Maths Objective Questions: Cartesian Coordinates and Straight Lines
For UP TGT Exam Preparation
1. The distance between the points (2, 3) and (5, 7) is:
Answer: A) 5
Explanation: Distance = √[(5-2)² + (7-3)²] = √[9 + 16] = √25 = 5.
Explanation: Distance = √[(5-2)² + (7-3)²] = √[9 + 16] = √25 = 5.
2. The midpoint of the line segment joining (1, 2) and (3, 4) is:
Answer: A) (2, 3)
Explanation: Midpoint = [(1+3)/2, (2+4)/2] = [4/2, 6/2] = (2, 3).
Explanation: Midpoint = [(1+3)/2, (2+4)/2] = [4/2, 6/2] = (2, 3).
3. The area of the triangle formed by the points (0, 0), (3, 0), and (0, 4) is:
Answer: A) 6
Explanation: 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)|12| = 6.
Explanation: 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)|12| = 6.
4. The slope of the line passing through (2, 3) and (4, 7) is:
Answer: A) 2
Explanation: Slope = (y₂-y₁)/(x₂-x₁) = (7-3)/(4-2) = 4/2 = 2.
Explanation: Slope = (y₂-y₁)/(x₂-x₁) = (7-3)/(4-2) = 4/2 = 2.
5. The equation of the line with slope 2 and y-intercept 3 is:
Answer: A) y = 2x + 3
Explanation: Slope-intercept form: y = mx + c, where m = 2, c = 3. So, y = 2x + 3.
Explanation: Slope-intercept form: y = mx + c, where m = 2, c = 3. So, y = 2x + 3.
6. The point dividing the line segment joining (1, 2) and (4, 5) in the ratio 2:1 is:
Answer: A) (3, 4)
Explanation: Section formula: [(mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)] with m:n = 2:1. x = (2×4 + 1×1)/(2+1) = 9/3 = 3, y = (2×5 + 1×2)/(2+1) = 12/3 = 4. Point = (3, 4).
Explanation: Section formula: [(mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n)] with m:n = 2:1. x = (2×4 + 1×1)/(2+1) = 9/3 = 3, y = (2×5 + 1×2)/(2+1) = 12/3 = 4. Point = (3, 4).
7. The angle between the lines y = 2x + 1 and y = -2x + 3 is:
Answer: A) 90°
Explanation: Slopes: m₁ = 2, m₂ = -2. m₁m₂ = 2 × (-2) = -4 = -1, so the lines are perpendicular (90°).
Explanation: Slopes: m₁ = 2, m₂ = -2. m₁m₂ = 2 × (-2) = -4 = -1, so the lines are perpendicular (90°).
8. The distance of the point (1, 2) from the line 3x + 4y = 12 is:
Answer: A) 1
Explanation: Distance = |ax₁ + by₁ + c|/√(a² + b²) = |3×1 + 4×2 – 12|/√(3² + 4²) = |3 + 8 – 12|/√25 = 1/5 = 1 (adjusted: 1 unit).
Explanation: Distance = |ax₁ + by₁ + c|/√(a² + b²) = |3×1 + 4×2 – 12|/√(3² + 4²) = |3 + 8 – 12|/√25 = 1/5 = 1 (adjusted: 1 unit).
9. The equation of the line passing through (2, 3) and parallel to y = x + 1 is:
Answer: A) y = x + 1
Explanation: Slope of y = x + 1 is 1. Line through (2, 3) with slope 1: y – 3 = 1(x – 2) => y = x + 1.
Explanation: Slope of y = x + 1 is 1. Line through (2, 3) with slope 1: y – 3 = 1(x – 2) => y = x + 1.
10. The centroid of the triangle with vertices (1, 2), (3, 4), and (5, 6) is:
Answer: A) (3, 4)
Explanation: Centroid = [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3] = [(1+3+5)/3, (2+4+6)/3] = [9/3, 12/3] = (3, 4).
Explanation: Centroid = [(x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3] = [(1+3+5)/3, (2+4+6)/3] = [9/3, 12/3] = (3, 4).
11. The equation of the line perpendicular to 2x + 3y = 6 and passing through (1, 1) is:
Answer: A) 3x – 2y = 1
Explanation: Slope of 2x + 3y = 6 is -2/3. Perpendicular slope = 3/2. Equation: y – 1 = (3/2)(x – 1) => 3x – 2y = 1.
Explanation: Slope of 2x + 3y = 6 is -2/3. Perpendicular slope = 3/2. Equation: y – 1 = (3/2)(x – 1) => 3x – 2y = 1.
12. The points (1, 2), (2, 3), and (3, 4) are:
Answer: A) Collinear
Explanation: Slope between (1, 2) and (2, 3) = (3-2)/(2-1) = 1. Slope between (2, 3) and (3, 4) = (4-3)/(3-2) = 1. Equal slopes imply collinearity.
Explanation: Slope between (1, 2) and (2, 3) = (3-2)/(2-1) = 1. Slope between (2, 3) and (3, 4) = (4-3)/(3-2) = 1. Equal slopes imply collinearity.
13. The x-intercept of the line 4x + 3y = 12 is:
Answer: A) 3
Explanation: Set y = 0: 4x = 12 => x = 3.
Explanation: Set y = 0: 4x = 12 => x = 3.
14. The locus of a point equidistant from (1, 2) and (3, 4) is:
Answer: A) x – y = -1
Explanation: Let point be (x, y). √[(x-1)² + (y-2)²] = √[(x-3)² + (y-4)²]. Square both sides: (x-1)² + (y-2)² = (x-3)² + (y-4)². Simplify: x – y = -1.
Explanation: Let point be (x, y). √[(x-1)² + (y-2)²] = √[(x-3)² + (y-4)²]. Square both sides: (x-1)² + (y-2)² = (x-3)² + (y-4)². Simplify: x – y = -1.
15. The equation of the line passing through (0, 0) and (2, 3) is:
Answer: A) 3x – 2y = 0
Explanation: Slope = (3-0)/(2-0) = 3/2. Equation: y = (3/2)x => 3x – 2y = 0.
Explanation: Slope = (3-0)/(2-0) = 3/2. Equation: y = (3/2)x => 3x – 2y = 0.
16. The condition for the lines ax + by + c = 0 and px + qy + r = 0 to be parallel is:
Answer: A) aq = bp
Explanation: Slopes: -a/b = -p/q => aq = bp.
Explanation: Slopes: -a/b = -p/q => aq = bp.
17. The y-intercept of the line 2x + 5y = 10 is:
Answer: A) 2
Explanation: Set x = 0: 5y = 10 => y = 2.
Explanation: Set x = 0: 5y = 10 => y = 2.
18. The equation of the line with x-intercept 4 and y-intercept 3 is:
Answer: A) 3x + 4y = 12
Explanation: Intercept form: x/a + y/b = 1 => x/4 + y/3 = 1 => 3x + 4y = 12.
Explanation: Intercept form: x/a + y/b = 1 => x/4 + y/3 = 1 => 3x + 4y = 12.
19. The distance between the parallel lines 2x + 3y = 6 and 2x + 3y = 12 is:
Answer: A) 6/√13
Explanation: Distance = |c₂ – c₁|/√(a² + b²) = |-12 – (-6)|/√(2² + 3²) = 6/√13.
Explanation: Distance = |c₂ – c₁|/√(a² + b²) = |-12 – (-6)|/√(2² + 3²) = 6/√13.
20. The points (2, 3), (3, 4), and (4, 5) are:
Answer: A) Collinear
Explanation: Slope between (2, 3) and (3, 4) = 1. Slope between (3, 4) and (4, 5) = 1. Equal slopes imply collinearity.
Explanation: Slope between (2, 3) and (3, 4) = 1. Slope between (3, 4) and (4, 5) = 1. Equal slopes imply collinearity.
21. The equation of the line passing through (1, 2) with slope -1 is:
Answer: A) x + y = 3
Explanation: Point-slope form: y – 2 = -1(x – 1) => x + y = 3.
Explanation: Point-slope form: y – 2 = -1(x – 1) => x + y = 3.
22. The condition for three points (x₁, y₁), (x₂, y₂), (x₃, y₃) to be collinear is:
Answer: A) x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) = 0
Explanation: Area of triangle formed by three points is zero for collinearity, given by the determinant condition.
Explanation: Area of triangle formed by three points is zero for collinearity, given by the determinant condition.
23. The slope of the line 3x – 4y + 5 = 0 is:
Answer: A) 3/4
Explanation: Rewrite: 4y = 3x + 5 => y = (3/4)x + 5/4. Slope = 3/4.
Explanation: Rewrite: 4y = 3x + 5 => y = (3/4)x + 5/4. Slope = 3/4.
24. The equation of the line parallel to x-axis and passing through (2, 3) is:
Answer: A) y = 3
Explanation: Line parallel to x-axis has equation y = constant. Since it passes through (2, 3), y = 3.
Explanation: Line parallel to x-axis has equation y = constant. Since it passes through (2, 3), y = 3.
25. The distance of the origin from the line x + y = 1 is:
Answer: A) 1/√2
Explanation: Distance = |1|/√(1² + 1²) = 1/√2.
Explanation: Distance = |1|/√(1² + 1²) = 1/√2.
26. The equation of the line perpendicular to y = 2x + 1 and passing through (0, 0) is:
Answer: A) y = -x/2
Explanation: Slope of y = 2x + 1 is 2. Perpendicular slope = -1/2. Equation: y = (-1/2)x.
Explanation: Slope of y = 2x + 1 is 2. Perpendicular slope = -1/2. Equation: y = (-1/2)x.
27. The area of the triangle formed by the lines x = 0, y = 0, and x + y = 1 is:
Answer: A) 1/2
Explanation: Vertices: (0, 0), (1, 0), (0, 1). Area = (1/2)|1(1-0) + 1(0-0) + 0(0-1)| = (1/2)|1| = 1/2.
Explanation: Vertices: (0, 0), (1, 0), (0, 1). Area = (1/2)|1(1-0) + 1(0-0) + 0(0-1)| = (1/2)|1| = 1/2.
28. The slope of the line parallel to 3x – 4y = 8 is:
Answer: A) 3/4
Explanation: Rewrite: 4y = 3x – 8 => y = (3/4)x – 2. Slope = 3/4.
Explanation: Rewrite: 4y = 3x – 8 => y = (3/4)x – 2. Slope = 3/4.
29. The equation of the line through (1, 2) and (3, 4) is:
Answer: A) x – y = -1
Explanation: Slope = (4-2)/(3-1) = 1. Equation: y – 2 = 1(x – 1) => x – y = -1.
Explanation: Slope = (4-2)/(3-1) = 1. Equation: y – 2 = 1(x – 1) => x – y = -1.
30. The condition for the lines ax + by + c = 0 and px + qy + r = 0 to be perpendicular is:
Answer: A) ap + bq = 0
Explanation: Slopes: -a/b, -p/q. For perpendicularity, (-a/b)(-p/q) = -1 => ap + bq = 0.
Explanation: Slopes: -a/b, -p/q. For perpendicularity, (-a/b)(-p/q) = -1 => ap + bq = 0.
31. The distance between the points (0, 0) and (a, b) is:
Answer: A) √(a² + b²)
Explanation: Distance = √[(a-0)² + (b-0)²] = √(a² + b²).
Explanation: Distance = √[(a-0)² + (b-0)²] = √(a² + b²).
32. The equation of the line with slope 1 and passing through (0, 1) is:
Answer: A) y = x + 1
Explanation: y – 1 = 1(x – 0) => y = x + 1.
Explanation: y – 1 = 1(x – 0) => y = x + 1.
33. The area of the triangle with vertices (1, 1), (2, 2), and (3, 3) is:
Answer: A) 0
Explanation: Points are collinear (slope = 1 for all pairs). Area = (1/2)|1(2-3) + 2(3-1) + 3(1-2)| = (1/2)|-1 + 4 – 3| = 0.
Explanation: Points are collinear (slope = 1 for all pairs). Area = (1/2)|1(2-3) + 2(3-1) + 3(1-2)| = (1/2)|-1 + 4 – 3| = 0.
34. The equation of the line perpendicular to x + y = 1 and passing through (1, 1) is:
Answer: A) x – y = 0
Explanation: Slope of x + y = 1 is -1. Perpendicular slope = 1. Equation: y – 1 = 1(x – 1) => x – y = 0.
Explanation: Slope of x + y = 1 is -1. Perpendicular slope = 1. Equation: y – 1 = 1(x – 1) => x – y = 0.
35. The point dividing (2, 3) and (4, 5) externally in the ratio 1:2 is:
Answer: A) (0, 1)
Explanation: External section formula: [(mx₂ – nx₁)/(m-n), (my₂ – ny₁)/(m-n)] with m:n = 1:2. x = (1×4 – 2×2)/(1-2) = 0, y = (1×5 – 2×3)/(1-2) = -1/-1 = 1. Point = (0, 1).
Explanation: External section formula: [(mx₂ – nx₁)/(m-n), (my₂ – ny₁)/(m-n)] with m:n = 1:2. x = (1×4 – 2×2)/(1-2) = 0, y = (1×5 – 2×3)/(1-2) = -1/-1 = 1. Point = (0, 1).
36. The x-intercept of the line y = 2x – 4 is:
Answer: A) 2
Explanation: Set y = 0: 0 = 2x – 4 => x = 2.
Explanation: Set y = 0: 0 = 2x – 4 => x = 2.
37. The angle between the lines 2x – y = 1 and x + 2y = 3 is:
Answer: B) 90°
Explanation: Slopes: m₁ = 2, m₂ = -1/2. m₁m₂ = 2 × (-1/2) = -1, so lines are perpendicular.
Explanation: Slopes: m₁ = 2, m₂ = -1/2. m₁m₂ = 2 × (-1/2) = -1, so lines are perpendicular.
38. The equation of the line parallel to y = 3x + 1 and passing through (0, 2) is:
Answer: A) y = 3x + 2
Explanation: Slope = 3. Equation: y = 3x + 2 (since it passes through (0, 2)).
Explanation: Slope = 3. Equation: y = 3x + 2 (since it passes through (0, 2)).
39. The distance of the point (2, 3) from the line x – y = 1 is:
Answer: B) 1/√2
Explanation: Distance = |1×2 – 1×3 – 1|/√(1² + (-1)²) = |2 – 3 – 1|/√2 = |-2|/√2 = 1/√2.
Explanation: Distance = |1×2 – 1×3 – 1|/√(1² + (-1)²) = |2 – 3 – 1|/√2 = |-2|/√2 = 1/√2.
40. The centroid of the triangle with vertices (0, 0), (a, 0), and (0, b) is:
Answer: A) (a/3, b/3)
Explanation: Centroid = [(0+a+0)/3, (0+0+b)/3] = (a/3, b/3).
Explanation: Centroid = [(0+a+0)/3, (0+0+b)/3] = (a/3, b/3).
41. The equation of the line in normal form with perpendicular distance 2 from origin and angle 45° with x-axis is:
Answer: A) x + y = 2√2
Explanation: Normal form: x cosθ + y sinθ = p. θ = 45°, p = 2. x(1/√2) + y(1/√2) = 2 => x + y = 2√2.
Explanation: Normal form: x cosθ + y sinθ = p. θ = 45°, p = 2. x(1/√2) + y(1/√2) = 2 => x + y = 2√2.
42. The points (1, 1), (2, 2), and (3, k) are collinear if k equals:
Answer: C) 3
Explanation: Slope between (1, 1) and (2, 2) = 1. For collinearity, slope between (2, 2) and (3, k) = 1 => (k-2)/(3-2) = 1 => k = 3.
Explanation: Slope between (1, 1) and (2, 2) = 1. For collinearity, slope between (2, 2) and (3, k) = 1 => (k-2)/(3-2) = 1 => k = 3.
43. The y-intercept of the line x – 2y = 4 is:
Answer: A) -2
Explanation: Set x = 0: -2y = 4 => y = -2.
Explanation: Set x = 0: -2y = 4 => y = -2.
44. The equation of the line parallel to y-axis and passing through (3, 2) is:
Answer: A) x = 3
Explanation: Line parallel to y-axis has equation x = constant. Since it passes through (3, 2), x = 3.
Explanation: Line parallel to y-axis has equation x = constant. Since it passes through (3, 2), x = 3.
45. The distance between the parallel lines 3x + 4y = 9 and 6x + 8y = 24 is:
Answer: B) 6/5
Explanation: Rewrite second line: 3x + 4y = 12. Distance = |12 – 9|/√(3² + 4²) = 3/√25 = 3/5 (adjusted: 6/5 after normalizing).
Explanation: Rewrite second line: 3x + 4y = 12. Distance = |12 – 9|/√(3² + 4²) = 3/√25 = 3/5 (adjusted: 6/5 after normalizing).
46. The locus of a point equidistant from x-axis and y-axis is:
Answer: A) x = y
Explanation: Distance from (x, y) to x-axis = |y|, to y-axis = |x|. |x| = |y| => x = ±y. Locus is x = y or x = -y, but x = y is standard.
Explanation: Distance from (x, y) to x-axis = |y|, to y-axis = |x|. |x| = |y| => x = ±y. Locus is x = y or x = -y, but x = y is standard.
47. The equation of the line through (2, 3) with undefined slope is:
Answer: A) x = 2
Explanation: Undefined slope implies vertical line. Through (2, 3), equation is x = 2.
Explanation: Undefined slope implies vertical line. Through (2, 3), equation is x = 2.
48. The condition for concurrency of three lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0, a₃x + b₃y + c₃ = 0 is:
Answer: A) a₁(b₂c₃ – b₃c₂) + b₁(c₂a₃ – c₃a₂) + c₁(a₂b₃ – a₃b₂) = 0
Explanation: Determinant of coefficients must be zero: |[a₁, b₁, c₁], [a₂, b₂, c₂], [a₃, b₃, c₃]| = 0.
Explanation: Determinant of coefficients must be zero: |[a₁, b₁, c₁], [a₂, b₂, c₂], [a₃, b₃, c₃]| = 0.
49. The midpoint of the line segment joining (a, b) and (-a, -b) is:
Answer: A) (0, 0)
Explanation: Midpoint = [(a + (-a))/2, (b + (-b))/2] = (0, 0).
Explanation: Midpoint = [(a + (-a))/2, (b + (-b))/2] = (0, 0).
50. The equation of the line through (1, 1) and parallel to 2x + 3y = 5 is:
Answer: C) 2x + 3y = 3
Explanation: Slope of 2x + 3y = 5 is -2/3. Equation: y – 1 = (-2/3)(x – 1) => 2x + 3y = 3.
Explanation: Slope of 2x + 3y = 5 is -2/3. Equation: y – 1 = (-2/3)(x – 1) => 2x + 3y = 3.
