CBSE Class 8 Mathematics Coordinate Geometry MCQs Set 29

Mathematics Objective Questions and Answers: Coordinate Geometry

Review structured MCQ sets for Class 8 Mathematics Coordinate Geometry. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Download Coordinate Geometry MCQs with Answers

Navigate directly to the 50 objective questions for 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: Two vertices of an equilateral triangle are (a, 0) and (−a, 0). The third vertex is
(a) (0, ±2a)
(b) (0, ±√3 a)
(c) (0, ± a√2)
(d) (±√3 a, 0)
Show Answer & Explanation

Answer: (b) (0, ±√3 a)

Explanation:
1. The third vertex lies on the perpendicular bisector of the base, the y-axis, so it is (0, y). Then a² + y² = (2a)², so y² = 3a² and y = ±√3 a.

Teacher's Note:
1. The base has length 2a.
2. The height of an equilateral triangle of side 2a is √3 a.

Question 2: The points A(5, 6), B(1, 5), C(2, 1) and D(6, 2) are the vertices of a square. Its side is
(a) √13
(b) √15
(c) 5
(d) √17
Show Answer & Explanation

Answer: (d) √17

Explanation:
1. AB = √((5 − 1)² + (6 − 5)²) = √17; similarly BC = CD = DA = √17, and the diagonals AC = BD = √34. All sides are equal and the diagonals are equal, so ABCD is a square of side √17.

Teacher's Note:
1. A quadrilateral with four equal sides and equal diagonals is a square.
2. Use the distance formula.

Question 3: The distance between the points (−5, 7) and (−1, 3) is
(a) √34
(b) 8
(c) 4√2
(d) 2√2
Show Answer & Explanation

Answer: (c) 4√2

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

Teacher's Note:
1. Square the differences of both coordinates.
2. Simplify the root at the end.

Question 4: AB is a diameter of a circle with centre (2, −3). If B is (1, 4), the coordinates of A are
(a) (3, 10)
(b) (1.5, 0.5)
(c) (3, −10)
(d) (−3, 10)
Show Answer & Explanation

Answer: (c) (3, −10)

Explanation:
1. The centre is the mid-point of AB. If A = (x, y), then (x + 1)/2 = 2 and (y + 4)/2 = −3, so x = 3 and y = −10.

Teacher's Note:
1. Use the mid-point formula backwards.
2. Check: the mid-point of (3, −10) and (1, 4) is (2, −3).

Question 5: The points A(2, −2), B(14, 10), C(11, 13) and D(−1, 1) are the vertices of a rectangle. Its sides are
(a) 12√2 and 3√2
(b) 6√2 and 3√2
(c) 12√2 and 6√2
(d) 4√2 and 3√2
Show Answer & Explanation

Answer: (a) 12√2 and 3√2

Explanation:
1. AB = CD = √(12² + 12²) = 12√2 and BC = DA = √(3² + 3²) = 3√2. The diagonals AC = BD = √(9² + 15²) = √306, so opposite sides are equal and the diagonals are equal: ABCD is a rectangle with sides 12√2 and 3√2.

Teacher's Note:
1. Show opposite sides equal and diagonals equal.
2. Then read the side lengths.

Question 6: The points (0, 0), (5, 5) and (−5, 5) are the vertices of a right isosceles triangle. Its sides are
(a) √40, √40 and 5
(b) √30, √30 and 6
(c) √30, √30 and 8
(d) √50, √50 and 10
Show Answer & Explanation

Answer: (d) √50, √50 and 10

Explanation:
1. The distance from (0, 0) to (5, 5) is √50, from (0, 0) to (−5, 5) is √50, and from (5, 5) to (−5, 5) is 10. Since 50 + 50 = 100 = 10², the triangle is right-angled at the origin.

Teacher's Note:
1. Compute all three distances.
2. Confirm the Pythagoras relation.

Question 7: The distance between the points (−6, 7) and (−1, −5) is
(a) 12
(b) 13
(c) 11
(d) √119
Show Answer & Explanation

Answer: (b) 13

Explanation:
1. Distance = √((−1 + 6)² + (−5 − 7)²) = √(25 + 144) = √169 = 13.

Teacher's Note:
1. The differences are 5 and −12.
2. 5, 12, 13 is a Pythagorean triple.

Question 8: The distance between the points (2, −3) and (10, y) is 10 units. A possible value of y is
(a) 3
(b) 6
(c) −3
(d) 9
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. (10 − 2)² + (y + 3)² = 100, so (y + 3)² = 36 and y + 3 = ±6. Hence y = 3 or y = −9; among the options only 3 is possible.

Teacher's Note:
1. Square both sides to remove the root.
2. Two values satisfy it; pick the one offered.

Question 9: The distance between the points (0, 3) and (−2, 0) is
(a) √13
(b) √14
(c) √15
(d) √5
Show Answer & Explanation

Answer: (a) √13

Explanation:
1. Distance = √((−2 − 0)² + (0 − 3)²) = √(4 + 9) = √13.

Teacher's Note:
1. Apply the distance formula.
2. Square each difference.

Question 10: The distance of the point (5, −2) from the x-axis is
(a) 5
(b) −2
(c) 3
(d) 2
Show Answer & Explanation

Answer: (d) 2

Explanation:
1. The distance from the x-axis is the absolute value of the y-coordinate, |−2| = 2. (Distance cannot be negative.)

Teacher's Note:
1. Distance from the x-axis is
2. y
3. .
4. Distance from the y-axis is
5. x
6. .

Question 11: In the figure, which point has the coordinates (−2, −3)?
X Y O −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 F A C P Q B D E (a) Q
(b) P
(c) A
(d) C
Show Answer & Explanation

Answer: (b) P

Explanation:
1. (−2, −3) is 2 units left of the y-axis and 3 units below the x-axis. The point P is at that position.

Teacher's Note:
1. Move 2 left, then 3 down from the origin.
2. Compare with the signs of the other points.

Question 12: In the figure, the coordinates of the point E are
X Y O −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 F A C P Q B D E (a) (5, −6)
(b) (1, 2)
(c) (5, 6)
(d) (−5, −6)
Show Answer & Explanation

Answer: (a) (5, −6)

Explanation:
1. E is 5 units to the right of the y-axis and 6 units below the x-axis, so E = (5, −6).

Teacher's Note:
1. Right gives a positive x, below gives a negative y.
2. Check the quadrant: IV.

Question 13: In the figure, the coordinates of the point B are
X Y O −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 M B L S (a) (−3, 4)
(b) (−5, −4)
(c) (3, −4)
(d) (4, 3)
Show Answer & Explanation

Answer: (d) (4, 3)

Explanation:
1. B is 4 units right of the y-axis and 3 units above the x-axis, so B = (4, 3).

Teacher's Note:
1. Follow the dotted lines to the axes.
2. The x-coordinate is read on the x-axis.

Question 14: In the figure, the coordinates of the point S are
X Y O −8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 8 −7 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 7 M B L S (a) (−3, 4)
(b) (3, −4)
(c) (4, 3)
(d) (−5, −4)
Show Answer & Explanation

Answer: (b) (3, −4)

Explanation:
1. S is 3 units right of the y-axis and 4 units below the x-axis, so S = (3, −4).

Teacher's Note:
1. Quadrant IV: x positive, y negative.
2. Follow the dotted lines.

Question 15: If the distance between the points (4, p) and (1, 0) is 5, then p is
(a) ±4
(b) 4
(c) −4
(d) 0
Show Answer & Explanation

Answer: (a) ±4

Explanation:
1. (4 − 1)² + p² = 25, so p² = 16 and p = ±4.

Teacher's Note:
1. Square both sides.
2. The square root gives both signs.

Question 16: The perimeter of the triangle formed by the points (0, 0), (1, 0) and (0, 1) is
(a) 1 + √2
(b) 3
(c) 2 + 2√2
(d) 2 + √2
Show Answer & Explanation

Answer: (d) 2 + √2

Explanation:
1. The sides are 1, 1 and √(1 + 1) = √2. The perimeter is 2 + √2.

Teacher's Note:
1. Two sides lie along the axes.
2. The third side is the hypotenuse.

Question 17: The point (−3, P) divides the line segment joining (−5, −4) and (−2, 3) in the ratio 2 : 1. The value of 3P is
(a) 1
(b) 2
(c) 3
(d) 6
Show Answer & Explanation

Answer: (b) 2

Explanation:
1. By the section formula, P = (2 × 3 + 1 × (−4))/(2 + 1) = 2/3. (The x-coordinate (2 × (−2) + 1 × (−5))/3 = −3 confirms the point.) So 3P = 2.

Teacher's Note:
1. Use the section formula with m : n = 2 : 1.
2. Check the x-coordinate matches −3.

Question 18: The perimeter of the triangle whose vertices are (0, 0), (3, 0) and (0, 4) is
(a) 12
(b) 7
(c) 10
(d) 14
Show Answer & Explanation

Answer: (a) 12

Explanation:
1. The sides are 3, 4 and √(9 + 16) = 5. The perimeter is 3 + 4 + 5 = 12.

Teacher's Note:
1. This is the 3-4-5 right triangle.
2. Add the three sides.

Question 19: The points (x, y), (1, 2) and (7, 0) are collinear. The relation between x and y is
(a) 3x + y = 0
(b) x + 3y = 7
(c) 2x + y = 0
(d) 2x − y = 0
Show Answer & Explanation

Answer: (b) x + 3y = 7

Explanation:
1. The area of the triangle is zero: x(2 − 0) + 1(0 − y) + 7(y − 2) = 0, so 2x + 6y − 14 = 0, i.e. x + 3y = 7. (Check: (1, 2) gives 1 + 6 = 7 and (7, 0) gives 7.)

Teacher's Note:
1. Set the area of the triangle formed by the three points to zero.
2. Verify with the two given points.

Question 20: The points (5, p), (1, 5), (2, 1) and (6, 2) are the vertices of a square. The value of p is
(a) 5
(b) 6
(c) 7
(d) 4
Show Answer & Explanation

Answer: (b) 6

Explanation:
1. Take A(1, 5), B(2, 1) and C(6, 2). AB = (1, −4) and BC = (4, 1) are equal in length (√17) and perpendicular, so they are consecutive sides. The fourth vertex is D = A + C − B = (1 + 6 − 2, 5 + 2 − 1) = (5, 6). So p = 6.

Teacher's Note:
1. The diagonals of a square bisect each other.
2. Use D = A + C − B.

Question 21: If the origin is the centroid of the triangle whose vertices are (a, b), (b, c) and (c, a), then a + b + c =
(a) 0
(b) 1
(c) 3
(d) abc
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. The centroid's x-coordinate is (a + b + c)/3 = 0, so a + b + c = 0.

Teacher's Note:
1. Use the centroid formula.
2. The same condition comes from the y-coordinates.

Question 22: The points (3, 2), (4, k) and (5, 3) are collinear. The value of 2k is
(a) 4
(b) 6
(c) 5
(d) 3
Show Answer & Explanation

Answer: (c) 5

Explanation:
1. The area is zero: 3(k − 3) + 4(3 − 2) + 5(2 − k) = 0, so −2k + 5 = 0 and k = 5/2. Hence 2k = 5.

Teacher's Note:
1. Set the triangle's area to zero.
2. Remember to double k at the end.

Question 23: If (x, 0), (0, y) and (1, 1) are collinear, then 1/x + 1/y =
(a) 0
(b) 1
(c) 2
(d) xy
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. The line through (x, 0) and (0, y) has the equation X/x + Y/y = 1. Since (1, 1) lies on it, 1/x + 1/y = 1.

Teacher's Note:
1. The line with x-intercept x and y-intercept y is X/x + Y/y = 1.
2. Substitute the point (1, 1).

Question 24: If (x, y), (y, z) and (z, x) are the vertices of a triangle whose centroid is the origin, then x³ + y³ + z³ =
(a) 3xyz
(b) 0
(c) xyz
(d) x + y + z
Show Answer & Explanation

Answer: (a) 3xyz

Explanation:
1. The centroid condition gives x + y + z = 0. If x + y + z = 0 then x³ + y³ + z³ = 3xyz (from the identity x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx)).

Teacher's Note:
1. Use the centroid formula to get x + y + z = 0.
2. Apply the standard cubic identity.

Question 25: The length of the median AD of the triangle formed by A(7, −3), B(5, 3) and C(3, −1) is
(a) 4
(b) √20
(c) 6
(d) 5
Show Answer & Explanation

Answer: (d) 5

Explanation:
1. D is the mid-point of BC = (4, 1). AD = √((7 − 4)² + (−3 − 1)²) = √(9 + 16) = 5.

Teacher's Note:
1. A median joins a vertex to the mid-point of the opposite side.
2. Use the distance formula.

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Practice MCQs for Class 8 Mathematics Coordinate Geometry

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