Class 10 Maths MCQs for Chapter 7 Coordinate Geometry with Answers PDF

MCQs on Coordinate Geometry Class 10 are available in this Maths article. These multiple‑choice questions help students practise the key concepts from Chapter 7 of the CBSE Maths syllabus in an exam‑oriented format. The MCQs with answers and detailed solutions cover distance formula, section formula, midpoint, area of a triangle, and applications in coordinate geometry to strengthen conceptual understanding and improve problem‑solving skills. By practising MCQs on Coordinate Geometry, students can improve accuracy, understand formulas better, and build confidence for board exams.

MCQS on Chapter 7: Coordinate Geometry for Class 10 With Answers

Question 1: The distance between points (−5, 7) and (−1, 3) is:

(a) 3√2

(b) 2√5

(c) 4√2

(d) 5√2

Answer: (c) 4√2

Explanation: Distance = √[(−1−(−5))² + (3−7)²] = √[16 + 16] = √32 = 4√2

Question 2: The distance between the points (a, b) and (−a, −b) is:

(a) √(a² + b²)

(b) a + b

(c) 2√(a² + b²)

(d) 2(a + b)

Answer: (c) 2√(a² + b²)

Explanation: Distance = √[(a−(−a))² + (b−(−b))²] = √[(2a)² + (2b)²] = √[4a² + 4b²] = 2√(a² + b²)

Question 3: If the distance between A(2, −2) and B(−1, x) is 5 units, the value of x is:

(a) 2

(b) −2

(c) 1

(d) −1

Answer: (a) 2

Explanation: √[(−1−2)² + (x+2)²] = 5 ⇒ 9 + (x+2)² = 25 ⇒ (x+2)² = 16 ⇒ x+2 = ±4. Taking the positive root: x = 2.

Question 4: The points (-1,3), (0,0) and (1,-3) are:

(a) Vertices of a triangle

(b) Collinear

(c) Vertices of a right triangle

(d) Vertices of a square

Answer: (b) Collinear

Explanation: The slope between (−1, 3) and (0, 0) is −3, and the slope between (0, 0) and (1, −3) is −3. Since the slopes are the same, the three points are collinear. 

Question 5: The points (5, −2), (6, 4) and (7, −2) are the vertices of:

(a) An equilateral triangle

(b) An isosceles triangle

(c) A right triangle

(d) A scalene triangle

Answer: (b) An isosceles triangle

Explanation: AB = √(1+36) = √37, BC = √(1+36) = √37, AC = √4 = 2. AB = BC, so the triangle is isosceles.

Question 6: The point which divides (−1, 7) and (4, −3) in the ratio 2:3 is:

(a) (−1, 3)

(b) (−1, −3)

(c) (1, −3)

(d) (1, 3)

Answer: (d) (1, 3)

Explanation: x = [2(4) + 3(−1)] / (2+3) = (8−3)/5 = 1

y = [2(−3) + 3(7)] / (2+3) = (−6+21)/5 = 3. Point = (1, 3).

Question 7: The ratio in which (−4, 6) divides A(−6, 10) and B(3, −8) is:

(a) 1:3

(b) 3:4

(c) 2:7

(d) 2:5

Answer: (c) 2:7

Explanation: Let ratio = k:1. Then −4 = (3k − 6)/(k + 1) ⇒ −4k − 4 = 3k − 6 ⇒ 7k = 2 ⇒ k = 2/7. Ratio = 2:7.

Question 8: If O(p/3, 4) is the midpoint of PQ where P(−6, 5) and Q(−2, 3), the value of p is:

(a) 7/2

(b) −12

(c) 4

(d) −4

Answer: (b) −12

Explanation: Midpoint x-coordinate = (−6 + (−2))/2 = −4. So p/3 = −4 ⇒ p = −12.

Question 9: The coordinates of the points of trisection of A (2, −2) and B (−7, 4) are:

(a) (0, 1) and (−4, 2)

(b) (−1, 0) and (−4, 2)

(c) (−1, 0) and (4, −2)

(d) (1, 0) and (4, 2)

Answer: (b) (−1, 0) and (−4, 2)

Explanation: P divides AB in ratio 1:2: x = [1(−7)+2(2)]/3 = −3/3 = −1; y = [1(4)+2(−2)]/3 = 0. P = (−1, 0).

Q divides AB in ratio 2:1: x = [2(−7)+1(2)]/3 = −12/3 = −4; y = [2(4)+1(−2)]/3 = 2. Q = (−4, 2).

Question 10: The area of a rhombus with vertices (3, 0), (4, 5), (−1, 4) and (−2, −1) is:

(a) 12 sq. units

(b) 24 sq. units

(c) 30 sq. units

(d) 32 sq. units

Answer: (b) 24 sq. units

Explanation: Diagonal d₁ = √[(3−(−1))²+(0−4)²] = √(16+16) = 4√2

Diagonal d₂ = √[(4−(−2))²+(5−(−1))²] = √(36+36) = 6√2

Area = ½ × 4√2 × 6√2 = ½ × 48 = 24 sq. units

Question 11: A point on the y-axis equidistant from A(6, 5) and B(−4, 3) is:

(a) (0, 5)

(b) (0, 7)

(c) (0, 3)

(d) (0, 9)

Answer: (d) (0, 9)

Explanation: Let P = (0, y). Set PA² = PB²: 36 + (5−y)² = 16 + (3−y)². 

Expanding: 36+25−10y+y² = 16+9−6y+y²

⇒ 61−10y = 25−6y ⇒ 4y = 36 ⇒ y = 9.

Question 12: A relay tower is placed at P between A(0, 0) and B(36, 15) such that PB = 2PA. The coordinates of P are:

(a) (18, 7.5)

(b) (24, 10)

(c) (12, 5)

(d) (9, 3.75)

Answer: (c) (12, 5)

Explanation: PA:PB = 1:2. By section formula: x = [1(36)+2(0)]/(1+2) = 12; y = [1(15)+2(0)]/3 = 5. P = (12, 5).

Question 13: The distance between towns A and B, where B is 36 km east and 15 km north of A, is:

(a) 39 km

(b) 39 km

(c) 41 km

(d) 45 km

Answer: 39 km

Explanation: Place A at origin, B at (36, 15). Distance = √(36² + 15²) = √(1296 + 225) = √1521 = 39 km. 

Question 14: The ratio in which the y-axis divides the segment joining (5, −6) and (−1, −4) is:

(a) 1:5

(b) 1:1

(c) 1:3

(d) 5:1

Answer: (d) 5:1

Explanation: Let ratio = k:1. The point lies on y-axis, so x = 0. Using section formula: (−k + 5)/(k+1) = 0 ⇒ k = 5. Ratio = 5:1.

Question 15: The coordinates of point P where PQ is the diameter of a circle with centre (2, −3) and Q(1, 4) are:

(a) (3, −10)

(b) (2, −10)

(c) (−3, 10)

(d) (−2, 10)

Answer: (a) (3, −10)

Explanation: Centre = midpoint of PQ. So (x+1)/2 = 2 ⇒ x = 3; (y+4)/2 = −3 ⇒ y = −10. P = (3, −10).

 

Click here to download the free PDF of MCQs worksheet on Chapter 7: Coordinate Geometry for Class 10 Maths based on the updated NCERT & CBSE pattern with important multiple-choice questions and answers.

MCQs Worksheet on Chapter 7: Coordinate Geometry for Class 10