CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 20

Practice MCQs for Class 10 Mathematics Chapter 07 Coordinate Geometry

Access targeted multiple-choice questions for Chapter 07 Coordinate Geometry designed to align with the latest CBSE academic syllabus for Class 10 Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Access Chapter 07 Coordinate Geometry Questions and Solutions

View or download the dedicated Chapter 07 Coordinate Geometry MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: A(4, 6), B(1, 5) and C(7, 2) are the vertices of a triangle. D lies on AB and E on AC with AD/AB = AE/AC = 1/4. The area of triangle ADE is:
(a) 15/8 square units
(b) 15/32 square units
(c) 15/16 square units
(d) 15/64 square units
Show Answer & Explanation

Answer: (b) 15/32 square units

Explanation:
1. Area ABC = ½ |12 − 4 + 7| = 15/2.
2. Area ADE = (1/4)² × 15/2 = 15/32.
3. Ratio area ADE : area ABC = 1 : 16.

Teacher's Note:
1. Find the area of ABC first.
2. ADE is similar to ABC with ratio 1/4.

Question 2: For a = 1, t = 2 let A = (at², 2at), B = (a/t², −2a/t) and C = (a, 0). The value of \( \frac{1}{AC} + \frac{1}{BC} \) is:
(a) 1
(b) 2
(c) 1/2
(d) 5/4
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. A = (4, 4), B = (1/4, −1), C = (1, 0). AC = √(9 + 16) = 5 and BC = √(9/16 + 1) = 5/4. So 1/5 + 4/5 = 1 = 1/a.

Teacher's Note:
1. Compute the coordinates for a = 1, t = 2.
2. Use the distance formula for AC and BC.

Question 3: The vertices of triangle ABC are A(−2, −5), B(−2, 11) and C(10, −1). D and E are the midpoints of AB and AC. The length DE is:
(a) \( 12\sqrt{2} \)
(b) \( 6 \)
(c) \( 12 \)
(d) \( 6\sqrt{2} \)
Show Answer & Explanation

Answer: (d) \( 6\sqrt{2} \)

Explanation:
1. D = (−2, 3) and E = (4, −3). DE = √(36 + 36) = 6√2. Also BC = √(144 + 144) = 12√2, so DE = BC/2.

Teacher's Note:
1. Find the midpoints first.
2. Apply the distance formula.

Question 4: A(1, 2) and B(1, 3) are two points and C(x, y) is a point other than A on the line AB with AB = BC. Then C is:
(a) (1, 1)
(b) (2, 3)
(c) (1, 4)
(d) (0, 3)
Show Answer & Explanation

Answer: (c) (1, 4)

Explanation:
1. A, B and C lie on x = 1. AB = 1, so C is 1 unit from B on the line x = 1, giving (1, 4) or (1, 2). Since C ≠ A, C = (1, 4).

Teacher's Note:
1. The line through A and B is x = 1.
2. B is the midpoint of AC.

Question 5: X, Y, Z are the midpoints of the sides PQ, QR, RP of triangle PQR with P(0, 0), Q(6, 0), R(0, 8). The perimeters of triangle XYZ and triangle PQR are respectively:
(a) 24 and 12
(b) 12 and 24
(c) 10 and 24
(d) 12 and 20
Show Answer & Explanation

Answer: (b) 12 and 24

Explanation:
1. PQ = 6, PR = 8, QR = 10, so the perimeter of PQR is 24. The medial triangle has half the perimeter: 12.

Teacher's Note:
1. The midpoint triangle has sides half those of PQR.
2. QR = 10.

Question 6: A(3, 2) and B(−2, 1) are two vertices of triangle ABC with centroid G(5/3, −1/3). The third vertex C is:
(a) (4, −4)
(b) (4, 4)
(c) (−4, 4)
(d) (2, −2)
Show Answer & Explanation

Answer: (a) (4, −4)

Explanation:
1. x: 3 − 2 + x₃ = 5 gives x₃ = 4. y: 2 + 1 + y₃ = −1 gives y₃ = −4.

Teacher's Note:
1. Sum of x-coordinates = 3 × centroid x.
2. Likewise for y.

Question 7: The incentre of the triangle with vertices A(0, 0), B(4, 0) and C(0, 3) is:
(a) (2, 1.5)
(b) (4/3, 1)
(c) (1, 2)
(d) (1, 1)
Show Answer & Explanation

Answer: (d) (1, 1)

Explanation:
1. The sides are BC = 5 (opposite A), CA = 3 (opposite B) and AB = 4 (opposite C). x = (5·0 + 3·4 + 4·0)/12 = 1 and y = (5·0 + 3·0 + 4·3)/12 = 1.

Teacher's Note:
1. Compute the side lengths and use the incentre formula.
2. For a right triangle, r = (3 + 4 − 5)/2 = 1.

Question 8: If (α, β) is the midpoint of the segment joining (3, 4) and (5, −9), the value of \( \alpha + 2\beta \) is:
(a) 1
(b) 0
(c) −1
(d) −2
Show Answer & Explanation

Answer: (c) −1

Explanation:
1. α = 4 and β = −5/2, so α + 2β = 4 − 5 = −1. (So α + 2β + 1 = 0.)

Teacher's Note:
1. Midpoint = average of the coordinates.

Question 9: The line 3x + y − 9 = 0 divides the segment joining (1, 3) and (2, 7) in the ratio:
(a) 4 : 3
(b) 3 : 4
(c) 3 : 2
(d) 2 : 3
Show Answer & Explanation

Answer: (b) 3 : 4

Explanation:
1. At (1, 3): 3 + 3 − 9 = −3. At (2, 7): 6 + 7 − 9 = 4.
2. k = −(−3)/4 = 3/4.
3. The ratio is 3 : 4.

Teacher's Note:
1. Use k = −(ax₁ + by₁ + c)/(ax₂ + by₂ + c).
2. Evaluate the line expression at both points.

Question 10: The ratio in which the x-axis divides the segment joining A(6, 3) and B(−2, −5) is:
(a) 3 : 5
(b) 5 : 3
(c) 3 : 2
(d) 1 : 1
Show Answer & Explanation

Answer: (a) 3 : 5

Explanation:
1. The x-axis divides in the ratio −y₁ : y₂ = −3 : −5 = 3 : 5.

Teacher's Note:
1. On the x-axis the y-coordinate is zero.

Question 11: If the line segment joining A(a, b) and B(c, d) subtends angle θ at the origin, then for A(3, 4) and B(4, 3) the value of cos θ is:
(a) 12/25
(b) 7/25
(c) 1
(d) 24/25
Show Answer & Explanation

Answer: (d) 24/25

Explanation:
1. cos θ = (ac + bd)/(√(a² + b²)√(c² + d²)) = (12 + 12)/(5 × 5) = 24/25.

Teacher's Note:
1. Use the dot product formula.
2. Both distances from the origin are 5.

Question 12: The centre of the circle passing through A(3, −7), B(6, 6) and C(3, 3) is:
(a) (3, −2)
(b) (6, 3)
(c) (11, −2)
(d) (4.5, 1)
Show Answer & Explanation

Answer: (c) (11, −2)

Explanation:
1. Centre (h, k): the perpendicular bisector of AC (vertical chord) is y = −2.
2. (h − 3)² + (−2 + 7)² = (h − 6)² + (−2 − 6)²: −6h + 34 = −12h + 100.
3. h = 11. Radius² = 64 + 25 = 89 (check with C).

Teacher's Note:
1. Perpendicular bisector of AC is the line y = −2.
2. Equate the distances to A and B.

Question 13: The midpoints of the sides of a triangle are (10, 5), (8, 4) and (6, 6). The vertices are:
(a) (4, 5), (8, 7), (12, 8)
(b) (12, 3), (8, 7), (4, 5)
(c) (12, 3), (8, 5), (4, 7)
(d) (10, 5), (6, 6), (8, 4)
Show Answer & Explanation

Answer: (b) (12, 3), (8, 7), (4, 5)

Explanation:
1. With midpoints M₁, M₂, M₃ the vertices are M₁ + M₂ − M₃, M₁ + M₃ − M₂, M₂ + M₃ − M₁: (12, 3), (8, 7), (4, 5).

Teacher's Note:
1. The medial triangle's vertices are the midpoints.
2. Each vertex = sum of two midpoints minus the third.

Question 14: The centroid of the triangle formed by P(a, b), Q(b, c), R(c, a) is the origin. The value of \( \frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} \) is:
(a) 3
(b) 0
(c) 1
(d) abc
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Centroid at the origin gives a + b + c = 0. The sum equals (a³ + b³ + c³)/(abc) = 3abc/abc = 3.

Teacher's Note:
1. a + b + c = 0.
2. Then a³ + b³ + c³ = 3abc.

Question 15: The point dividing the line joining (a + b, a − b) to (a − b, a + b) internally in the ratio a : b is:
(a) \( \left(\frac{a^2-b^2}{a+b},\ \frac{a^2+2ab+b^2}{a+b}\right) \)
(b) \( \left(\frac{a^2+b^2}{a-b},\ \frac{a^2+2ab-b^2}{a-b}\right) \)
(c) \( \left(\frac{a^2+2ab-b^2}{a+b},\ \frac{a^2+b^2}{a+b}\right) \)
(d) \( \left(\frac{a^2+b^2}{a+b},\ \frac{a^2+2ab-b^2}{a+b}\right) \)
Show Answer & Explanation

Answer: (d) \( \left(\frac{a^2+b^2}{a+b},\ \frac{a^2+2ab-b^2}{a+b}\right) \)

Explanation:
1. x = [a(a − b) + b(a + b)]/(a + b) = (a² + b²)/(a + b). y = [a(a + b) + b(a − b)]/(a + b) = (a² + 2ab − b²)/(a + b).

Teacher's Note:
1. Use the section formula with m = a, n = b.
2. Simplify the numerators.

Question 16: The perimeter of the triangle with vertices (−1, 4), (−4, −2) and (3, −4) is:
(a) 38
(b) 16
(c) \( 7\sqrt{5} + \sqrt{53} \)
(d) 42
Show Answer & Explanation

Answer: (c) \( 7\sqrt{5} + \sqrt{53} \)

Explanation:
1. The sides are √(9 + 36) = 3√5, √(49 + 4) = √53 and √(16 + 64) = 4√5. The perimeter is 7√5 + √53 ≈ 22.9.

Teacher's Note:
1. Apply the distance formula to each pair.

Question 17: The incentre of the triangle with vertices (1, √3), (0, 0) and (2, 0) is:
(a) \( \left(1,\ \frac{\sqrt{3}}{2}\right) \)
(b) \( \left(1,\ \frac{1}{\sqrt{3}}\right) \)
(c) \( \left(\frac{2}{3},\ \frac{1}{\sqrt{3}}\right) \)
(d) \( \left(\frac{2}{3},\ \frac{\sqrt{3}}{2}\right) \)
Show Answer & Explanation

Answer: (b) \( \left(1,\ \frac{1}{\sqrt{3}}\right) \)

Explanation:
1. All sides equal 2, so the triangle is equilateral and the incentre equals the centroid ((1 + 0 + 2)/3, (√3 + 0 + 0)/3) = (1, 1/√3).

Teacher's Note:
1. Check that all sides are 2.
2. In an equilateral triangle the incentre is the centroid.

Question 18: P(1, 2), Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS. Then:
(a) a = 2, b = 3
(b) a = 2, b = 4
(c) a = 3, b = 4
(d) a = 3, b = 5
Show Answer & Explanation

Answer: (a) a = 2, b = 3

Explanation:
1. The diagonals PR and QS bisect each other: (1 + 5)/2 = (4 + a)/2 gives a = 2, and (2 + 7)/2 = (6 + b)/2 gives b = 3.

Teacher's Note:
1. Equate the midpoints of the diagonals PR and QS.

Question 19: The points (a, 0), (0, b) and (1, 1) are collinear if:
(a) \( \frac{1}{a^2} + \frac{1}{b^2} = 1 \)
(b) \( \frac{1}{a} - \frac{1}{b} = 1 \)
(c) \( \frac{1}{a^2} - \frac{1}{b^2} = 1 \)
(d) \( \frac{1}{a} + \frac{1}{b} = 1 \)
Show Answer & Explanation

Answer: (d) \( \frac{1}{a} + \frac{1}{b} = 1 \)

Explanation:
1. The line through (a, 0) and (0, b) is x/a + y/b = 1. The point (1, 1) lies on it if 1/a + 1/b = 1.

Teacher's Note:
1. Write the intercept form of the line.
2. Substitute (1, 1).

Question 20: (3, −4) and (−6, 5) are the ends of a diagonal of a parallelogram and (−2, 1) is a third vertex. The fourth vertex is:
(a) (0, −1)
(b) (−1, 1)
(c) (−1, 0)
(d) (1, 0)
Show Answer & Explanation

Answer: (c) (−1, 0)

Explanation:
1. The midpoint of the given diagonal is (−3/2, 1/2). The fourth vertex is the reflection of (−2, 1) in it: (−3 + 2, 1 − 1) = (−1, 0).

Teacher's Note:
1. Diagonals bisect each other.

Question 21: The area of the triangle with vertices (a, b + c), (b, c + a), (c, a + b) is:
(a) a + b + c
(b) 0
(c) ab + bc + ca
(d) abc
Show Answer & Explanation

Answer: (b) 0

Explanation:
1. The points satisfy x + y = a + b + c, so they lie on one line and the area is 0.

Teacher's Note:
1. Add the coordinates of each point.

Question 22: Point A lies on the x-axis with abscissa p + q and point B lies on the y-axis with ordinate p − q. The distance AB is:
(a) \( \sqrt{2(p^2+q^2)} \)
(b) \( 2(p^2+q^2) \)
(c) \( \sqrt{2p^2+q^2} \)
(d) \( \sqrt{p^2+q^2} \)
Show Answer & Explanation

Answer: (a) \( \sqrt{2(p^2+q^2)} \)

Explanation:
1. A = (p + q, 0), B = (0, p − q). AB² = (p + q)² + (p − q)² = 2(p² + q²).

Teacher's Note:
1. Write the coordinates of A and B.

Question 23: The centroid, circumcentre and orthocentre of a triangle are:
(a) always coincident
(b) always forming a triangle
(c) always collinear but never coincident
(d) coincident in an equilateral triangle, otherwise collinear
Show Answer & Explanation

Answer: (d) coincident in an equilateral triangle, otherwise collinear

Explanation:
1. They lie on the Euler line in every triangle, and they coincide only when the triangle is equilateral.

Teacher's Note:
1. The Euler line contains all three.

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Chapter 07 Coordinate Geometry Objective Questions & Solutions for Class 10 Mathematics

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