Coordinate Geometry for Class 10 Maths: Important Questions and Answers

Important questions on Coordinate Geometry for Class 10 are available in this article to help students prepare effectively for the CBSE board exams. These questions are designed according to the latest CBSE syllabus and exam pattern, covering important concepts such as the distance formula, section formula, and area of a triangle. This chapter carries good weightage in the Maths exam and includes important application-based and long-answer questions. To support effective preparation, our subject experts have curated well-structured questions with clear and detailed solutions. With well-structured questions and detailed step-by-step solutions, students can revise the chapter thoroughly and gain confidence for the examination.

Table of Contents

• Key Formulas of Coordinate Geometry for Class 10
• Important Questions on Distance Formula
• Important Questions on Section Formula
• Important Questions on Area of Triangle
• Collinearity & Special Geometry Questions
• HOTS & Previous-Year Questions
• Practice Questions on Coordinate Geometry for Class 10

Key Formulas of Coordinate Geometry for Class 10

Every solved question in this chapter flows from one of these three.

Distance Formula:

d = √[(x₂ − x₁)² + (y₂ − y₁) ²]

The distance formula gives the straight-line distance between points A(x₁, y₁) and B(x₂, y₂).

Special case: distance from origin O(0,0) to P(x, y) = √(x² + y²).

Section Formula (Internal Division): 

Point P divides the segment joining A (x₁, y₁) and B (x₂, y₂) internally in the ratio m : n. The section formula gives, 

P(x, y) = [(mx₂ + nx₁)/(m+n) , (my₂ + ny₁)/(m+n)]

Midpoint special case (m = n = 1): M = [(x₁ + x₂)/2, (y₁ + y₂)/2]

Area of a triangle:

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

For a triangle with vertices A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).

If Area = 0, the three points are collinear (lie on the same line).

Important Questions on Distance Formula

Board exams commonly test the distance formula through the identification of triangle types, equidistant point problems, and collinearity.

Q1: Find the distance of the point P(2, 3) from the x-axis.

Solution: For any point P(x, y), the perpendicular distance from the x-axis is simply the absolute value of the y-coordinate.

P = (2, 3) 

⇒ y = 3

Therefore, the distance from the x-axis = 3 units

Q2: Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

Solution: Let P(x, y) be equidistant from A(7, 1) and B(3, 5). 

So PA = PB, which means PA² = PB².

PA² = (x − 7)² + (y − 1)²

PB² = (x − 3)² + (y − 5)²

Setting PA² = PB²:

x² − 14x + 49 + y² − 2y + 1 = x² − 6x + 9 + y² − 10y + 25

⇒ −14x − 2y + 50 = −6x − 10y + 34

⇒ −14x + 6x − 2y + 10y = 34 − 50 

⇒ −8x + 8y = −16 

⇒ x − y = 2

Therefore, the required relation is x − y = 2

Q3: Name the type of triangle formed by the points A (−5, 6), B (−4, −2) and C (7, 5).

Solution: Calculate all three side lengths using the distance formula:

AB = √[(−4−(−5))² + (−2−6)²] = √[1 + 64] = √65

BC = √[(7−(−4))² + (5−(−2)) ²] = √[121 + 49] = √170

CA = √[(7−(−5))² + (5−6)²] = √[144 + 1] = √145

All three sides are different (√65 ≠ √170 ≠ √145) and no two are equal.

Therefore, the triangle ABC is a scalene triangle

Q 4: Find the distance of a point P(x, y) from the origin.

Solution: Origin = O(0, 0). 

Apply the distance formula between O (0, 0) and P (x, y):

OP = √[(x − 0)² + (y − 0) ²]

∴ OP = √(x² + y²) units

Important Questions on Section Formula

Q 5: Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by the point (−1, 6).

Solution: Let the ratio be k: 1. Using the section formula for the x-coordinate:

x = (k×x₂ + x₁) / (k + 1) = (6k − 3) / (k + 1) = −1

⟹ 6k − 3 = −k − 1 

⟹ 7k = 2 

⟹ k = 2/7

Verify using y-coordinate: y = (−8k + 10)/(k + 1) = (−16/7 + 10)/(2/7 + 1) = (54/7)/(9/7) = 6

The point (−1, 6) divides the segment in the ratio 2 : 7

Q 6: Find the coordinates of the trisection points of the line segment joining A(2, −2) and B(−7, 4).

Solution: Trisection means two points P and Q divide AB into three equal parts, so AP = PQ = QB.

Point P divides AB in the ratio 1 : 2:

P = [(1×(−7) + 2×2)/(1+2) , (1×4 + 2×(−2))/(1+2)]

⇒ P = [(−7 + 4)/3 , (4 − 4)/3] 

= (−3/3, 0/3) = (−1, 0)

Point Q divides AB in the ratio 2 : 1:

Q = [(2×(−7) + 1×2)/(2+1) , (2×4 + 1×(−2))/(2+1)]

⇒ Q = [(−14 + 2)/3 , (8 − 2)/3]

= (−12/3, 6/3) = (−4, 2)

Trisection points are P(−1, 0) and Q(−4, 2)

Q 7: If C (−1, 2) divides internally the line segment joining A (2, 5) and B (x, y) in the ratio 3 : 4, find the coordinates of B.

Solution: Using the section formula with m : n = 3 : 4

Given A(2, 5), B(x, y), and C(−1, 2):

For x-coordinate: (3x + 4×2)/(3+4) = −1 

⟹ 3x + 8 = −7 

⟹ x = −5

For y-coordinate: (3y + 4 × 5)/(3 + 4) = 2 

⟹ 3y + 20 = 14 

⟹ y = −2

Coordinates of B = (−5, −2)

Q8: If A (−2, 1), B (a, 0), C (4, b) and D (1, 2) are the vertices of parallelogram ABCD, find a, b and the length of each side.

Solution: In a parallelogram, the diagonals bisect each other. 

So, the midpoint of diagonal AC = midpoint of diagonal BD.

Midpoint of AC = [(-2+4)/2, (1+b)/2] = [1, (1+b)/2]

Midpoint of BD = [(a+1)/2, (0+2)/2] = [(a+1)/2, 1]

Equating: 1 = (a + 1)/2 

⟹ a = 1 and (1+b)/2 = 1 

⟹ b = 1

AB = √[(1+2)² + (0−1)²] = √[9+1] = √10 units

Therefore, a = 1, b = 1 and length of each side of parallelogram = √10 units

Important Questions on Area of Triangle

Q9: Find the value of k if the points A(2, 3), B(4, k) and C(6, −3) are collinear.

Solution: Three points are collinear when the area of the triangle formed by them equals zero.

Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| = 0

⟹ ½ |2(k − (−3)) + 4((−3) − 3) + 6(3 − k)| = 0

⟹ |2(k+3) + 4(−6) + 6(3−k)| = 0

⟹ |2k + 6 − 24 + 18 − 6k| = 0

⟹ |−4k + 0| = 0 

⟹ 4k = 0 

⟹ k = 0

Q10: Find the area of triangle PQR with vertices P(−5, 7), Q(−4, −5) and R(4, 5).

Solution: Apply the area formula with x₁=−5, y₁=7, x₂=−4, y₂=−5, x₃=4, y₃=5:

Area = ½ |−5(−5 − 5) + (−4)(5 − 7) + 4(7 − (−5))|

= ½ |−5(−10) + (−4)(−2) + 4(12)|

= ½ |50 + 8 + 48|

= ½ × 106 = 53

Therefore, the area of triangle PQR = 53 sq. units

Q 11: Find the area of the triangle formed by the midpoints of the sides of the triangle with vertices (0, −1), (2, 1) and (0, 3). Also find the ratio of this area to the original triangle's area.

Solution: Let A(0,−1), B(2,1), C(0,3). First find midpoints D, E, F using the midpoint formula:

D (midpoint of AB) = [(0+2)/2, (−1+1)/2] = (1, 0)

E (midpoint of AC) = [(0+0)/2, (−1+3)/2] = (0, 1)

F (midpoint of BC) = [(2+0)/2, (1+3)/2] = (1, 2)

Area of △DEF = ½ |1(1−2) + 0(2−0) + 1(0−1)| = ½ |−1 + 0 − 1| = ½ × 2 = 1 sq. unit

Area of △ABC = ½ |0(1−3) + 2(3−(−1)) + 0(−1−1)| = ½ |0 + 8 + 0| = 4 sq. units

Ratio of Area (△DEF) : Area (△ABC) = 1 : 4

Q12: Find the area of quadrilateral ABCD with vertices A(−5, 7), B(−4, −5), C(−1, −6) and D(4, 5).

Solution: Divide quadrilateral ABCD into two triangles: △ABD and △BCD along the diagonal BD.

Area of △ABD = ½ |−5(−5−5) + (−4)(5−7) + 4(7−(−5))| = ½ |50 + 8 + 48| = 53 sq. units

Area of △BCD = ½ |(−4)(−6−5) + (−1)(5−(−5)) + 4(−5−(−6))| = ½ |44 − 10 + 4| = 19 sq. units

Area of quadrilateral = 53 + 19 = 72 sq. units

Area of quadrilateral ABCD = 72 sq. units

Collinearity & Special Geometry Questions

Q13: Find the ratio in which the line x − 3y = 0 divides the line segment joining (−2, −5) and (6, 3). Find the point of intersection.

Solution: Let the line x − 3y = 0 divide the segment in ratio k : 1. Using section formula, the point of division is:

P(x, y) = [(6k − 2)/(k+1), (3k − 5)/(k+1)]

Since P lies on x − 3y = 0:

(6k − 2)/(k + 1) − 3 × (3k − 5)/(k + 1) = 0

6k − 2 − 9k + 15 = 0 

⟹ −3k + 13 = 0 

⟹ k = 13/3

Ratio = 13 : 3

x = (6 × 13/3 − 2)/(13/3 + 1) = (26 − 2)/(16/3) × (1/3) = 72/16 = 9/2

y = (3 × 13/3 − 5)/(16/3) = (13 − 5)/(16/3) = 24/16 = 3/2

Ratio = 13:3, and point of intersection = (9/2, 3/2)

Q 14: Find the ratio in which P(4, m) divides the segment joining A(2, 3) and B(6, −3). Hence, find m.

Solution: Let the ratio be k: 1. Using the section formula for the x-coordinate:

(6k + 2)/(k + 1) = 4 

⟹ 6k + 2 = 4k + 4 

⟹ 2k = 2 

⟹ k = 1

Ratio = 1:1 (P is the midpoint of AB)

For m, equate y: (−3×1 + 3)/(1+1) = (−3+3)/2 = 0/2 = 0

Ratio = 1 : 1 and m = 0

HOTS & Previous-Year Questions

HOTS 1: Show that the points (1, 7), (4, 2), (−1, −1) and (−4, 4) are the vertices of a square.

Solution: For a square, all four sides must be equal and both diagonals must be equal.

Let A(1,7), B(4,2), C(−1,−1), D(−4,4).

AB = √[(4−1)²+(2−7)²] = √[9+25] = √34

BC = √[(−1−4)²+(−1−2)²] = √[25+9] = √34

CD = √[(−4+1)²+(4+1)²] = √[9+25] = √34

DA = √[(1+4)²+(7−4)²] = √[25+9] = √34

AC (diagonal) = √[(−1−1) ²+(−1−7)²] = √[4+64] = √68

BD (diagonal) = √[(−4−4)²+(4−2)²] = √[64+4] = √68

All four sides = √34; both diagonals = √68 (= √2 × √34)

Since all sides are equal and both diagonals are equal, ABCD is a square.

HOTS 2: Prove that if the distances of P(x, y) from A(5, 1) and B(−1, 5) are equal, then 3x = 2y.

Solution: Given PA = PB, so PA² = PB²:

PA² = (x−5)² + (y−1)²

PB² = (x+1)² + (y−5)²

Setting equal: x²−10x+25 + y²−2y+1 = x²+2x+1 + y²−10y+25

−10x − 2y + 26 = 2x − 10y + 26

−10x − 2y = 2x − 10y

8y = 12x 

⟹ 2y = 3x 

⟹ 3x = 2y

Hence proved: 3x = 2y

HOTS 3: The centre of a circle is (2a, a−7). Find the value of a if the circle passes through (11, −9) and has diameter 10√2 units.

Solution: Radius = diameter/2 = 10√2/2 = 5√2. 

Distance from centre to point (11, −9) = radius.

⟹ √[(11−2a)² + (−9−(a−7)) ²] = 5√2

⟹ √[(11−2a)² + (−2−a)²] = 5√2

Squaring: (11−2a)² + (a+2)² = 50

121 − 44a + 4a² + a² + 4a + 4 = 50

5a² − 40a + 125 = 50 

⟹ 5a² − 40a + 75 = 0 

⟹ a² − 8a + 15 = 0

(a−3)(a−5) = 0 

⟹ a = 3 or a = 5

Therefore, a = 3 or a = 5.

Practice Questions on Coordinate Geometry for Class 10

1. Write the coordinates of a point on the x-axis equidistant from A(−2, 0) and B(6, 0).

2. Find the midpoint of the segment joining (−3, 4) and (5, −8).

3. Find the relation between x and y if A(x, y), B(−4, 6), and C(−2, 3) are collinear.

4. Point P divides the segment joining A (2, 1) and B (5, −8) in the ratio 1:2. Is P on the line 2x − y + k = 0? Find k.

5. Prove that the diagonals of a rectangle bisect each other and are equal using coordinate geometry.

6. Find the area of a rhombus with vertices (3, 0), (4, 5), (−1, 4) and (−2, −1). [Hint: Area = ½ × d₁ × d₂]

7. The vertices A(1, −2), B(2, 3), C(a, 2) and D(−4, −3) form a parallelogram. Find a and the height, taking AB as the base.