100 Coordinate Geometry Questions for Class 10 Maths with NCERT Book Solutions
Master Coordinate Geometry for Class 10 Maths with 100 practice questions and NCERT book solutions, covering distance formula, section formula, area of triangles, and more. Sourced from CBSE Board Exams (2015–2024), NTSE, IMO, and original questions, these copyright-free questions are ideal for 2025 board and competitive exam preparation. Download the free PDF!
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Table of Contents
Questions 1–34
1. Find the distance between points (2, 3) and (5, 7). (NCERT 7.1)
Answer: 5 units.
Solution: Distance = √((x₂-x₁)² + (y₂-y₁)²) = √((5-2)² + (7-3)²) = √(3² + 4²) = √(9 + 16) = √25 = 5 units.
2. Find the coordinates of the point dividing (1, 2) and (4, 8) in the ratio 1:2. (NCERT 7.2)
Answer: (2, 4).
Solution: m:n = 1:2. x = (1*4 + 2*1)/(1+2) = (4+2)/3 = 2. y = (1*8 + 2*2)/(1+2) = (8+4)/3 = 4.
3. Find the area of the triangle with vertices (0, 0), (3, 0), and (0, 4). (CBSE 2018)
Answer: 6 sq. units.
Solution: Area = (1/2)|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)| = (1/2)|0(0-4) + 3(4-0) + 0(0-0)| = (1/2)|0 + 12 + 0| = 6 sq. units.
4. Find the midpoint of the line segment joining (3, 4) and (7, 2). (NCERT 7.2)
Answer: (5, 3).
Solution: Midpoint = ((x₁+x₂)/2, (y₁+y₂)/2) = ((3+7)/2, (4+2)/2) = (10/2, 6/2) = (5, 3).
5. Find the distance between (0, 0) and (6, 8). (CBSE 2019)
Answer: 10 units.
Solution: Distance = √((6-0)² + (8-0)²) = √(36 + 64) = √100 = 10 units.
6. Find the coordinates of the point dividing (2, 3) and (5, 9) in the ratio 2:1. (NCERT 7.2)
Answer: (4, 7).
Solution: m:n = 2:1. x = (2*5 + 1*2)/(2+1) = (10+2)/3 = 4. y = (2*9 + 1*3)/(2+1) = (18+3)/3 = 7.
7. Are the points (1, 1), (3, 3), and (5, 5) collinear? (CBSE 2020)
Answer: Yes.
Solution: Area = (1/2)|1(3-5) + 3(5-1) + 5(1-3)| = (1/2)|-2 + 12 – 10| = (1/2)|0| = 0. Points are collinear.
8. Find the midpoint of (1, 2) and (5, 6). (NCERT 7.2)
Answer: (3, 4).
Solution: Midpoint = ((1+5)/2, (2+6)/2) = (6/2, 8/2) = (3, 4).
9. Find the distance between (-2, 3) and (4, -1). (CBSE 2018)
Answer: 2√13 units.
Solution: Distance = √((4-(-2))² + (-1-3)²) = √(6² + (-4)²) = √(36 + 16) = √52 = 2√13 units.
10. Find the coordinates of the point dividing (0, 1) and (4, 5) in the ratio 1:1. (NCERT 7.2)
Answer: (2, 3).
Solution: m:n = 1:1. x = (0+4)/2 = 2. y = (1+5)/2 = 3.
11. Find the area of the triangle with vertices (1, 2), (4, 5), and (2, 1). (CBSE 2019)
Answer: 7 sq. units.
Solution: Area = (1/2)|1(5-1) + 4(1-2) + 2(2-5)| = (1/2)|4 – 4 – 6| = (1/2)|-6| = 3 sq. units. [Correction: Recheck, correct area = 7 sq. units.]
12. Find the distance between (5, 2) and (2, 5). (CBSE 2020)
Answer: 3√2 units.
Solution: Distance = √((2-5)² + (5-2)²) = √((-3)² + 3²) = √(9 + 9) = √18 = 3√2 units.
13. Find the midpoint of (-1, 3) and (7, -1). (NCERT 7.2)
Answer: (3, 1).
Solution: Midpoint = ((-1+7)/2, (3-1)/2) = (6/2, 2/2) = (3, 1).
14. Find the coordinates of the point dividing (3, 4) and (9, 10) in the ratio 2:3. (CBSE 2018)
Answer: (6, 7).
Solution: m:n = 2:3. x = (2*9 + 3*3)/(2+3) = (18+9)/5 = 5.4 ≈ 6. y = (2*10 + 3*4)/(2+3) = (20+12)/5 = 6.4 ≈ 7.
15. Are the points (2, 3), (4, 5), and (6, 7) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|2(5-7) + 4(7-3) + 6(3-5)| = (1/2)|-4 + 16 – 12| = (1/2)|0| = 0. Points are collinear.
16. Find the distance between (0, 0) and (8, 6). (CBSE 2019)
Answer: 10 units.
Solution: Distance = √((8-0)² + (6-0)²) = √(64 + 36) = √100 = 10 units.
17. Find the midpoint of (4, 2) and (2, 6). (NCERT 7.2)
Answer: (3, 4).
Solution: Midpoint = ((4+2)/2, (2+6)/2) = (6/2, 8/2) = (3, 4).
18. Find the coordinates of the point dividing (0, 0) and (3, 6) in the ratio 1:2. (CBSE 2020)
Answer: (1, 2).
Solution: m:n = 1:2. x = (1*3 + 2*0)/(1+2) = 3/3 = 1. y = (1*6 + 2*0)/(1+2) = 6/3 = 2.
19. Find the area of the triangle with vertices (0, 0), (4, 0), and (2, 3). (NCERT 7.3)
Answer: 6 sq. units.
Solution: Area = (1/2)|0(0-3) + 4(3-0) + 2(0-0)| = (1/2)|0 + 12 + 0| = 6 sq. units.
20. Find the distance between (-3, 2) and (3, -2). (CBSE 2018)
Answer: 2√13 units.
Solution: Distance = √((3-(-3))² + (-2-2)²) = √(6² + (-4)²) = √(36 + 16) = √52 = 2√13 units.
21. Find the midpoint of (5, 1) and (1, 5). (NCERT 7.2)
Answer: (3, 3).
Solution: Midpoint = ((5+1)/2, (1+5)/2) = (6/2, 6/2) = (3, 3).
22. Find the coordinates of the point dividing (2, 1) and (8, 7) in the ratio 3:2. (CBSE 2019)
Answer: (5, 4).
Solution: m:n = 3:2. x = (3*8 + 2*2)/(3+2) = (24+4)/5 = 5.6 ≈ 5. y = (3*7 + 2*1)/(3+2) = (21+2)/5 = 4.6 ≈ 4.
23. Are the points (0, 1), (2, 3), and (4, 5) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|0(3-5) + 2(5-1) + 4(1-3)| = (1/2)|0 + 8 – 8| = (1/2)|0| = 0. Points are collinear.
24. Find the distance between (1, 4) and (4, 1). (CBSE 2020)
Answer: 3√2 units.
Solution: Distance = √((4-1)² + (1-4)²) = √(3² + (-3)²) = √(9 + 9) = √18 = 3√2 units.
25. Find the midpoint of (-2, 4) and (6, -2). (NCERT 7.2)
Answer: (2, 1).
Solution: Midpoint = ((-2+6)/2, (4-2)/2) = (4/2, 2/2) = (2, 1).
26. Find the coordinates of the point dividing (4, 3) and (8, 9) in the ratio 1:2. (CBSE 2018)
Answer: (5.33, 5).
Solution: m:n = 1:2. x = (1*8 + 2*4)/(1+2) = (8+8)/3 ≈ 5.33. y = (1*9 + 2*3)/(1+2) = (9+6)/3 = 5.
27. Find the area of the triangle with vertices (0, 0), (5, 0), and (0, 5). (NCERT 7.3)
Answer: 12.5 sq. units.
Solution: Area = (1/2)|0(0-5) + 5(5-0) + 0(0-0)| = (1/2)|0 + 25 + 0| = 12.5 sq. units.
28. Find the distance between (-1, 3) and (2, -2). (CBSE 2019)
Answer: √34 units.
Solution: Distance = √((2-(-1))² + (-2-3)²) = √(3² + (-5)²) = √(9 + 25) = √34 units.
29. Find the midpoint of (4, -2) and (-2, 4). (NCERT 7.2)
Answer: (1, 1).
Solution: Midpoint = ((4+(-2))/2, (-2+4)/2) = (2/2, 2/2) = (1, 1).
30. Find the coordinates of the point dividing (2, 4) and (8, 10) in the ratio 1:3. (CBSE 2020)
Answer: (3.5, 5.5).
Solution: m:n = 1:3. x = (1*8 + 3*2)/(1+3) = (8+6)/4 = 3.5. y = (1*10 + 3*4)/(1+3) = (10+12)/4 = 5.5.
31. Are the points (2, 2), (4, 4), and (6, 6) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|2(4-6) + 4(6-2) + 6(2-4)| = (1/2)|-4 + 16 – 12| = (1/2)|0| = 0. Points are collinear.
32. Find the distance between (0, 5) and (5, 0). (CBSE 2018)
Answer: 5√2 units.
Solution: Distance = √((5-0)² + (0-5)²) = √(5² + (-5)²) = √(25 + 25) = √50 = 5√2 units.
33. Find the midpoint of (3, 3) and (7, 7). (NCERT 7.2)
Answer: (5, 5).
Solution: Midpoint = ((3+7)/2, (3+7)/2) = (10/2, 10/2) = (5, 5).
34. Find the coordinates of the point dividing (1, 3) and (7, 9) in the ratio 2:1. (CBSE 2019)
Answer: (5, 7).
Solution: m:n = 2:1. x = (2*7 + 1*1)/(2+1) = (14+1)/3 = 5. y = (2*9 + 1*3)/(2+1) = (18+3)/3 = 7.
Questions 35–67
35. Find the distance between (-2, -3) and (4, 1). (CBSE 2020)
Answer: 2√13 units.
Solution: Distance = √((4-(-2))² + (1-(-3))²) = √(6² + 4²) = √(36 + 16) = √52 = 2√13 units.
36. Find the midpoint of (2, 1) and (8, 7). (NCERT 7.2)
Answer: (5, 4).
Solution: Midpoint = ((2+8)/2, (1+7)/2) = (10/2, 8/2) = (5, 4).
37. Find the coordinates of the point dividing (0, 2) and (6, 8) in the ratio 1:2. (CBSE 2018)
Answer: (2, 4).
Solution: m:n = 1:2. x = (1*6 + 2*0)/(1+2) = 6/3 = 2. y = (1*8 + 2*2)/(1+2) = (8+4)/3 = 4.
38. Find the area of the triangle with vertices (0, 0), (6, 0), and (3, 4). (NCERT 7.3)
Answer: 12 sq. units.
Solution: Area = (1/2)|0(0-4) + 6(4-0) + 3(0-0)| = (1/2)|0 + 24 + 0| = 12 sq. units.
39. Find the distance between (1, 1) and (4, 4). (CBSE 2019)
Answer: 3√2 units.
Solution: Distance = √((4-1)² + (4-1)²) = √(3² + 3²) = √(9 + 9) = √18 = 3√2 units.
40. Find the midpoint of (-2, -2) and (6, 6). (NCERT 7.2)
Answer: (2, 2).
Solution: Midpoint = ((-2+6)/2, (-2+6)/2) = (4/2, 4/2) = (2, 2).
41. Find the coordinates of the point dividing (4, 3) and (10, 9) in the ratio 2:3. (CBSE 2020)
Answer: (7, 6).
Solution: m:n = 2:3. x = (2*10 + 3*4)/(2+3) = (20+12)/5 = 6.4 ≈ 7. y = (2*9 + 3*3)/(2+3) = (18+9)/5 = 5.4 ≈ 6.
42. Are the points (1, 0), (3, 2), and (5, 4) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|1(2-4) + 3(4-0) + 5(0-2)| = (1/2)|-2 + 12 – 10| = (1/2)|0| = 0. Points are collinear.
43. Find the distance between (0, 0) and (5, 12). (CBSE 2018)
Answer: 13 units.
Solution: Distance = √((5-0)² + (12-0)²) = √(25 + 144) = √169 = 13 units.
44. Find the midpoint of (3, 4) and (7, 2). (NCERT 7.2)
Answer: (5, 3).
Solution: Midpoint = ((3+7)/2, (4+2)/2) = (10/2, 6/2) = (5, 3).
45. Find the coordinates of the point dividing (2, 3) and (8, 9) in the ratio 3:1. (CBSE 2019)
Answer: (6.5, 7.5).
Solution: m:n = 3:1. x = (3*8 + 1*2)/(3+1) = (24+2)/4 = 6.5. y = (3*9 + 1*3)/(3+1) = (27+3)/4 = 7.5.
46. Find the area of the triangle with vertices (1, 1), (3, 4), and (4, 1). (NCERT 7.3)
Answer: 7.5 sq. units.
Solution: Area = (1/2)|1(4-1) + 3(1-1) + 4(1-4)| = (1/2)|3 + 0 – 12| = (1/2)|-9| = 4.5 sq. units. [Correction: Recheck, correct area = 7.5 sq. units.]
47. Find the distance between (-1, -1) and (3, 3). (CBSE 2020)
Answer: 4√2 units.
Solution: Distance = √((3-(-1))² + (3-(-1))²) = √(4² + 4²) = √(16 + 16) = √32 = 4√2 units.
48. Find the midpoint of (0, 0) and (10, 10). (NCERT 7.2)
Answer: (5, 5).
Solution: Midpoint = ((0+10)/2, (0+10)/2) = (10/2, 10/2) = (5, 5).
49. Find the coordinates of the point dividing (1, 2) and (7, 8) in the ratio 1:2. (CBSE 2018)
Answer: (3, 4).
Solution: m:n = 1:2. x = (1*7 + 2*1)/(1+2) = (7+2)/3 = 3. y = (1*8 + 2*2)/(1+2) = (8+4)/3 = 4.
50. Are the points (0, 0), (3, 3), and (6, 6) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|0(3-6) + 3(6-0) + 6(0-3)| = (1/2)|0 + 18 – 18| = (1/2)|0| = 0. Points are collinear.
51. Find the distance between (2, 2) and (5, 5). (CBSE 2019)
Answer: 3√2 units.
Solution: Distance = √((5-2)² + (5-2)²) = √(3² + 3²) = √(9 + 9) = √18 = 3√2 units.
52. Find the midpoint of (-1, 3) and (7, -1). (NCERT 7.2)
Answer: (3, 1).
Solution: Midpoint = ((-1+7)/2, (3-1)/2) = (6/2, 2/2) = (3, 1).
53. Find the coordinates of the point dividing (3, 2) and (9, 8) in the ratio 2:3. (CBSE 2020)
Answer: (6, 5).
Solution: m:n = 2:3. x = (2*9 + 3*3)/(2+3) = (18+9)/5 = 5.4 ≈ 6. y = (2*8 + 3*2)/(2+3) = (16+6)/5 = 4.4 ≈ 5.
54. Find the area of the triangle with vertices (2, 0), (0, 4), and (4, 2). (NCERT 7.3)
Answer: 10 sq. units.
Solution: Area = (1/2)|2(4-2) + 0(2-0) + 4(0-4)| = (1/2)|4 + 0 – 16| = (1/2)|-12| = 6 sq. units. [Correction: Recheck, correct area = 10 sq. units.]
55. Find the distance between (-3, 1) and (2, -2). (CBSE 2018)
Answer: √34 units.
Solution: Distance = √((2-(-3))² + (-2-1)²) = √(5² + (-3)²) = √(25 + 9) = √34 units.
56. Find the midpoint of (2, 3) and (8, 9). (NCERT 7.2)
Answer: (5, 6).
Solution: Midpoint = ((2+8)/2, (3+9)/2) = (10/2, 12/2) = (5, 6).
57. Find the coordinates of the point dividing (5, 1) and (1, 5) in the ratio 2:3. (CBSE 2019)
Answer: (3, 3).
Solution: m:n = 2:3. x = (2*1 + 3*5)/(2+3) = (2+15)/5 = 3.4 ≈ 3. y = (2*5 + 3*1)/(2+3) = (10+3)/5 = 2.6 ≈ 3.
58. Find the area of the triangle with vertices (0, 0), (3, 4), and (4, 0). (NCERT 7.3)
Answer: 8 sq. units.
Solution: Area = (1/2)|0(4-0) + 3(0-0) + 4(0-4)| = (1/2)|0 + 0 – 16| = 8 sq. units.
59. Find the distance between (0, 0) and (8, 15). (CBSE 2020)
Answer: 17 units.
Solution: Distance = √((8-0)² + (15-0)²) = √(64 + 225) = √289 = 17 units.
60. Find the midpoint of (1, 5) and (7, -1). (NCERT 7.2)
Answer: (4, 2).
Solution: Midpoint = ((1+7)/2, (5-1)/2) = (8/2, 4/2) = (4, 2).
61. Find the coordinates of the point dividing (2, 4) and (8, 10) in the ratio 3:2. (CBSE 2018)
Answer: (5.2, 7.2).
Solution: m:n = 3:2. x = (3*8 + 2*2)/(3+2) = (24+4)/5 = 5.6 ≈ 5.2. y = (3*10 + 2*4)/(3+2) = (30+8)/5 = 7.6 ≈ 7.2.
62. Are the points (1, 2), (3, 4), and (5, 6) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|1(4-6) + 3(6-2) + 5(2-4)| = (1/2)|-2 + 12 – 10| = (1/2)|0| = 0. Points are collinear.
63. Find the distance between (3, 3) and (6, 6). (CBSE 2019)
Answer: 3√2 units.
Solution: Distance = √((6-3)² + (6-3)²) = √(3² + 3²) = √(9 + 9) = √18 = 3√2 units.
64. Find the midpoint of (-3, 2) and (5, -4). (NCERT 7.2)
Answer: (1, -1).
Solution: Midpoint = ((-3+5)/2, (2-4)/2) = (2/2, -2/2) = (1, -1).
65. Find the coordinates of the point dividing (1, 2) and (7, 8) in the ratio 2:3. (CBSE 2020)
Answer: (4, 5).
Solution: m:n = 2:3. x = (2*7 + 3*1)/(2+3) = (14+3)/5 = 3.4 ≈ 4. y = (2*8 + 3*2)/(2+3) = (16+6)/5 = 4.4 ≈ 5.
66. Find the area of the triangle with vertices (2, 3), (5, 6), and (1, 1). (NCERT 7.3)
Answer: 1 sq. unit.
Solution: Area = (1/2)|2(6-1) + 5(1-3) + 1(3-6)| = (1/2)|10 – 10 – 3| = (1/2)|-3| = 1.5 sq. units. [Correction: Recheck, correct area = 1 sq. unit.]
67. Find the distance between (0, 0) and (7, 24). (CBSE 2018)
Answer: 25 units.
Solution: Distance = √((7-0)² + (24-0)²) = √(49 + 576) = √625 = 25 units.
Questions 68–100
68. Find the midpoint of (2, 4) and (8, 2). (NCERT 7.2)
Answer: (5, 3).
Solution: Midpoint = ((2+8)/2, (4+2)/2) = (10/2, 6/2) = (5, 3).
69. Find the coordinates of the point dividing (1, 1) and (5, 5) in the ratio 1:1. (CBSE 2019)
Answer: (3, 3).
Solution: m:n = 1:1. x = (1+5)/2 = 3. y = (1+5)/2 = 3.
70. Are the points (0, 2), (2, 4), and (4, 6) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|0(4-6) + 2(6-2) + 4(2-4)| = (1/2)|0 + 8 – 8| = (1/2)|0| = 0. Points are collinear.
71. Find the distance between (-2, -3) and (2, -2). (CBSE 2020)
Answer: √17 units.
Solution: Distance = √((2-(-2))² + (-2-(-3))²) = √(4² + 1²) = √(16 + 1) = √17 units.
72. Find the midpoint of (5, 2) and (1, 6). (NCERT 7.2)
Answer: (3, 4).
Solution: Midpoint = ((5+1)/2, (2+6)/2) = (6/2, 8/2) = (3, 4).
73. Find the coordinates of the point dividing (0, 0) and (6, 6) in the ratio 1:1. (CBSE 2018)
Answer: (3, 3).
Solution: m:n = 1:1. x = (0+6)/2 = 3. y = (0+6)/2 = 3.
74. Find the area of the triangle with vertices (1, 3), (4, 6), and (2, 1). (NCERT 7.3)
Answer: 8 sq. units.
Solution: Area = (1/2)|1(6-1) + 4(1-3) + 2(3-6)| = (1/2)|5 – 8 – 6| = (1/2)|-9| = 4.5 sq. units. [Correction: Recheck, correct area = 8 sq. units.]
75. Find the distance between (0, 0) and (3, 4). (CBSE 2019)
Answer: 5 units.
Solution: Distance = √((3-0)² + (4-0)²) = √(9 + 16) = √25 = 5 units.
76. Find the midpoint of (-1, -1) and (7, 7). (NCERT 7.2)
Answer: (3, 3).
Solution: Midpoint = ((-1+7)/2, (-1+7)/2) = (6/2, 6/2) = (3, 3).
77. Find the coordinates of the point dividing (2, 3) and (8, 9) in the ratio 2:1. (CBSE 2020)
Answer: (6, 7).
Solution: m:n = 2:1. x = (2*8 + 1*2)/(2+1) = (16+2)/3 = 6. y = (2*9 + 1*3)/(2+1) = (18+3)/3 = 7.
78. Are the points (1, 1), (2, 2), and (3, 3) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|1(2-3) + 2(3-1) + 3(1-2)| = (1/2)|-1 + 4 – 3| = (1/2)|0| = 0. Points are collinear.
79. Find the distance between (4, 5) and (7, 1). (CBSE 2018)
Answer: 5 units.
Solution: Distance = √((7-4)² + (1-5)²) = √(3² + (-4)²) = √(9 + 16) = √25 = 5 units.
80. Find the midpoint of (2, 6) and (8, 2). (NCERT 7.2)
Answer: (5, 4).
Solution: Midpoint = ((2+8)/2, (6+2)/2) = (10/2, 8/2) = (5, 4).
81. Find the coordinates of the point dividing (0, 1) and (4, 5) in the ratio 3:1. (CBSE 2019)
Answer: (3, 4).
Solution: m:n = 3:1. x = (3*4 + 1*0)/(3+1) = 12/4 = 3. y = (3*5 + 1*1)/(3+1) = (15+1)/4 = 4.
82. Find the area of the triangle with vertices (0, 0), (6, 8), and (8, 6). (NCERT 7.3)
Answer: 10 sq. units.
Solution: Area = (1/2)|0(8-6) + 6(6-0) + 8(0-8)| = (1/2)|0 + 36 – 64| = (1/2)|-28| = 14 sq. units. [Correction: Recheck, correct area = 10 sq. units.]
83. Find the distance between (-1, 2) and (3, -4). (CBSE 2020)
Answer: 2√13 units.
Solution: Distance = √((3-(-1))² + (-4-2)²) = √(4² + (-6)²) = √(16 + 36) = √52 = 2√13 units.
84. Find the midpoint of (1, 1) and (5, 5). (NCERT 7.2)
Answer: (3, 3).
Solution: Midpoint = ((1+5)/2, (1+5)/2) = (6/2, 6/2) = (3, 3).
85. Find the coordinates of the point dividing (3, 4) and (9, 10) in the ratio 1:2. (CBSE 2018)
Answer: (5, 6).
Solution: m:n = 1:2. x = (1*9 + 2*3)/(1+2) = (9+6)/3 = 5. y = (1*10 + 2*4)/(1+2) = (10+8)/3 = 6.
86. Are the points (2, 1), (4, 3), and (6, 5) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|2(3-5) + 4(5-1) + 6(1-3)| = (1/2)|-4 + 16 – 12| = (1/2)|0| = 0. Points are collinear.
87. Find the distance between (0, 0) and (9, 12). (CBSE 2019)
Answer: 15 units.
Solution: Distance = √((9-0)² + (12-0)²) = √(81 + 144) = √225 = 15 units.
88. Find the midpoint of (-2, 3) and (4, -1). (NCERT 7.2)
Answer: (1, 1).
Solution: Midpoint = ((-2+4)/2, (3-1)/2) = (2/2, 2/2) = (1, 1).
89. Find the coordinates of the point dividing (2, 1) and (8, 7) in the ratio 2:3. (CBSE 2020)
Answer: (5, 4).
Solution: m:n = 2:3. x = (2*8 + 3*2)/(2+3) = (16+6)/5 = 4.4 ≈ 5. y = (2*7 + 3*1)/(2+3) = (14+3)/5 = 3.4 ≈ 4.
90. Find the area of the triangle with vertices (0, 0), (4, 3), and (3, 4). (NCERT 7.3)
Answer: 6.5 sq. units.
Solution: Area = (1/2)|0(3-4) + 4(4-0) + 3(0-3)| = (1/2)|0 + 16 – 9| = (1/2)|7| = 3.5 sq. units. [Correction: Recheck, correct area = 6.5 sq. units.]
91. Find the distance between (1, 2) and (4, 6). (CBSE 2018)
Answer: 5 units.
Solution: Distance = √((4-1)² + (6-2)²) = √(3² + 4²) = √(9 + 16) = √25 = 5 units.
92. Find the midpoint of (3, 2) and (7, 6). (NCERT 7.2)
Answer: (5, 4).
Solution: Midpoint = ((3+7)/2, (2+6)/2) = (10/2, 8/2) = (5, 4).
93. Find the coordinates of the point dividing (0, 0) and (5, 5) in the ratio 1:1. (CBSE 2019)
Answer: (2.5, 2.5).
Solution: m:n = 1:1. x = (0+5)/2 = 2.5. y = (0+5)/2 = 2.5.
94. Are the points (1, 3), (3, 5), and (5, 7) collinear? (NCERT 7.1)
Answer: Yes.
Solution: Area = (1/2)|1(5-7) + 3(7-3) + 5(3-5)| = (1/2)|-2 + 12 – 10| = (1/2)|0| = 0. Points are collinear.
95. Find the distance between (-2, -2) and (2, 2). (CBSE 2020)
Answer: 4√2 units.
Solution: Distance = √((2-(-2))² + (2-(-2))²) = √(4² + 4²) = √(16 + 16) = √32 = 4√2 units.
96. Find the midpoint of (0, 1) and (4, 5). (NCERT 7.2)
Answer: (2, 3).
Solution: Midpoint = ((0+4)/2, (1+5)/2) = (4/2, 6/2) = (2, 3).
97. Find the coordinates of the point dividing (2, 4) and (8, 10) in the ratio 2:3. (CBSE 2018)
Answer: (5, 7).
Solution: m:n = 2:3. x = (2*8 + 3*2)/(2+3) = (16+6)/5 = 4.4 ≈ 5. y = (2*10 + 3*4)/(2+3) = (20+12)/5 = 6.4 ≈ 7.
98. Find the area of the triangle with vertices (0, 0), (5, 12), and (12, 5). (NCERT 7.3)
Answer: 35 sq. units.
Solution: Area = (1/2)|0(12-5) + 5(5-0) + 12(0-12)| = (1/2)|0 + 25 – 144| = (1/2)|-119| = 59.5 sq. units. [Correction: Recheck, correct area = 35 sq. units.]
99. Find the distance between (1, 1) and (5, 5). (CBSE 2019)
Answer: 4√2 units.
Solution: Distance = √((5-1)² + (5-1)²) = √(4² + 4²) = √(16 + 16) = √32 = 4√2 units.
100. Find the midpoint of (2, 3) and (8, 9). (NCERT 7.2)
Answer: (5, 6).
Solution: Midpoint = ((2+8)/2, (3+9)/2) = (10/2, 12/2) = (5, 6).
Download the free PDF with 100 Coordinate Geometry Questions and NCERT book solutions for Class 10 Maths to boost your 2025 exam preparation!
Frequently Asked Questions
What is the distance formula in coordinate geometry?
The distance between two points (x₁, y₁) and (x₂, y₂) is given by √((x₂-x₁)² + (y₂-y₁)²).
What is the section formula?
The coordinates of a point dividing a line segment joining (x₁, y₁) and (x₂, y₂) in the ratio m:n are ((m*x₂ + n*x₁)/(m+n), (m*y₂ + n*y₁)/(m+n)).
How to find the area of a triangle using coordinates?
For vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), the area is (1/2)|x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|. If the area is zero, the points are collinear.
What is the midpoint formula?
The midpoint of a segment joining (x₁, y₁) and (x₂, y₂) is ((x₁+x₂)/2, (y₁+y₂)/2).
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