15
Maths_Basic/430_C_1 Mathematics Basic.pdfOfficial marking scheme · PDF p. 8
Mathematics_Basic/241_C_1.pdf

Question 20 sources

2023 · Basic · Supplementary · 430/C/2 · Q24Official question paper · PDF p. 15
Maths_Basic/430_C_2 Mathematics Basic.pdfOfficial marking scheme · PDF p. 9
Mathematics_Basic/241_C_2.pdf

Question 21 sources

2023 · Basic · Supplementary · 430/C/3 · Q23Official question paper · PDF p. 15
Maths_Basic/430_C_3 Mathematics Basic.pdfOfficial marking scheme · PDF p. 7
Mathematics_Basic/241_C_3.pdf

Question 22 sources

2023 · Standard · Supplementary · 30/C/1 · Q25Official question paper · PDF p. 13
Maths_Standard/30_C_1 Mathematics Standard.pdfOfficial marking scheme · PDF p. 10
Mathematics_Standard/041_30-C-1.pdf
2023 · Standard · Supplementary · 30/C/2 · Q25Official question paper · PDF p. — see Q25
Maths_Standard/30_C_2 Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q25
Mathematics_Standard/041_30-C-2.pdf
2023 · Standard · Supplementary · 30/C/3 · Q24Official question paper · PDF p. — see Q24
Maths_Standard/30_C_3 Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q24
Mathematics_Standard/041_30-C-3.pdf

Question 23 sources

2024 · Basic · Main · 430/3/1 · Q24Official question paper · PDF p. 15
MATHEMATICS_BASIC/430-3-1MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF pp. 8, 9
Mathematics/X_241_MS_430_3_1_Unsigned.pdf

Question 24 sources

2024 · Basic · Main · 430/3/3 · Q22Official question paper · PDF p. 13
MATHEMATICS_BASIC/430-3-3MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF p. 8
Mathematics/X_241_MS_430_3_3_Unsigned.pdf

Question 25 sources

2024 · Basic · Main · 430/5/1 · Q22Official question paper · PDF p. 13
MATHEMATICS_BASIC/430-5-1MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF p. 8
Mathematics/X_241_MS_430_5_1_Unsigned.pdf
2024 · Basic · Main · 430/5/2 · Q25Official question paper · PDF p. — see Q25
MATHEMATICS_BASIC/430-5-2MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF p. — see Q25
Mathematics/X_241_MS_430_5_2_Unsigned.pdf
2024 · Basic · Main · 430/5/3 · Q25Official question paper · PDF p. — see Q25
MATHEMATICS_BASIC/430-5-3MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF p. — see Q25
Mathematics/X_241_MS_430_5_3_Unsigned.pdf

Question 26 sources

2024 · Standard · Main · 30/2/1 · Q25Official question paper · PDF p. 15
MATHEMATICS_STANDARD/30-2-1(Mathematics Standard).pdfOfficial marking scheme · PDF p. 11
Mathematics_Standard/MS 041_30-2-1 Mathematics 2023-24.pdf
2024 · Standard · Main · 30/2/2 · Q22Official question paper · PDF p. — see Q22
MATHEMATICS_STANDARD/30-2-2(Mathematics Standard).pdfOfficial marking scheme · PDF p. — see Q22
Mathematics_Standard/MS 041_30-2-2 Mathematics 2023-24.pdf
2024 · Standard · Main · 30/2/3 · Q21Official question paper · PDF p. — see Q21
MATHEMATICS_STANDARD/30-2-3(Mathematics Standard).pdfOfficial marking scheme · PDF p. — see Q21
Mathematics_Standard/MS 041_30-2-3 Mathematics 2023-24.pdf

Question 27 sources

2024 · Standard · Main · 30/5/1 · Q24Official question paper · PDF p. 13
MATHEMATICS_STANDARD/30-5-1(Mathematics Standard).pdfOfficial marking scheme · PDF pp. 8, 9
Mathematics_Standard/MS 041_30-5-1 Mathematics 2023-24.pdf
2024 · Standard · Main · 30/5/2 · Q22Official question paper · PDF p. — see Q22
MATHEMATICS_STANDARD/30-5-2(Mathematics Standard).pdfOfficial marking scheme · PDF p. — see Q22
Mathematics_Standard/MS 041_30-5-2 Mathematics 2023-24.pdf
2024 · Standard · Main · 30/5/3 · Q23Official question paper · PDF p. — see Q23
MATHEMATICS_STANDARD/30-5-3(Mathematics Standard).pdfOfficial marking scheme · PDF p. — see Q23
Mathematics_Standard/MS 041_30-5-3 Mathematics 2023-24.pdf

Question 28 sources

2025 · Basic · Main · 430/5/1 · Q23Official question paper · PDF p. 15
241_Mathematics_Basic/430-5-1_Mathematics Basic.pdfOfficial marking scheme · PDF p. 8
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_1.pdf
2025 · Basic · Main · 430/5/3 · Q22Official question paper · PDF p. — see Q22
241_Mathematics_Basic/430-5-3_Mathematics Basic.pdfOfficial marking scheme · PDF p. — see Q22
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_3.pdf

Question 29 sources

2025 · Basic · Main · 430/4/1 · Q22Official question paper · PDF p. 9
241_Mathematics_Basic/430-4-1_Mathematics Basic.pdfOfficial marking scheme · PDF p. 7
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_4_1.pdf

Question 30 sources

2025 · Basic · Main · 430/4/2 · Q23Official question paper · PDF p. 9
241_Mathematics_Basic/430-4-2_Mathematics Basic.pdfOfficial marking scheme · PDF p. 8
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_4_2.pdf

Question 31 sources

2025 · Basic · Main · 430/4/3 · Q22Official question paper · PDF p. 9
241_Mathematics_Basic/430-4-3_Mathematics Basic.pdfOfficial marking scheme · PDF p. 8
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_4_3.pdf

Question 32 sources

2025 · Standard · Main · 30/4/1 · Q23Official question paper · PDF p. 9
041_Mathematics_Standard/30-4-1_Mathematics Standard.pdfOfficial marking scheme · PDF p. 7
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard _30-4-1_2024-25_.pdf
2025 · Standard · Main · 30/4/2 · Q25Official question paper · PDF p. — see Q25
041_Mathematics_Standard/30-4-2_Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q25
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-4-2_2024-25_.pdf
2025 · Standard · Main · 30/4/3 · Q22Official question paper · PDF p. — see Q22
041_Mathematics_Standard/30-4-3_Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q22
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-4-3_2024-25_.pdf

Question 33 sources

2025 · Standard · Main · 30/5/1 · Q22Official question paper · PDF p. 13
041_Mathematics_Standard/30-5-1_Mathematics Standard.pdfOfficial marking scheme · PDF p. 7
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-5-1_2024-25.pdf
2025 · Standard · Main · 30/5/2 · Q24Official question paper · PDF p. — see Q24
041_Mathematics_Standard/30-5-2_Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q24
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-5-2_2024-25.pdf
2025 · Standard · Main · 30/5/3 · Q21Official question paper · PDF p. — see Q21
041_Mathematics_Standard/30-5-3_Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q21
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-5-3_2024-25.pdf

Question 34 sources

2025 · Basic · Supplementary · 430/S/1 · Q23Official question paper · PDF p. 13
Maths_Basic/430-S-1Math Basic.pdfOfficial marking scheme · PDF p. 9
MATHEMATICS_BASIC/MS_X_Mathematics(Basic)_241_430_S_1.pdf
2025 · Basic · Supplementary · 430/S/3 · Q24Official question paper · PDF p. — see Q24
Maths_Basic/430-S-3Math Basic.pdfOfficial marking scheme · PDF p. — see Q24
MATHEMATICS_BASIC/MS_X_Mathematics(Basic)_241_430_S_3.pdf

Question 35 sources

2026 · Basic · Main · 430/4/3 · Q22Official question paper · PDF p. 13
Mathematics_Standard/1158-3_430-4-3 Mathematics (Basic).pdfOfficial marking scheme · PDF pp. 18, 19
241_MATHEMATICS BASIC\X_241_MS_430_4_3_unsigned.pdf

Question 36 sources

2026 · Standard · Second board examination · 30/7/1 · Q23Official question paper · PDF p. 15
Mathmatics_Std/30-7-1_Maths Std_E2_10_2026.pdfOfficial marking scheme · PDF p. 10
Mathematics_Standard/Mathematics Standard-041 E2 (Set 30-7-1).pdf

Question 37 sources

2026 · Standard · Second board examination · 30/7/2 · Q24Official question paper · PDF p. 15
Mathmatics_Std/30-7-2_Maths Std_E2_10_2026.pdfOfficial marking scheme · PDF p. 11
Mathematics_Standard/Mathematics Standard-041 E2 (Set 30-7-2).pdf

Question 38 sources

2026 · Standard · Second board examination · 30/8/1 · Q24Official question paper · PDF p. 17
Mathmatics_Std/30-8-1_Maths Std_E2_10_2026.pdfOfficial marking scheme · PDF p. 9
Mathematics_Standard/MS_X_041_Mathematics_30-8-1_2025-26 F (1).pdf

Question 39 sources

2023 · Basic · Main · 430/2/1 · Q23Official question paper · PDF p. 15
MATHEMATICS_BASIC/430_2_1_Maths Basic.pdfOfficial marking scheme · PDF p. 10
Math_Basic/430_2_1.pdf
2023 · Basic · Main · 430/2/2 · Q21Official question paper · PDF p. — see Q21
MATHEMATICS_BASIC/430_2_2_Maths Basic.pdfOfficial marking scheme · PDF p. — see Q21
Math_Basic/430_2_2.pdf
2023 · Basic · Main · 430/2/3 · Q25Official question paper · PDF p. — see Q25
MATHEMATICS_BASIC/430_2_3_Maths Basic.pdfOfficial marking scheme · PDF p. — see Q25
Math_Basic/430_2_3.pdf

Question 40 sources

2023 · Basic · Supplementary · 430/C/B · Q23Official question paper · PDF p. 15
Maths_Basic/430_C_B Mathematics Basic.pdfOfficial marking scheme · PDF p. 5
Mathematics_Basic/241_C_B.pdf

Question 41 sources

2023 · Standard · Supplementary · 30/C/B · Q23Official question paper · PDF p. 13
Maths_Standard/30_C_B Mathematics Standard.pdfOfficial marking scheme · PDF pp. 8, 9
Mathematics_Standard/041_30-C-B.pdf

Question 42 sources

2024 · Basic · Main · 430/3/1 · Q21Official question paper · PDF p. 13
MATHEMATICS_BASIC/430-3-1MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF pp. 7, 8
Mathematics/X_241_MS_430_3_1_Unsigned.pdf
2024 · Basic · Main · 430/3/2 · Q24Official question paper · PDF p. — see Q24
MATHEMATICS_BASIC/430-3-2MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF p. — see Q24
Mathematics/X_241_MS_430_3_2_Unsigned.pdf
2024 · Basic · Main · 430/3/3 · Q23Official question paper · PDF p. — see Q23
MATHEMATICS_BASIC/430-3-3MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF p. — see Q23
Mathematics/X_241_MS_430_3_3_Unsigned.pdf

Question 43 sources

2024 · Basic · Main · 430/3/2 · Q23Official question paper · PDF p. 15
MATHEMATICS_BASIC/430-3-2MATHEMATICS (BASIC).pdfOfficial marking scheme · PDF p. 9
Mathematics/X_241_MS_430_3_2_Unsigned.pdf

Question 44 sources

2024 · Standard · Main · 30/3/1 · Q23Official question paper · PDF p. 15
MATHEMATICS_STANDARD/30-3-1(Mathematics Standard).pdfOfficial marking scheme · PDF p. 9
Mathematics_Standard/MS 041_30-3-1 Mathematics 2023-24.pdf
2024 · Standard · Main · 30/3/2 · Q22Official question paper · PDF p. — see Q22
MATHEMATICS_STANDARD/30-3-2(Mathematics Standard).pdfOfficial marking scheme · PDF p. — see Q22
Mathematics_Standard/MS 041_30-3-2 Mathematics 2023-24.pdf
2024 · Standard · Main · 30/3/3 · Q25Official question paper · PDF p. — see Q25
MATHEMATICS_STANDARD/30-3-3(Mathematics Standard).pdfOfficial marking scheme · PDF p. — see Q25
Mathematics_Standard/MS 041_30-3-3 Mathematics 2023-24.pdf

Question 45 sources

2025 · Basic · Main · 430/5/2 · Q21Official question paper · PDF p. 13
241_Mathematics_Basic/430-5-2_Mathematics Basic.pdfOfficial marking scheme · PDF p. 7
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_2.pdf

Question 46 sources

2025 · Standard · Main · 30/6/1 · Q23Official question paper · PDF p. 13
041_Mathematics_Standard/30-6-1_Mathematics Standard.pdfOfficial marking scheme · PDF p. 8
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-6-1_2024-25.pdf
2025 · Standard · Main · 30/6/3 · Q22Official question paper · PDF p. — see Q22
041_Mathematics_Standard/30-6-3_Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q22
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-6-3_2024-25.pdf
2025 · Standard · Main · 30/6/2 · Q25Official question paper · PDF p. — see Q25
041_Mathematics_Standard/30-6-2_Mathematics Standard.pdfOfficial marking scheme · PDF p. — see Q25
Math/041 Mathematics Standard/X_MS_041_Mathematics Standard_30-6-2_2024-25.pdf

Question 47 sources

2026 · Standard · Second board examination · 30/7/3 · Q21Official question paper · PDF p. 15
Mathmatics_Std/30-7-3_Maths Std_E2_10_2026.pdfOfficial marking scheme · PDF p. 8
Mathematics_Standard/Mathematics Standard-041 E2 (Set 30-7-3).pdf

Question 48 sources

2026 · Basic · Second board examination · 430/8/3 · Q25Official question paper · PDF p. 15
Mathmatics_Basic/430-8-3_Maths Basic_E2_10_2026.pdfOfficial marking scheme · PDF p. 9
Mathematics_Basic/241_430_8_3 Mathematics Basic.pdf
2026 · Basic · Second board examination · 430/8/1 · Q21Official question paper · PDF p. — see Q21
Mathmatics_Basic/430-8-1_Maths Basic_E2_10_2026.pdfOfficial marking scheme · PDF p. — see Q21
Mathematics_Basic/241_430_8_1 Mathematics Basic.pdf
2026 · Basic · Second board examination · 430/8/2 · Q23Official question paper · PDF p. — see Q23
Mathmatics_Basic/430-8-2_Maths Basic_E2_10_2026.pdfOfficial marking scheme · PDF p. — see Q23
Mathematics_Basic/241_430_8_2 Mathematics Basic.pdf

Question 49 sources

2024–25 · Basic · Official sample · Basic sample · Q21Official question paper · PDF p. 4
2024-Basic-SQP.pdfOfficial marking scheme · PDF p. 2
2024-Basic-MS.pdf

Question 50 sources

2023 · Standard · Main · 30/2/1 · Q23Official question paper · PDF p. 13
MATHEMATICS_STANDARD/30_2_1_Maths Standard.pdfOfficial marking scheme · PDF pp. 8, 9
Maths_Standard/MS 041_30-2-1 Mathematics 2022-23.pdf
2023 · Standard · Main · 30/2/2 · Q24Official question paper · PDF p. — see Q24
MATHEMATICS_STANDARD/30_2_2_Maths Standard.pdfOfficial marking scheme · PDF p. — see Q24
Maths_Standard/MS 041_30-2-2 Mathematics 2022-23.pdf
2023 · Standard · Main · 30/2/3 · Q22Official question paper · PDF p. — see Q22
MATHEMATICS_STANDARD/30_2_3_Maths Standard.pdfOfficial marking scheme · PDF p. — see Q22
Maths_Standard/MS 041_30-2-3 Mathematics 2022-23.pdf

Question 51 sources

2023 · Basic · Main · 430/5/1 · Q24Official question paper · PDF p. 13
MATHEMATICS_BASIC/430_5_1_Maths Basic.pdfOfficial marking scheme · PDF pp. 8, 9
Math_Basic/430_5_1.pdf
2023 · Basic · Main · 430/5/2 · Q22Official question paper · PDF p. — see Q22
MATHEMATICS_BASIC/430_5_2_Maths Basic.pdfOfficial marking scheme · PDF p. — see Q22
Math_Basic/430_5_2.pdf
2023 · Basic · Main · 430/5/3 · Q25Official question paper · PDF p. — see Q25
MATHEMATICS_BASIC/430_5_3_Maths Basic.pdfOfficial marking scheme · PDF p. — see Q25
Math_Basic/430_5_3.pdf

Question 52 sources

2023 · Standard · Supplementary · 30/C/1 · Q23Official question paper · PDF p. 13
Maths_Standard/30_C_1 Mathematics Standard.pdfOfficial marking scheme · PDF p. 9
Mathematics_Standard/041_30-C-1.pdf

Question 53 sources

2023 · Standard · Supplementary · 30/C/2 · Q21Official question paper · PDF p. 11
Maths_Standard/30_C_2 Mathematics Standard.pdfOfficial marking scheme · PDF p. 8
Mathematics_Standard/041_30-C-2.pdf

Question 54 sources

2024 · Standard · Main · 30/2/1 · Q22Official question paper · PDF p. 15
MATHEMATICS_STANDARD/30-2-1(Mathematics Standard).pdfOfficial marking scheme · PDF p. 9
Mathematics_Standard/MS 041_30-2-1 Mathematics 2023-24.pdf

Question 55 sources

2025 · Basic · Main · 430/5/1 · Q25Official question paper · PDF p. 15
241_Mathematics_Basic/430-5-1_Mathematics Basic.pdfOfficial marking scheme · PDF pp. 8, 9
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_1.pdf
2025 · Basic · Main · 430/5/2 · Q22Official question paper · PDF p. — see Q22
241_Mathematics_Basic/430-5-2_Mathematics Basic.pdfOfficial marking scheme · PDF p. — see Q22
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_2.pdf

Question 56 sources

2025 · Standard · Supplementary · 30/B · Q24Official question paper · PDF p. 15
Maths_Standard/30(B)-S Maths Standard Blind.pdfOfficial marking scheme · PDF p. 8
MATHEMATICS_STANDARD/MS_X_041_Mathematics_30(B)-S_2024-25.pdf

Question 57 sources

2024–25 · Standard · Official sample · Standard sample · Q25Official question paper · PDF p. 4
2024-Standard-SQP.pdfOfficial marking scheme · PDF p. 3
2024-Standard-MS.pdf

Question 58 sources

2026–27 · Basic · Official sample · Basic sample · Q25Official question paper · PDF p. 4
2026-Basic-SQP.pdfOfficial marking scheme · PDF p. 6
2026-Basic-MS.pdf

Question 59 sources

2025 · Basic · Main · 430/5/3 · Q23Official question paper · PDF p. 13
241_Mathematics_Basic/430-5-3_Mathematics Basic.pdfOfficial marking scheme · PDF p. 8
Math/241 Mathematics Basic/X_241_Mathematics Basic_MS_430_5_3.pdf

Question 60 sources

2026 · Standard · Main · 30/1/3 · Q21Official question paper · PDF p. 7
Mathematics_Standard/30-1-3P-Maths (Std)_X.pdfOfficial marking scheme · PDF p. 7
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-1-3_2025-26 (1).pdf

Question 61 sources

2026 · Standard · Main · 30/2/1 · Q23Official question paper · PDF p. 17
Mathematics_Standard/30_2_1_Mathematics Std._cl.10.pdfOfficial marking scheme · PDF p. 9
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-2-1_2025-26 (1).pdf
2026 · Standard · Main · 30/2/2 · Q25Official question paper · PDF p. — see Q25
Mathematics_Standard/30_2_2_Mathematics Std._cl.10.pdfOfficial marking scheme · PDF p. — see Q25
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-2-2_2025-26 (1).pdf

Question 62 sources

2026 · Standard · Main · 30/3/1 · Q23Official question paper · PDF p. 15
Mathematics_Standard/30_3_1_Mathematics Std._cl.10.pdfOfficial marking scheme · PDF pp. 8, 9
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-3-1_2025-26 (1).pdf
2026 · Standard · Main · 30/3/2 · Q21Official question paper · PDF p. — see Q21
Mathematics_Standard/30_3_2_Mathematics Std._cl.10.pdfOfficial marking scheme · PDF p. — see Q21
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-3-2_2025-26 (1).pdf
2026 · Standard · Main · 30/3/3 · Q25Official question paper · PDF p. — see Q25
Mathematics_Standard/30_3_3_Mathematics Std._cl.10.pdfOfficial marking scheme · PDF p. — see Q25
041_MATHEMATICS STANDARD\MS_X_041_Mathematics_30-3-3_2025-26 (1).pdf

Question 63 sources

2026 · Standard · Second board examination · 30/8/2 · Q25Official question paper · PDF p. 17
Mathmatics_Std/30-8-2_Maths Std_E2_10_2026.pdfOfficial marking scheme · PDF p. 10
Mathematics_Standard/MS_X_041_Mathematics_30-8-2_2025-26 F (1).pdf

NCERT Class 10, Chapter 7: Coordinate Geometry
CBSE Mathematics syllabus, 2026–27

Need help with Coordinate Geometry?

Get one-to-one help with coordinate formulae, geometric proofs and clear exam working.

Book a Free Assessment →