Coordinate Geometry PYQ Bank | Class X Maths

Questions

2-Mark Qs

3-Mark Qs

Years Covered

📋 Key Topics Covered (As per CBSE 2025-26 Syllabus)

Distance Formula — √[(x₂–x₁)²+(y₂–y₁)²]

Section Formula — (m₁x₂+m₂x₁)/(m₁+m₂)

Midpoint Formula — ((x₁+x₂)/2, (y₁+y₂)/2)

Prove Geometric Shapes using distance/section

Equidistant Points — PA = PB type

Division by Axes — ratio on X/Y axis

🚫

DELETED for 2025-26: Area of a Triangle using coordinates has been completely removed from the syllabus. Questions on area of triangle will NOT appear in the board exam. Focus only on Distance Formula and Section Formula.

✅

Unit Weightage: Coordinate Geometry unit carries 6 marks in the board exam. Expect 1-2 MCQs + 1-2 SA questions from this chapter.

📝 SECTION A — 2-MARK QUESTIONS

15 Questions • Very Short Answer (VSA) Type • Board PYQs 2018–2025

CBSE 2025

The distance of the point (–1, 7) from the x-axis is: (a) –1 (b) 7 (c) 6 (d) √50

2 Marks Distance from Axis ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Key Distance of a point (x, y) from the x-axis = |y|

Step 1 Point = (–1, 7) → Distance from x-axis = |7| = 7

∴ Answer: (b) 7

CBSE 2024

The point which divides the line segment joining the points A(1, 2) and B(–1, 1) internally in the ratio 1 : 2 is ___.

2 Marks Section Formula ⭐ MOST REPEATED

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✅ DETAILED SOLUTION

Section Formula: P = ((m₁x₂+m₂x₁)/(m₁+m₂), (m₁y₂+m₂y₁)/(m₁+m₂))

Step 1 m₁ : m₂ = 1 : 2, A(1, 2), B(–1, 1)

Step 2 x = (1×(–1) + 2×1)/(1+2) = (–1+2)/3 = 1/3

Step 3 y = (1×1 + 2×2)/(1+2) = (1+4)/3 = 5/3

∴ Point = (1/3, 5/3)

CBSE 2024

Find the ratio in which the line segment joining the points (6, 4) and (1, –7) is divided by the x-axis.

2 Marks Division by Axis ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 Let x-axis divide in ratio k : 1 at point (a, 0).

Step 2 y-coordinate = 0 ⟹ [k(–7) + 1(4)] / (k+1) = 0

Step 3 –7k + 4 = 0 ⟹ k = 4/7

∴ Ratio = 4 : 7

CBSE 2023

If the line segment joining A(3, –4) and B(1, 2) is trisected at points P(p, –2) and Q(5/3, q), find p and q.

2 Marks Trisection

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✅ DETAILED SOLUTION

Key Trisection ⟹ P divides AB in 1:2, Q divides AB in 2:1

Step 1 P divides A(3,–4) and B(1,2) in 1 : 2:

p = (1×1 + 2×3)/3 = 7/3

Step 2 Q divides A(3,–4) and B(1,2) in 2 : 1:

q = (2×2 + 1×(–4))/3 = 0

∴ p = 7/3, q = 0

CBSE 2023

Find the coordinates of a point P on the x-axis which is equidistant from A(7, 6) and B(–3, 4).

2 Marks Equidistant Point ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 P is on x-axis ⟹ P = (x, 0). PA = PB.

Step 2 PA² = PB² ⟹ (x–7)²+(0–6)² = (x+3)²+(0–4)²

Step 3 x²–14x+49+36 = x²+6x+9+16

–14x+85 = 6x+25 ⟹ 60 = 20x ⟹ x = 3

∴ P = (3, 0)

CBSE 2022

In which quadrant does the point P lie that divides the line segment joining A(2, –5) and B(5, 2) in the ratio 2 : 3?

2 Marks Section Formula + Quadrant

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✅ DETAILED SOLUTION

Step 1 x = (2×5 + 3×2)/5 = 16/5

Step 2 y = (2×2 + 3×(–5))/5 = (4–15)/5 = –11/5

Step 3 P = (16/5, –11/5) → x > 0, y < 0

∴ P lies in Quadrant IV

CBSE 2022

If C(1, –1) is the mid-point of line segment AB joining A(4, x) and B(–2, 4), find the value of x.

2 Marks Midpoint

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✅ DETAILED SOLUTION

Step 1 Midpoint y-coordinate: (x + 4)/2 = –1

Step 2 x + 4 = –2 ⟹ x = –6

Verify x-coordinate: (4 + (–2))/2 = 2/2 = 1 ✓

∴ x = –6

CBSE 2021

Find the distance between the points A(2, –3) and B(–2, 0).

2 Marks Distance Formula

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✅ DETAILED SOLUTION

Distance = √[(x₂–x₁)² + (y₂–y₁)²]

Step 1 AB = √[(–2–2)² + (0–(–3))²] = √[(–4)² + 3²]

Step 2 AB = √[16 + 9] = √25 = 5

∴ AB = 5 units

CBSE 2021

If coordinates of one end of a diameter of a circle are (2, 3) and the centre is (–2, 5), find the coordinates of the other end.

2 Marks Midpoint (Reverse)

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✅ DETAILED SOLUTION

Key Centre = midpoint of diameter.

Step 1 Let other end = (x, y). Centre = (–2, 5).

Step 2 (2+x)/2 = –2 ⟹ x = –6

Step 3 (3+y)/2 = 5 ⟹ y = 7

∴ Other end = (–6, 7)

Q10

CBSE 2020

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3), D(0, 3). Find the length of diagonal AC.

2 Marks Distance in Shapes

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✅ DETAILED SOLUTION

Step 1 In rectangle ABCD: A = (0, 0), B(4, 0), C(4, 3), D(0, 3)

Step 2 AC = √[(4–0)² + (3–0)²] = √[16+9] = √25 = 5

∴ Diagonal AC = 5 units

Q11

CBSE 2020

Find the value(s) of x for which the distance between P(x, 4) and Q(9, 10) is 10 units.

2 Marks Distance Formula → Find x

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✅ DETAILED SOLUTION

Step 1 PQ = 10 ⟹ PQ² = 100

Step 2 (9–x)² + (10–4)² = 100 ⟹ (9–x)² + 36 = 100

Step 3 (9–x)² = 64 ⟹ 9–x = ±8

Step 4 x = 1 or x = 17

∴ x = 1 or x = 17

Q12

CBSE 2019

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

2 Marks Distance Formula → Find y

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✅ DETAILED SOLUTION

Step 1 AB² = 100 ⟹ (11–3)² + (y+1)² = 100

Step 2 64 + (y+1)² = 100 ⟹ (y+1)² = 36

Step 3 y + 1 = ±6 ⟹ y = 5 or y = –7

∴ y = 5 or y = –7

Q13

CBSE 2019

If A(0, 2) is equidistant from B(3, p) and C(p, 5), find the value of p.

2 Marks Equidistant

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✅ DETAILED SOLUTION

Step 1 AB² = AC² ⟹ (3–0)² + (p–2)² = (p–0)² + (5–2)²

Step 2 9 + p²–4p+4 = p² + 9

Step 3 13 – 4p = 9 ⟹ 4p = 4 ⟹ p = 1

∴ p = 1

Q14

CBSE 2018

Find the distance of the point (–3, 4) from the x-axis.

2 Marks Distance from Axis

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✅ DETAILED SOLUTION

Step 1 Distance from x-axis = |y-coordinate| = |4| = 4

∴ Distance = 4 units

Q15

CBSE 2018

Find the midpoint of the line segment joining A(–1, 7) and B(4, –3).

2 Marks Midpoint

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✅ DETAILED SOLUTION

Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2)

Step 1 x = (–1+4)/2 = 3/2 ; y = (7+(–3))/2 = 4/2 = 2

∴ Midpoint = (3/2, 2)

📝 SECTION B — 3-MARK QUESTIONS

15 Questions • Short Answer (SA) Type • Board PYQs 2018–2025

Q16

CBSE 2024

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).

3 Marks Equidistant → Relation ⭐ MOST IMPORTANT

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✅ DETAILED SOLUTION

Step 1 PA = PB ⟹ PA² = PB²

Step 2 (x–7)²+(y–1)² = (x–3)²+(y–5)²

Step 3 x²–14x+49+y²–2y+1 = x²–6x+9+y²–10y+25

Step 4 –14x–2y+50 = –6x–10y+34

Step 5 –8x + 8y + 16 = 0 ⟹ –x + y + 2 = 0 ⟹ x – y = 2

∴ x – y = 2

Q17

CBSE 2024

Find the ratio in which the line segment joining the points (5, 3) and (–1, 6) is divided by the y-axis.

3 Marks Division by Y-axis ⭐ HOTS

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✅ DETAILED SOLUTION

Step 1 Y-axis ⟹ point is (0, y). Let ratio = k : 1

Step 2 x-coordinate: [k(–1) + 1(5)]/(k+1) = 0

Step 3 –k + 5 = 0 ⟹ k = 5

Step 4 y = [5(6) + 1(3)]/(5+1) = 33/6 = 11/2

∴ Ratio = 5 : 1; Point = (0, 11/2)

Q18

CBSE 2024

Line 4x + y = 4 divides the line segment joining the points (–2, –1) and (3, 5) in a certain ratio. Find the ratio.

3 Marks Line divides Segment ⭐ CHALLENGING

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✅ DETAILED SOLUTION

Step 1 Let line 4x+y=4 divide segment in ratio k:1. Using section formula:

Step 2 x = (3k–2)/(k+1) , y = (5k–1)/(k+1)

Step 3 Substitute in 4x + y = 4:

4 × (3k–2)/(k+1) + (5k–1)/(k+1) = 4

(12k–8+5k–1)/(k+1) = 4 ⟹ 17k–9 = 4k+4

Step 4 13k = 13 ⟹ k = 1

∴ Ratio = 1 : 1 (i.e., the line bisects the segment)

Q19

CBSE 2023

Prove that the points A(–2, 3), B(1, 2) and C(7, 0) are collinear using the distance formula.

3 Marks Collinearity ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 AB = √[(1+2)²+(2–3)²] = √[9+1] = √10

Step 2 BC = √[(7–1)²+(0–2)²] = √[36+4] = √40 = 2√10

Step 3 AC = √[(7+2)²+(0–3)²] = √[81+9] = √90 = 3√10

Step 4 AB + BC = √10 + 2√10 = 3√10 = AC ✓

∴ AB + BC = AC. Hence A, B, C are collinear.

Q20

CBSE 2023

Show that the points A(1, 2), B(5, 4), C(3, 8) and D(–1, 6) are the vertices of a square.

3 Marks Prove Square ⭐ MOST ASKED TYPE

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✅ DETAILED SOLUTION

Step 1 AB = √[(5–1)²+(4–2)²] = √[16+4] = √20

Step 2 BC = √[(3–5)²+(8–4)²] = √[4+16] = √20

Step 3 CD = √[(–1–3)²+(6–8)²] = √[16+4] = √20

Step 4 DA = √[(1+1)²+(2–6)²] = √[4+16] = √20

All 4 sides equal: AB = BC = CD = DA = √20 ✓ (could be rhombus)

Step 5 Diagonal AC = √[(3–1)²+(8–2)²] = √[4+36] = √40

Step 6 Diagonal BD = √[(–1–5)²+(6–4)²] = √[36+4] = √40

Diagonals equal: AC = BD = √40 ✓ ⟹ It's a square (not just rhombus)

∴ All sides equal + Diagonals equal ⟹ ABCD is a square.

Q21

CBSE 2022

The point A(3, y) is equidistant from the points P(6, 5) and Q(0, –3). Find the value of y.

3 Marks Equidistant

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✅ DETAILED SOLUTION

Step 1 AP = AQ ⟹ AP² = AQ²

Step 2 (3–6)²+(y–5)² = (3–0)²+(y+3)²

Step 3 9 + y²–10y+25 = 9 + y²+6y+9

Step 4 34 – 10y = 18 + 6y ⟹ 16 = 16y ⟹ y = 1

∴ y = 1

Q22

CBSE 2022

Find the ratio in which the y-axis divides the line segment joining the points (–4, –6) and (10, 12). Also find the coordinates of the point of division.

3 Marks Division by Y-axis ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 Y-axis ⟹ point (0, y). Let ratio = k : 1

Step 2 x: [k(10) + 1(–4)]/(k+1) = 0 ⟹ 10k–4 = 0 ⟹ k = 2/5

Step 3 Ratio = 2/5 : 1 = 2 : 5

Step 4 y = [2(12)+5(–6)]/7 = (24–30)/7 = –6/7

∴ Ratio = 2 : 5; Point = (0, –6/7)

Q23

CBSE 2021

Prove that the points (7, 10), (–2, 5) and (3, –4) are the vertices of an isosceles right triangle.

3 Marks Prove Triangle Type ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 Let A(7,10), B(–2,5), C(3,–4)

Step 2 AB² = (7+2)²+(10–5)² = 81+25 = 106

Step 3 BC² = (–2–3)²+(5+4)² = 25+81 = 106

Step 4 AC² = (7–3)²+(10+4)² = 16+196 = 212

Check Isosceles: AB² = BC² = 106 ⟹ AB = BC ✓

Check Right angle: AB² + BC² = 106+106 = 212 = AC² ✓ (Pythagoras)

∴ AB = BC (isosceles) and AB²+BC²=AC² (right angle). Hence, isosceles right triangle.

Q24

CBSE 2020

If A(4, 3), B(–1, y) and C(3, 4) are the vertices of a right triangle ABC right-angled at A, find the value of y.

3 Marks Right Triangle → Find y

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✅ DETAILED SOLUTION

Key Right-angled at A ⟹ BC² = AB² + AC² (Pythagoras)

Step 1 AB² = (4+1)²+(3–y)² = 25+(3–y)²

Step 2 AC² = (4–3)²+(3–4)² = 1+1 = 2

Step 3 BC² = (–1–3)²+(y–4)² = 16+(y–4)²

Step 4 BC² = AB² + AC²:

16+(y–4)² = 25+(3–y)²+2

16+y²–8y+16 = 27+9–6y+y²

32–8y = 36–6y ⟹ –2y = 4 ⟹ y = –2

∴ y = –2

Q25

CBSE 2021

Find the point on the y-axis which is equidistant from the points (–5, –2) and (3, 2).

3 Marks Equidistant on Axis

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✅ DETAILED SOLUTION

Step 1 Point on y-axis = (0, y). PA = PB.

Step 2 (0+5)²+(y+2)² = (0–3)²+(y–2)²

Step 3 25+y²+4y+4 = 9+y²–4y+4

Step 4 29+4y = 13–4y ⟹ 8y = –16 ⟹ y = –2

∴ Point = (0, –2)

Q26

CBSE 2020

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.

3 Marks Parallelogram

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✅ DETAILED SOLUTION

Key In a parallelogram, diagonals bisect each other. Midpoint of AC = Midpoint of BD.

Step 1 Mid(AC) = ((3+a)/2, (1+b)/2)

Step 2 Mid(BD) = ((5+4)/2, (1+3)/2) = (9/2, 2)

Step 3 (3+a)/2 = 9/2 ⟹ a = 6

Step 4 (1+b)/2 = 2 ⟹ b = 3

∴ a = 6, b = 3 → C = (6, 3)

Q27

CBSE 2019

Find the centre of a circle passing through the points (5, –8), (2, –9) and (2, 1).

3 Marks Centre of Circle ⭐ HOTS

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✅ DETAILED SOLUTION

Key Centre H(x,y) is equidistant from all three points. AH = BH = CH.

Step 1 AH² = BH² ⟹ (x–5)²+(y+8)² = (x–2)²+(y+9)²

x²–10x+25+y²+16y+64 = x²–4x+4+y²+18y+81

–10x+16y+89 = –4x+18y+85 ⟹ –6x–2y = –4 ⟹ 3x+y = 2 … (i)

Step 2 BH² = CH² ⟹ (x–2)²+(y+9)² = (x–2)²+(y–1)²

y²+18y+81 = y²–2y+1 ⟹ 20y = –80 ⟹ y = –4

Step 3 From (i): 3x + (–4) = 2 ⟹ x = 2

∴ Centre = (2, –4)

Q28

CBSE 2019

F lies on the line segment joining E(–3, 2) and G(4, 5). F divides EG in the ratio 2 : 1. Find the coordinates of F.

3 Marks Section Formula

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✅ DETAILED SOLUTION

Step 1 m₁:m₂ = 2:1, E(–3,2), G(4,5)

Step 2 x = (2×4 + 1×(–3))/3 = (8–3)/3 = 5/3

Step 3 y = (2×5 + 1×2)/3 = (10+2)/3 = 4

∴ F = (5/3, 4)

Q29

CBSE 2018

If A(5, 2), B(2, –2) and C(–2, t) are vertices of a right-angled triangle with ∠B = 90°, find the value of t.

3 Marks Right Triangle → Find t

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✅ DETAILED SOLUTION

Key ∠B = 90° ⟹ AC² = AB² + BC²

Step 1 AB² = (5–2)²+(2+2)² = 9+16 = 25

Step 2 BC² = (2+2)²+(–2–t)² = 16+(t+2)²

Step 3 AC² = (5+2)²+(2–t)² = 49+(2–t)²

Step 4 AC² = AB² + BC²:

49+(2–t)² = 25+16+(t+2)²

49+4–4t+t² = 41+4+4t+t²

53–4t = 45+4t ⟹ 8 = 8t ⟹ t = 1

∴ t = 1

Q30

CBSE 2018

Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

3 Marks Trisection ⭐ CLASSIC

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✅ DETAILED SOLUTION

Key Trisection ⟹ Two points P (1:2) and Q (2:1).

Step 1 P divides A(4,–1) & B(–2,–3) in 1:2:

P = ((1×(–2)+2×4)/3, (1×(–3)+2×(–1))/3) = (6/3, –5/3) = (2, –5/3)

Step 2 Q divides A(4,–1) & B(–2,–3) in 2:1:

Q = ((2×(–2)+1×4)/3, (2×(–3)+1×(–1))/3) = (0/3, –7/3) = (0, –7/3)

∴ Points of trisection: P = (2, –5/3) and Q = (0, –7/3)

📐 KEY FORMULAS TO REMEMBER

Distance Formula: d = √[(x₂–x₁)² + (y₂–y₁)²]

Section Formula: P = ( (m₁x₂ + m₂x₁)/(m₁+m₂) , (m₁y₂ + m₂y₁)/(m₁+m₂) )

Midpoint Formula: M = ( (x₁+x₂)/2 , (y₁+y₂)/2 )

Distance from x-axis: |y-coordinate|
Distance from y-axis: |x-coordinate|
Distance from origin: √(x² + y²)
Division by x-axis: Set y = 0 in section formula
Division by y-axis: Set x = 0 in section formula
Collinear: AB + BC = AC (largest distance)
Square: 4 sides equal + diagonals equal
Rhombus: 4 sides equal (diagonals may differ)
Rectangle: Opp sides equal + diagonals equal
Parallelogram: Midpoint of diagonals same

💡 BOARD EXAM TIPS — COORDINATE GEOMETRY

PA = PB trick: Always square both sides (PA² = PB²) — avoids square root calculations entirely!

"Division by axis" shortcut: For x-axis, set y=0. For y-axis, set x=0. Always assume ratio as k:1.

Square vs Rhombus: Proving 4 equal sides only proves RHOMBUS. You must ALSO prove diagonals equal for SQUARE.

Collinearity: Find all 3 distances, then check if the two smaller add up to the largest.

🚫 Area of Triangle DELETED: Do NOT study the area formula ½|x₁(y₂–y₃)+…| — it won't be asked.

Expected marks: 6 marks (typically 2 MCQs + 1 SA from this chapter).