CBSE Class 10 Mathematics Coordinate Geometry MCQs Set 21

Mathematics Objective Questions and Answers: 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.

Download Chapter 07 Coordinate Geometry MCQs with Answers

Access the complete set of multiple-choice questions for Chapter 07 Coordinate Geometry below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: The distance between the points (-6, 7) and (-1, -5) is:
(a) \( 13 \)
(b) \( 11 \)
(c) \( 12 \)
(d) \( 15 \)
Show Answer & Explanation

Answer: (a) \( 13 \)

Explanation:
1. d = √[(-1 + 6)² + (-5 - 7)²] = √(25 + 144) = √169 = 13.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 2: The distance between the points (a + b, b + c) and (a - b, c - b) is:
(a) \( 2b \)
(b) \( b\sqrt2 \)
(c) \( 4b \)
(d) \( 2b\sqrt2 \)
Show Answer & Explanation

Answer: (d) \( 2b\sqrt2 \)

Explanation:
1. The difference in x is 2b and the difference in y is also 2b. So d = √(4b² + 4b²) = 2b√2.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 3: The distance between the points (a, 0) and (0, b) is:
(a) \( a+b \)
(b) \( \sqrt{a^2-b^2} \)
(c) \( \sqrt{a^2+b^2} \)
(d) \( a^2+b^2 \)
Show Answer & Explanation

Answer: (c) \( \sqrt{a^2+b^2} \)

Explanation:
1. d = √[(0 - a)² + (b - 0)²] = √(a² + b²).

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 4: The distance between the points (a cos t + b sin t, 0) and (0, a sin t - b cos t) is:
(a) \( a+b \)
(b) \( \sqrt{a^2+b^2} \)
(c) \( \sqrt{a^2-b^2} \)
(d) \( ab \)
Show Answer & Explanation

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

Explanation:
1. d² = (a cos t + b sin t)² + (a sin t - b cos t)² = a²(cos²t + sin²t) + b²(sin²t + cos²t) = a² + b².

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 5: The distance between the points (cos t, sin t) and (sin t, -cos t) is:
(a) \( 1 \)
(b) \( \sqrt2 \)
(c) \( 2 \)
(d) \( 2\sqrt2 \)
Show Answer & Explanation

Answer: (b) \( \sqrt2 \)

Explanation:
1. d² = (cos t - sin t)² + (sin t + cos t)² = 2(cos²t + sin²t) = 2, so d = √2.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 6: If the distance between the points (3, a) and (4, 1) is √10, a possible value of a is:
(a) \( 4 \)
(b) \( 2 \)
(c) \( 3 \)
(d) \( -4 \)
Show Answer & Explanation

Answer: (a) \( 4 \)

Explanation:
1. 1 + (a - 1)² = 10, so (a - 1)² = 9 and a = 4 or a = -2. Of the options, only 4 is a possible value.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 7: If the distance between the points (4, p) and (1, 0) is 5, a possible value of p is:
(a) \( 3 \)
(b) \( 5 \)
(c) \( 25 \)
(d) \( 4 \)
Show Answer & Explanation

Answer: (d) \( 4 \)

Explanation:
1. (4 - 1)² + p² = 25, so p² = 16 and p = 4 or -4. Of the options only 4 is possible.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 8: The point on the x-axis equidistant from (5, 9) and (-4, 6) is:
(a) \( (-3, 0) \)
(b) \( (0, 3) \)
(c) \( (3, 0) \)
(d) \( (2, 0) \)
Show Answer & Explanation

Answer: (c) \( (3, 0) \)

Explanation:
1. Let the point be (x, 0). (x - 5)² + 81 = (x + 4)² + 36, so 106 - 10x = 52 + 8x, giving x = 3.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 9: The point on the x-axis equidistant from (-3, 4) and (2, 5) is:
(a) \( (\frac{5}{2}, 0) \)
(b) \( (- \frac{2}{5}, 0) \)
(c) \( (\frac{2}{5}, 0) \)
(d) \( (0, \frac{2}{5}) \)
Show Answer & Explanation

Answer: (c) \( (\frac{2}{5}, 0) \)

Explanation:
1. Let the point be (x, 0). (x + 3)² + 16 = (x - 2)² + 25 gives (x + 3)² - (x - 2)² = 9, i.e. 10x + 5 = 9, so x = 2/5.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 10: The points (-2, 5), (0, 1) and (2, -3) are:
(a) the vertices of an equilateral triangle
(b) collinear
(c) the vertices of a right-angled triangle
(d) the vertices of an isosceles triangle that is not right-angled
Show Answer & Explanation

Answer: (b) collinear

Explanation:
1. The distances are 2√5 (first to second), 2√5 (second to third) and 4√5 (first to third). Since 2√5 + 2√5 = 4√5, the points lie on one straight line.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 11: Three consecutive vertices of a parallelogram are (a + b, a - b), (2a + b, 2a - b) and (a - b, a + b). The fourth vertex is:
(a) \( (-b, b) \)
(b) \( (b, -b) \)
(c) \( (3a + b, 3a - b) \)
(d) \( (-b, -b) \)
Show Answer & Explanation

Answer: (a) \( (-b, b) \)

Explanation:
1. The diagonals bisect each other, so D = A + C - B. x: (a + b) + (a - b) - (2a + b) = -b. y: (a - b) + (a + b) - (2a - b) = b. So D = (-b, b).

Teacher's Note:
1. In a parallelogram ABCD the midpoints of the diagonals AC and BD coincide.

Question 12: If the mid-points of the sides of a triangle are (1, 1), (2, -3) and (3, 4), the vertices of the triangle are:
(a) \( (2, 10), (0, -4), (4, 2) \)
(b) \( (1, 1), (2, -3), (3, 4) \)
(c) \( (4, 6), (-2, 2), (0, -2) \)
(d) \( (2, 8), (0, -6), (4, 0) \)
Show Answer & Explanation

Answer: (d) \( (2, 8), (0, -6), (4, 0) \)

Explanation:
1. Each vertex equals the sum of two mid-points minus the third. (1, 1) + (3, 4) - (2, -3) = (2, 8); (1, 1) + (2, -3) - (3, 4) = (0, -6); (2, -3) + (3, 4) - (1, 1) = (4, 0). Check: the midpoint of (2, 8) and (0, -6) is (1, 1).

Teacher's Note:
1. The vertices of a triangle are the sums of two mid-points minus the third mid-point.

Question 13: The centroid of the triangle with vertices (1, 4), (-1, -1) and (3, -2) is:
(a) \( (\frac{1}{3}, 1) \)
(b) \( (1, 3) \)
(c) \( (3, \frac{1}{3}) \)
(d) \( (1, \frac{1}{3}) \)
Show Answer & Explanation

Answer: (d) \( (1, \frac{1}{3}) \)

Explanation:
1. Centroid = ((1 - 1 + 3)/3, (4 - 1 - 2)/3) = (1, 1/3).

Teacher's Note:
1. The centroid of a triangle is the mean of the x-coordinates and the mean of the y-coordinates of the vertices.

Question 14: The centroid of the triangle with vertices (-2, 3), (2, -1) and (4, 0) is:
(a) \( (\frac{2}{3}, \frac{4}{3}) \)
(b) \( (4, 2) \)
(c) \( (\frac{4}{3}, \frac{2}{3}) \)
(d) \( (\frac{4}{3}, \frac{5}{3}) \)
Show Answer & Explanation

Answer: (c) \( (\frac{4}{3}, \frac{2}{3}) \)

Explanation:
1. Centroid = ((-2 + 2 + 4)/3, (3 - 1 + 0)/3) = (4/3, 2/3).

Teacher's Note:
1. The centroid of a triangle is the mean of the x-coordinates and the mean of the y-coordinates of the vertices.

Question 15: Two vertices of a triangle are (1, 2) and (3, 5) and its centroid is at the origin. The third vertex is:
(a) \( (4, 7) \)
(b) \( (-4, -7) \)
(c) \( (-4, 7) \)
(d) \( (-3, -7) \)
Show Answer & Explanation

Answer: (b) \( (-4, -7) \)

Explanation:
1. If the third vertex is (x, y), (1 + 3 + x)/3 = 0 and (2 + 5 + y)/3 = 0, so x = -4 and y = -7.

Teacher's Note:
1. The centroid of a triangle is the mean of the x-coordinates and the mean of the y-coordinates of the vertices.

Question 16: If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, its centroid is:
(a) \( (2, \frac{5}{3}) \)
(b) \( (\frac{5}{3}, 2) \)
(c) \( (2, 5) \)
(d) \( (\frac{4}{3}, \frac{5}{3}) \)
Show Answer & Explanation

Answer: (a) \( (2, \frac{5}{3}) \)

Explanation:
1. The centroid of the triangle formed by the midpoints is the same point as the centroid of the original triangle. It is ((-2 + 4 + 4)/3, (3 - 3 + 5)/3) = (2, 5/3).

Teacher's Note:
1. The medial triangle has the same centroid as the original triangle.

Question 17: The area of the triangle with vertices (6, 3), (-3, 5) and (4, -2) is:
(a) \( \frac{49}{2} \) sq. units
(b) \( \frac{47}{2} \) sq. units
(c) \( \frac{51}{2} \) sq. units
(d) \( 49 \) sq. units
Show Answer & Explanation

Answer: (a) \( \frac{49}{2} \) sq. units

Explanation:
1. Area = (1/2)[6(5 + 2) + (-3)(-2 - 3) + 4(3 - 5)] = (1/2)(42 + 15 - 8) = 49/2 square units.

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Question 18: The area of the triangle with vertices (a, c + a), (a, c) and (-a, c - a) is:
(a) \( 2a^2 \) sq. units
(b) \( \frac{a^2}{2} \) sq. units
(c) \( a^2+c^2 \) sq. units
(d) \( a^2 \) sq. units
Show Answer & Explanation

Answer: (d) \( a^2 \) sq. units

Explanation:
1. Area = (1/2)[a(c - (c - a)) + a((c - a) - (c + a)) + (-a)((c + a) - c)] = (1/2)[a² - 2a² - a²]; taking the absolute value gives (1/2)(2a²) = a².

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Question 19: The vertices of a square ABCD of side 2a are placed with its centre at the origin and its sides parallel to the axes. A possible set of vertices in order A, B, C, D is:
(a) \( A(-a,-a), B(a,a), C(a,-a), D(-a,a) \)
(b) \( A(0,0), B(a,0), C(a,a), D(0,a) \)
(c) \( A(-a,-a), B(a,-a), C(a,a), D(-a,a) \)
(d) \( A(-a,0), B(0,-a), C(a,0), D(0,a) \)
Show Answer & Explanation

Answer: (c) \( A(-a,-a), B(a,-a), C(a,a), D(-a,a) \)

Explanation:
1. Since the centre is the origin and the side is 2a, each vertex is a distance a from both axes. Going around in order gives (-a, -a), (a, -a), (a, a), (-a, a).

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 20: The points (-4, -1), (-2, -4), (4, 0) and (2, 3), taken in order, are the vertices of a:
(a) square
(b) rectangle
(c) rhombus that is not a square
(d) parallelogram that is not a rectangle
Show Answer & Explanation

Answer: (b) rectangle

Explanation:
1. With A(-4, -1), B(-2, -4), C(4, 0), D(2, 3): AB = √13, BC = √(36 + 16) = √52, CD = √(4 + 9) = √13 and DA = √(36 + 16) = √52. The diagonals AC = √(64 + 1) = √65 and BD = √(16 + 49) = √65 are equal. Opposite sides are equal and the diagonals are equal, so it is a rectangle; it is not a square because AB differs from BC.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 21: The points A(1, -2), B(3, 6), C(5, 10) and D(3, 2), taken in order, form a:
(a) rectangle
(b) parallelogram that is not a rectangle
(c) rhombus
(d) square
Show Answer & Explanation

Answer: (b) parallelogram that is not a rectangle

Explanation:
1. The vector AB = (2, 8) and DC = (2, 8), so AB and DC are equal and parallel. Also AB = √68 and BC = √(4 + 16) = √20, which differ, and the diagonals AC = √(16 + 144) and BD = √(0 + 16) are unequal. So it is a parallelogram that is not a rectangle.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 22: The points A(0, 1), B(1, 4), C(4, 3) and D(3, 0), taken in order, form a:
(a) square
(b) rectangle that is not a square
(c) rhombus that is not a square
(d) parallelogram that is not a rhombus
Show Answer & Explanation

Answer: (a) square

Explanation:
1. AB = BC = CD = DA = √10, and the diagonals AC = √(16 + 4)= √20 and BD = √(4 + 16) = √20 are equal. Equal sides and equal diagonals give a square.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 23: The points (3, 0), (6, 4) and (-1, 3) are the vertices of:
(a) an equilateral triangle
(b) a right-angled triangle that is not isosceles
(c) an isosceles triangle that is not right-angled
(d) a right-angled isosceles triangle
Show Answer & Explanation

Answer: (d) a right-angled isosceles triangle

Explanation:
1. With A(3, 0), B(6, 4), C(-1, 3): AB = 5, AC = √(16 + 9) = 5, BC = √(49 + 1) = √50. AB = AC and AB² + AC² = 50 = BC², so the triangle is right-angled and isosceles.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 24: The points (2, -2), (-2, 1) and (5, 2) are the vertices of a right-angled triangle. Its area and hypotenuse are:
(a) \( 25 \) sq. units and 5√2
(b) \( \frac{25}{2} \) sq. units and 5
(c) \( \frac{25}{2} \) sq. units and 5√2
(d) \( \frac{35}{2} \) sq. units and 5√2
Show Answer & Explanation

Answer: (c) \( \frac{25}{2} \) sq. units and 5√2

Explanation:
1. Let A(2, -2), B(-2, 1), C(5, 2). AB = 5, AC = 5, BC = √(49 + 1) = 5√2. Right angle at A (25 + 25 = 50), BC is the hypotenuse and area = (1/2)(5)(5) = 25/2.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 25: The points (2, 3), (-4, -6) and (1, 3/2) are:
(a) the vertices of a right-angled triangle
(b) the vertices of an isosceles triangle
(c) collinear
(d) the vertices of a triangle of area 3 sq. units
Show Answer & Explanation

Answer: (c) collinear

Explanation:
1. Area = (1/2)[2(-6 - 3/2) + (-4)(3/2 - 3) + 1(3 + 6)] = (1/2)[-15 + 6 + 9] = 0. The area is zero, so the points are collinear.

Teacher's Note:
1. Area of a triangle = the absolute value of (1/2)[x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)]. Three points are collinear exactly when this area is zero.

Question 26: The points (2, -1), (3, 4), (-2, 3) and (-3, -2), taken in order, form a:
(a) square
(b) rhombus that is not a square
(c) rectangle that is not a square
(d) trapezium that is not a parallelogram
Show Answer & Explanation

Answer: (b) rhombus that is not a square

Explanation:
1. Sides: √(1 + 25) = √26 for each of AB, BC, CD, DA. Diagonals: AC = √(16 + 16) = 4√2 and BD = √(36 + 36) = 6√2, which are unequal. Equal sides with unequal diagonals give a rhombus but not a square.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

Question 27: The points A(5, 6), B(1, 5), C(2, 1) and D(6, 2) are the vertices of a:
(a) square
(b) rhombus that is not a square
(c) rectangle that is not a square
(d) kite
Show Answer & Explanation

Answer: (a) square

Explanation:
1. AB = BC = CD = DA = √17 and AC = √(9 + 25) = √34 = BD. Equal sides and equal diagonals make it a square.

Teacher's Note:
1. The distance between (x1, y1) and (x2, y2) is the square root of (x2 - x1)^2 + (y2 - y1)^2.

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

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