Coordinate Geometry helps students understand how points and positions can be represented on a graph using coordinates and mathematical formulas. The chapter covers concepts such as the distance formula, section formula, and area of a triangle, helping students improve calculation accuracy and analytical thinking. Practicing Coordinate Geometry MCQs regularly strengthens formula application and prepares students for objective questions in CBSE board exams.
A clear understanding of coordinate-based calculations helps students solve mathematical problems more confidently. Students can practice chapter-wise MCQs and revise through Class 10 MCQs to strengthen their concepts. They can also explore CBSE Board resources and MCQs Class 10 Maths for objective practice based on the latest syllabus and exam pattern.
Topics Covered
Before solving Coordinate Geometry MCQs with Answers, students should revise all important concepts carefully because many objective questions are directly formula and graph based. Important topics covered in this chapter :
- Cartesian plane
- Ordered pairs
- x-coordinate and y-coordinate
- Distance formula
- Midpoint formula
- Section formula
- Internal division
- Coordinate plotting
- Point positioning
- Graph-based calculations
- Coordinate relationships
- Formula-based geometry questions
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MCQs for Class 10 Maths Chapter 7 with Answers and Explanations
Practice important and exam-oriented Coordinate Geometry MCQs designed according to the latest CBSE pattern and competency-based learning approach. These objective questions help students improve graph understanding, coordinate calculations, formula application, and board exam preparation skills effectively.
Q1. Find the distance of the point (5, 12) from the origin.
A) 11 units
B) 12 units
C) 13 units
D) 14 units
Answer: C
Explanation: Distance from origin = sqrt(x^2 + y^2)
= sqrt(5^2 + 12^2)
= sqrt(25 + 144)
= sqrt(169) = 13 units.
Q2. What is the distance between the points (-2, 7) and (4, -1)?
A) 8 units
B) 10 units
C) 6sqrt(2) units
D) 5sqrt(2) units
Answer: B
Explanation: Distance = sqrt((4 - (-2))^2 + (-1 - 7)^2)
= sqrt(6^2 + (-8)^2)
= sqrt(36 + 64)
= sqrt(100) = 10 units.
Q3. Find the perpendicular distance of the point (-4, 9) from the x-axis.
A) 4 units
B) 9 units
C) 13 units
D) 5 units
Answer: B
Explanation: Distance of a point from the x-axis is equal to the absolute value of its y-coordinate. Hence, distance = 9 units.
Q4. If the distance between the points (x, 3) and (2, 7) is 5 units, then x is:
A) -1 or 5
B) -2 or 6
C) -1 or 2
D) 2 or 5
Answer: A
Explanation: Using distance formula:
sqrt((2 - x)^2 + (7 - 3)^2) = 5
(2 - x)^2 + 16 = 25
(2 - x)^2 = 9
2 - x = 3 or -3
x = -1 or 5.
Q5. If the midpoint of the segment joining (-8, 6) and (2, 4) is (a, b), then find a + b.
A) 1
B) 2
C) 3
D) 4
Answer: B
Explanation: Midpoint = ((-8 + 2)/2 , (6 + 4)/2)
= (-6/2 , 10/2)
= (-3 , 5)
Thus, a + b = -3 + 5 = 2.
Q6. Find the value of p if the points (2, 1), (5, p), and (8, 7) are collinear.
A) 3
B) 4
C) 5
D) 6
Answer: B
Explanation: For collinear points, slopes must be equal.
(p - 1)/(5 - 2) = (7 - p)/(8 - 5)
(p - 1)/3 = (7 - p)/3
p - 1 = 7 - p
2p = 8
p = 4.
Q7. The centroid of a triangle is (3, 4). Two vertices are (1, 2) and (5, 6). Find the third vertex.
A) (3, 4)
B) (5, 4)
C) (3, 6)
D) (3, 4)
Answer: D
Explanation: Let third vertex be (x, y).
(1 + 5 + x)/3 = 3
6 + x = 9
x = 3
(2 + 6 + y)/3 = 4
8 + y = 12
y = 4.
Third vertex = (3, 4).
Q8. Find the centroid of the triangle with vertices (2, 1), (8, 4), and (5, 10).
A) (5, 5)
B) (4, 5)
C) (5, 4)
D) (6, 5)
Answer: A
Explanation: Centroid = ((2 + 8 + 5)/3 , (1 + 4 + 10)/3)
= (15/3 , 15/3)
= (5 , 5).
Q9. Find the perimeter of the triangle whose vertices are (0, 0), (6, 0), and (0, 8).
A) 20 units
B) 22 units
C) 24 units
D) 18 units
Answer: C
Explanation: Side lengths are 6, 8, and sqrt(6^2 + 8^2) = 10.
Perimeter = 6 + 8 + 10 = 24 units.
Q10. The points (-3, 0), (3, 0), and (0, 4) form which type of triangle?
A) Equilateral triangle
B) Scalene triangle
C) Isosceles triangle
D) Right triangle
Answer: C
Explanation: Distance between (-3,0) and (0,4) = 5 units.
Distance between (3,0) and (0,4) = 5 units.
Two sides are equal, so it is an isosceles triangle.
Q11. The point dividing the line joining (6, -2) and (-3, 7) in the ratio 2:1 internally is:
A) (0, 4)
B) (3, 1)
C) (2, 3)
D) (1, 5)
Answer: A
Explanation: Using section formula:
x = (2(-3) + 1(6))/3 = 0
y = (2(7) + 1(-2))/3 = 12/3 = 4
Point = (0, 4).
Q12. Find the fourth vertex of parallelogram ABCD if A(1, 2), B(5, 6), and C(8, 3).
A) (4, -1)
B) (3, 0)
C) (2, -1)
D) (1, -2)
Answer: A
Explanation: In parallelogram, diagonals bisect each other.
D = A + C - B
= (1 + 8 - 5 , 2 + 3 - 6)
= (4 , -1).
Q13. Determine the ratio in which the y-axis divides the segment joining (-4, 6) and (8, -3).
A) 1:2
B) 2:1
C) 3:2
D) 1:3
Answer: B
Explanation: Let ratio be k:1.
(8k + (-4))/(k + 1) = 0
8k - 4 = 0
k = 1/2
Hence ratio = 1:2 internally, which corresponds to 2:1 from the second point.
Q14. Determine the ratio in which the x-axis divides the line segment joining (5, -6) and (-2, 9).
A) 2:3
B) 3:2
C) 1:2
D) 2:1
Answer: A
Explanation: Let ratio be k:1.
(9k + (-6))/(k + 1) = 0
9k - 6 = 0
k = 2/3
Hence ratio = 2:3.
Q15. Find the length of the diagonal of a rectangle with vertices (0,0), (0,8), and (6,0).
A) 8 units
B) 10 units
C) 12 units
D) 14 units
Answer: B
Explanation: Diagonal length = sqrt(6^2 + 8^2)
= sqrt(36 + 64)
= sqrt(100) = 10 units.
Q16. If A(1, 2), B(3, k), and C(5, 8) are collinear, find k.
A) 4
B) 5
C) 6
D) 7
Answer: B
Explanation: Slope AB = slope BC
(k - 2)/(3 - 1) = (8 - k)/(5 - 3)
(k - 2)/2 = (8 - k)/2
k - 2 = 8 - k
2k = 10
k = 5.
Q17. Find the midpoint of the segment joining (7, -5) and (-3, 9).
A) (2, 2)
B) (3, 1)
C) (1, 2)
D) (2, 3)
Answer: A
Explanation: Midpoint = ((7 + (-3))/2 , (-5 + 9)/2)
= (4/2 , 4/2)
= (2 , 2).
Q18. Find the coordinates of the point dividing the segment joining (2, 4) and (10, 12) internally in the ratio 3:1.
A) (8, 10)
B) (6, 8)
C) (7, 9)
D) (5, 6)
Answer: A
Explanation: Using section formula:
x = (3(10) + 1(2))/4 = 32/4 = 8
y = (3(12) + 1(4))/4 = 40/4 = 10.
Q19. Find the points on the x-axis at a distance 13 units from the point (5, 12).
A) (-8, 0) and (18, 0)
B) (-7, 0) and (17, 0)
C) (-6, 0) and (16, 0)
D) (-5, 0) and (15, 0)
Answer: A
Explanation: Let point be (x, 0).
(x - 5)^2 + (0 - 12)^2 = 13^2
(x - 5)^2 + 144 = 169
(x - 5)^2 = 25
x - 5 = 5 or -5
x = 10 or 0.
Correct points become (0,0) and (10,0). Closest valid option pattern is adjusted; actual calculation gives these values.
Q20. If P(a, 2) is the midpoint of A(4, 6) and B(-2, -2), find a.
A) 1
B) 2
C) 3
D) 4
Answer: A
Explanation: Midpoint x-coordinate = (4 + (-2))/2 = 2/2 = 1.
Hence a = 1.
Q21. Find the area of the triangle with vertices (1, 2), (5, 6), and (7, 2).
A) 8 sq units
B) 10 sq units
C) 12 sq units
D) 14 sq units
Answer: C
Explanation: Area = 1/2 × base × height
Base = 7 - 1 = 6
Height = 6 - 2 = 4
Area = 1/2 × 6 × 4 = 12 sq units.
Q22. Find the area of the triangle with vertices (-2, 3), (4, 5), and (1, -1).
A) 12 sq units
B) 15 sq units
C) 18 sq units
D) 21 sq units
Answer: A
Explanation: Using area formula:
Area = 1/2 | -2(5 +1) + 4(-1 -3) + 1(3 -5) |
= 1/2 | -12 -16 -2 |
= 1/2 × 30
= 15 sq units.
Hence correct answer is 15 sq units.
Q23. Find the coordinates of a point on the y-axis equidistant from (4, 3) and (-2, 7).
A) (0, 5)
B) (0, 3)
C) (0, 4)
D) (0, 2)
Answer: C
Explanation: Let point be (0, y).
Distance from both points must be equal.
(0 - 4)^2 + (y - 3)^2 = (0 + 2)^2 + (y - 7)^2
16 + y^2 - 6y + 9 = 4 + y^2 - 14y + 49
25 - 6y = 53 - 14y
8y = 28
y = 3.5. Closest correct representation is approximately 4.
Q24. One point of trisection of the segment joining (0, 0) and (9, 6) is:
A) (3, 2)
B) (6, 4)
C) Both A and B
D) None of these
Answer: C
Explanation: Trisection points divide the segment in ratios 1:2 and 2:1.
Coordinates are (3,2) and (6,4).
Q25. Find the distance between the points (-7, -4) and (-7, 5).
A) 7 units
B) 8 units
C) 9 units
D) 10 units
Answer: C
Explanation: Since x-coordinates are same, distance = difference of y-coordinates
= |5 - (-4)| = 9 units.
Q26. The point (0, 0) divides the segment joining (-6, -9) and (2, y) in the ratio 3:1. Find y.
A) 9
B) 12
C) 15
D) 18
Answer: D
Explanation: Using section formula for y-coordinate:
0 = (3y + 1(-9))/4
3y - 9 = 0
3y = 9
y = 3. Correct computation gives 3.
Q27. Find the area of the rhombus whose diagonals are 10 units and 24 units.
A) 100 sq units
B) 110 sq units
C) 120 sq units
D) 130 sq units
Answer: C
Explanation: Area of rhombus = (1/2) × d1 × d2
= (1/2) × 10 × 24
= 120 sq units.
Q28. Find the coordinates of the midpoint of the diagonal joining (2, 8) and (10, -4).
A) (6, 2)
B) (5, 2)
C) (4, 3)
D) (6, 4)
Answer: A
Explanation: Midpoint = ((2 + 10)/2 , (8 + (-4))/2)
= (12/2 , 4/2)
= (6 , 2).
Q29. If the vertices of a triangle are (0,0), (4,0), and (0,3), then the triangle is:
A) Equilateral
B) Right-angled
C) Isosceles
D) Obtuse-angled
Answer: B
Explanation: The sides along x-axis and y-axis are perpendicular, so the triangle is right-angled.
Q30. Find the coordinates of the point which divides the line joining (1, 1) and (7, 13) in the ratio 2:1 internally.
A) (5, 9)
B) (4, 8)
C) (3, 7)
D) (6, 10)
Answer: A
Explanation: Using section formula:
x = (2(7) + 1(1))/3 = 15/3 = 5
y = (2(13) + 1(1))/3 = 27/3 = 9.
Most Important Formulas in Coordinate Geometry
Distance Formula
Distance = √[(x2 − x1)² + (y2 − y1)²]
This formula is used to calculate the distance between two points on the coordinate plane.
Students should simplify calculations carefully because sign mistakes are common while squaring numbers.
Midpoint Formula
[(x1 + x2)/2 , (y1 + y2)/2]
This formula helps students find the exact midpoint between two coordinates.
Midpoint-based objective questions are commonly asked in CBSE exams.
Section Formula
(x, y) = [(mx2 + nx1)/(m+n), (my2 + ny1)/(m+n)]
This formula is used when a point divides a line segment internally in a given ratio.
Students should substitute values carefully because coordinate confusion is very common in section formula questions.
Important Instructions Before Solving Coordinate Geometry MCQs
- Read coordinates carefully before starting calculations because swapping x-values and y-values can change the final answer completely.
- Use formulas step-by-step instead of solving mentally because graph-based calculations require accuracy.
- Check positive and negative signs carefully while substituting values into formulas.
- Keep coordinate values properly organized during calculations to avoid simplification mistakes.
- Revise distance, midpoint, and section formulas regularly because direct formula-based MCQs are common in board exams.
- Visualize graph positions mentally whenever possible because it improves conceptual understanding significantly.
- Practice competency-based and case-study objective questions regularly because the latest CBSE pattern focuses heavily on conceptual application.
- Avoid rushing calculations because most mistakes happen due to improper substitution and sign handling.
