CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 25

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 07 Coordinate Geometry

Review structured MCQ sets for Class 10 Mathematics Chapter 07 Coordinate Geometry. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Practice Chapter 07 Coordinate Geometry MCQs for Class 10 Mathematics

Navigate directly to the 50 objective questions for Chapter 07 Coordinate Geometry using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: The points A(−4, −1), B(−2, −4), C(4, 0) and D(2, 3) are the vertices of a (choose the most specific description)
(a) parallelogram
(b) rectangle
(c) rhombus
(d) None of these
Show Answer & Explanation

Answer: (b) rectangle

Explanation:
1. AB = CD = √13 and BC = DA = 2√13, so opposite sides are equal.
2. The diagonals AC = BD = √65 are equal, so the parallelogram is a rectangle (the sides are unequal, so it is not a rhombus).

Teacher's Note:
1. A parallelogram with equal diagonals is a rectangle.

Question 2: The points (−2, 2), (8, −2) and (−4, −3) are the vertices of
(a) an isosceles triangle
(b) an equilateral triangle
(c) a right-angled triangle
(d) None of these
Show Answer & Explanation

Answer: (c) a right-angled triangle

Explanation:
1. The squared sides are 116, 145 and 29.
2. 29 + 116 = 145, so the triangle is right-angled.

Teacher's Note:
1. Test the Pythagoras relation on squared distances.

Question 3: If the y-axis divides the line joined by (2, 4) and (−3, 5), the ratio is
(a) 2 : 3
(b) 2 : 5
(c) 3 : 2
(d) 5 : 2
Show Answer & Explanation

Answer: (a) 2 : 3

Explanation:
1. On the y-axis x = 0: (m(−3) + n(2))/(m + n) = 0, so 3m = 2n and m : n = 2 : 3.

Teacher's Note:
1. Use the section formula: the point dividing A(x₁, y₁) and B(x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).

Question 4: The ratio in which the line segment joining (3, 4) and (−2, −1) is divided by the x-axis is
(a) 3 : 2
(b) 4 : 1
(c) 4 : 3
(d) None of these
Show Answer & Explanation

Answer: (b) 4 : 1

Explanation:
1. On the x-axis y = 0: (m(−1) + n(4))/(m + n) = 0, so m = 4n and the ratio is 4 : 1.

Teacher's Note:
1. Use the section formula: the point dividing A(x₁, y₁) and B(x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).

Question 5: The line segment joining (3, −4) and (1, 2) is trisected at P and Q. If P = (p, −2) and Q = (5/3, q), then
(a) p = 7/3, q = −3
(b) p = 7/3, q = 0
(c) p = 5/3, q = 0
(d) None of these
Show Answer & Explanation

Answer: (b) p = 7/3, q = 0

Explanation:
1. P divides in 1 : 2: x = (1 + 6)/3 = 7/3, y = (2 − 8)/3 = −2 ✓.
2. Q divides in 2 : 1: x = (2 + 3)/3 = 5/3, y = (4 − 4)/3 = 0.

Teacher's Note:
1. Use the section formula: the point dividing A(x₁, y₁) and B(x₂, y₂) in the ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)).

Question 6: One end of a diameter of a circle is (5, −7) and the centre is (7, 3). The other end is
(a) (6, −2)
(b) (9, 13)
(c) (−2, 6)
(d) (13, 9)
Show Answer & Explanation

Answer: (b) (9, 13)

Explanation:
1. The centre is the midpoint: (5 + x)/2 = 7 and (−7 + y)/2 = 3, so x = 9 and y = 13.

Teacher's Note:
1. Use the midpoint formula in reverse.

Question 7: The point (11, 10) divides the line segment joining (5, −2) and (9, 6) in the ratio
(a) 1 : 3 internally
(b) 1 : 3 externally
(c) 3 : 1 internally
(d) 3 : 1 externally
Show Answer & Explanation

Answer: (d) 3 : 1 externally

Explanation:
1. Along x: (11 − 5)/(9 − 11) = 6/(−2) = −3, a negative ratio, so the division is external with ratio 3 : 1.
2. Check y: (10 + 2)/(6 − 10) = −3 ✓.

Teacher's Note:
1. A point beyond the end of the segment divides it externally.

Question 8: A and B are the points (−3, 4) and (2, 1). The coordinates of the point C on AB produced such that AC = 2BC are
(a) (2, 4)
(b) (3, 7)
(c) (7, −2)
(d) (−1/2, 5/2)
Show Answer & Explanation

Answer: (c) (7, −2)

Explanation:
1. AC = 2BC with C beyond B means B is the midpoint of AC.
2. C = 2B − A = (4 + 3, 2 − 4) = (7, −2).

Teacher's Note:
1. If AC = 2BC and C lies beyond B, then B is the midpoint.

Question 9: Three vertices of a parallelogram are (a + b, a − b), (2a + b, 2a − b) and (a − b, a + b). The fourth vertex is
(a) (−a, a)
(b) (−a, −a)
(c) (−b, −b)
(d) None of these
Show Answer & Explanation

Answer: (d) None of these

Explanation:
1. The possible fourth vertices are (−b, b), (2a + 3b, 2a − 3b) and (2a − b, 2a + b), obtained from P + R − Q for each choice of the opposite pair.
2. None of the options matches.

Teacher's Note:
1. The fourth vertex is the sum of two vertices minus the third (the one opposite to it).

Question 10: (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) (−1, 0)
(b) (0, −1)
(c) (−1, 1)
(d) None of these
Show Answer & Explanation

Answer: (a) (−1, 0)

Explanation:
1. The diagonal's midpoint is (−3/2, 1/2).
2. The fourth vertex = 2(−3/2, 1/2) − (−2, 1) = (−1, 0).

Teacher's Note:
1. Diagonals of a parallelogram bisect each other.

Question 11: The midpoint of the segment joining (−7, 14) and (K, 4) is (a, b), where 2a + 3b = 5. Then K is
(a) 15
(b) −7
(c) −15
(d) 10
Show Answer & Explanation

Answer: (c) −15

Explanation:
1. a = (K − 7)/2 and b = 9. So K − 7 + 27 = 5, giving K = −15.

Teacher's Note:
1. Substitute a and b into the given relation.

Question 12: The points which trisect the line segment joining (0, 0) and (9, 12) are
(a) (3, 4) only
(b) (8, 6) and (4, 0)
(c) (3, 4) and (6, 8)
(d) (4, 0) only
Show Answer & Explanation

Answer: (c) (3, 4) and (6, 8)

Explanation:
1. The trisection points divide the segment in 1 : 2 and 2 : 1.
2. They are (9/3, 12/3) = (3, 4) and (18/3, 24/3) = (6, 8).

Teacher's Note:
1. Two trisection points exist, one from each end.

Question 13: The value of m for which the point P(m, 6) divides the join of A(−4, 3) and B(2, 8) is
(a) 5
(b) 3/2
(c) −2/5
(d) None
Show Answer & Explanation

Answer: (c) −2/5

Explanation:
1. With ratio k : 1, the y-coordinate is (8k + 3)/(k + 1) = 6, so 2k = 3 and k = 3/2.
2. m = (2k − 4)/(k + 1) = (3 − 4)/(5/2) = −2/5.

Teacher's Note:
1. Find the ratio from y first.

Question 14: A(1, 1) and B(2, −3) are two points and D is a point on AB produced such that AD = 3AB. The coordinates of D are
(a) (4, 11)
(b) (4, −11)
(c) (−2, 5)
(d) (−4, −11)
Show Answer & Explanation

Answer: (b) (4, −11)

Explanation:
1. D = A + 3(B − A) = (1 + 3, 1 − 12) = (4, −11).

Teacher's Note:
1. Move three times the vector AB from A.

Question 15: Four points lie on a line segment in the order A, B, C, D with AB : BC = 1 : 2 and BC : CD = 8 : 5. Then AB : BD is
(a) 4 : 13
(b) 1 : 13
(c) 1 : 7
(d) 3 : 13
Show Answer & Explanation

Answer: (a) 4 : 13

Explanation:
1. Take BC = 8, so AB = 4 and CD = 5.
2. BD = 8 + 5 = 13, so AB : BD = 4 : 13.

Teacher's Note:
1. Make BC equal in both ratios.

Question 16: The line joining A(2, 3) and B(−5, 7) is cut by the x-axis at P. Then AP : PB is
(a) 3 : 7
(b) −3 : 7
(c) 7 : 3
(d) 7 : −3
Show Answer & Explanation

Answer: (b) −3 : 7

Explanation:
1. Ratio k : 1 with y = 0: (7k + 3)/(k + 1) = 0 gives k = −3/7.
2. AP : PB = −3 : 7 (external division).

Teacher's Note:
1. A negative ratio means external division.

Question 17: 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 = 4
(b) a = 3, b = 4
(c) a = 2, b = 3
(d) a = 3, b = 5
Show Answer & Explanation

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

Explanation:
1. 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.

Question 18: A(2, 2), B(−4, −4) and C(5, −8) are the vertices of a triangle. The length of the median through C is
(a) √65
(b) √117
(c) √85
(d) √113
Show Answer & Explanation

Answer: (c) √85

Explanation:
1. The midpoint of AB is (−1, −1).
2. Distance to C(5, −8) = √(36 + 49) = √85.

Teacher's Note:
1. A median goes to the midpoint of the opposite side.

Question 19: The midpoints of the sides of a triangle are (−2, 3), (4, −3) and (4, 5). The centroid is
(a) (5/3, 2)
(b) (5/6, 1)
(c) (1, 5/6)
(d) (2, 5/3)
Show Answer & Explanation

Answer: (d) (2, 5/3)

Explanation:
1. The medial triangle has the same centroid as the original triangle: ((−2 + 4 + 4)/3, (3 − 3 + 5)/3) = (2, 5/3).

Teacher's Note:
1. The centroid is the average of any three points that are midpoints of the sides.

Question 20: The circumcentre of the triangle with vertices (2, 1), (5, 2) and (3, 4) is
(a) (13/2, 9/2)
(b) (13/4, 9/4)
(c) (9/4, 13/4)
(d) None
Show Answer & Explanation

Answer: (b) (13/4, 9/4)

Explanation:
1. Equate squared distances: (x − 2)² + (y − 1)² = (x − 5)² + (y − 2)² gives 6x + 2y = 28.
2. (x − 2)² + (y − 1)² = (x − 3)² + (y − 4)² gives 2x + 6y = 20.
3. Solving: x = 13/4, y = 9/4.

Teacher's Note:
1. The circumcentre is equidistant from all three vertices.

Question 21: The centroid of a triangle is (1, 3) and two of its vertices are (−7, 6) and (8, 5). The third vertex is
(a) (2/3, 14/3)
(b) (−2/3, −14/3)
(c) (2, −2)
(d) (−2, 2)
Show Answer & Explanation

Answer: (c) (2, −2)

Explanation:
1. x: 3 − (−7 + 8) = 2; y: 9 − (6 + 5) = −2.

Teacher's Note:
1. Sum of vertex coordinates = 3 × centroid.

Question 22: The incentre of the triangle with vertices (0, 0), (5, 12) and (16, 12) is
(a) (7, 9)
(b) (9, 7)
(c) (−9, 7)
(d) (−7, 9)
Show Answer & Explanation

Answer: (a) (7, 9)

Explanation:
1. The side lengths opposite to (0, 0), (5, 12), (16, 12) are 11, 20 and 13 respectively.
2. Incentre = (11·0 + 20·5 + 13·16, 11·0 + 20·12 + 13·12)/44 = (308/44, 396/44) = (7, 9).

Teacher's Note:
1. Incentre = weighted average of the vertices with the opposite side lengths as weights.

Question 23: The centroid and circumcentre of a triangle are (3, 3) and (6, 2). The orthocentre is
(a) (−3, 5)
(b) (−3, 1)
(c) (3, −1)
(d) (9, 5)
Show Answer & Explanation

Answer: (a) (−3, 5)

Explanation:
1. The centroid divides the line from orthocentre to circumcentre in 2 : 1, so H = 3G − 2S.
2. H = (9 − 12, 9 − 4) = (−3, 5).

Teacher's Note:
1. Euler line: H = 3G − 2S.

Question 24: The incentre of the triangle with vertices (1, √3), (0, 0) and (2, 0) is
(a) (1, √3/2)
(b) (2/3, 1/√3)
(c) (2/3, √3/2)
(d) (1, 1/√3)
Show Answer & Explanation

Answer: (d) (1, 1/√3)

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

Teacher's Note:
1. For an equilateral triangle all the centres coincide.

Question 25: The orthocentre of the triangle with vertices (0, 0), (3, 4) and (4, 0) is
(a) (−3, 3/4)
(b) (3, 12)
(c) (3, 3/4)
(d) (3, 9)
Show Answer & Explanation

Answer: (c) (3, 3/4)

Explanation:
1. The altitude from (3, 4) is x = 3 (perpendicular to the base on the x-axis).
2. The side from (3, 4) to (4, 0) has slope −4, so the altitude from the origin has slope 1/4: y = x/4.
3. At x = 3, y = 3/4.

Teacher's Note:
1. Intersect two altitudes.

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Multiple Choice Questions (MCQs) for Class 10 Mathematics Chapter 07 Coordinate Geometry

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