IOE entrance Mathematics · Chapter 4
Coordinate Geometry
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81 questions in 4 syllabus topics · 30 are 2-mark questions.
4.1 Straight lines and pair of lines
20 questions
1. The slope of the line joining the points (2, 3) and (5, 9) is:
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Slope = (y₂ − y₁)/(x₂ − x₁).
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Answer: B. 2
Slope = (9 − 3)/(5 − 2) = 6/3 = 2.
2. The intercepts made by the line 3x − 4y + 12 = 0 on the x-axis and y-axis respectively are:
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Convert to the intercept form x/a + y/b = 1.
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Answer: C. −4 and 3
Write as 3x − 4y = −12, i.e. x/(−4) + y/3 = 1. So x-intercept = −4 and y-intercept = 3.
3. The acute angle between the lines y = x and y = √3x is:
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Find the inclination of each line.
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Answer: C. 15°
The lines make 45° and 60° with the x-axis, so the angle between them is 60° − 45° = 15°.
4. The perpendicular distance of the point (3, 4) from the line 3x + 4y − 5 = 0 is:
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d = |ax₁ + by₁ + c|/√(a² + b²).
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Answer: D. 4
d = |3·3 + 4·4 − 5|/√(3² + 4²) = |20|/5 = 4.
5. The distance between the parallel lines 3x + 4y − 7 = 0 and 6x + 8y + 6 = 0 is:
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Make the x and y coefficients identical before using |c₁ − c₂|/√(a² + b²).
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Answer: A. 2
Divide the second equation by 2: 3x + 4y + 3 = 0. Distance = |−7 − 3|/√(9 + 16) = 10/5 = 2.
6. The lines 2x + ky = 5 and 3x − 6y = 1 are perpendicular when k equals:
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Use a₁a₂ + b₁b₂ = 0.
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Answer: D. 1
For perpendicular lines a₁a₂ + b₁b₂ = 0: 2·3 + k(−6) = 0, so k = 1. (k = −4 would make them parallel.)
7. The equation of the line through (1, 2) perpendicular to 2x − y + 3 = 0 is:
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Swap the coefficients of x and y and change one sign.
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Answer: B. x + 2y − 5 = 0
A perpendicular line has the form x + 2y + k = 0. Through (1, 2): 1 + 4 + k = 0, so k = −5, giving x + 2y − 5 = 0.
8. The perpendicular from the origin to a line has length 4 and makes an angle of 30° with the positive x-axis. The equation of the line is:
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Use the normal form x cos α + y sin α = p.
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Answer: A. √3x + y = 8
Normal form: x cos30° + y sin30° = 4, i.e. (√3/2)x + (1/2)y = 4, so √3x + y = 8.
9. The line passing through the point of intersection of x + y − 3 = 0 and x − y + 1 = 0 and parallel to 2x + 3y = 0 is:
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Find the intersection point, then use 2x + 3y = k.
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Answer: C. 2x + 3y − 8 = 0
Solving, the lines meet at (1, 2). A parallel line is 2x + 3y = k; through (1, 2), k = 2 + 6 = 8. So 2x + 3y − 8 = 0.
10. The lines x + y = 2, x − y = 0 and 2x + ky = 5 are concurrent if k equals:
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Substitute the intersection of the first two lines into the third.
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Answer: C. 3
The first two lines meet at (1, 1). For concurrency (1, 1) must lie on the third line: 2 + k = 5, so k = 3.
11. The area of the triangle formed by the line 3x + 4y = 12 and the coordinate axes is:
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Area = ½ × |x-intercept| × |y-intercept|.
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Answer: C. 6 sq. units
Intercepts are 4 on the x-axis and 3 on the y-axis, so area = ½ × 4 × 3 = 6.
12. The foot of the perpendicular drawn from the origin to the line x + y = 4 is:
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The perpendicular from the origin to x + y = 4 is the line y = x.
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Answer: C. (2, 2)
The perpendicular from the origin is y = x. Solving with x + y = 4 gives (2, 2).
13. The reflection (image) of the point (2, 1) in the line x + y − 5 = 0 is:
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Image: (x − x₁)/a = (y − y₁)/b = −2(ax₁ + by₁ + c)/(a² + b²).
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Answer: A. (4, 3)
(x − 2)/1 = (y − 1)/1 = −2(2 + 1 − 5)/(1² + 1²) = 2, so the image is (4, 3). Check: the midpoint (3, 2) lies on the line.
14. The equation of the line through (2, −3) with inclination 135° is:
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Slope = tan(inclination); then use point-slope form.
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Answer: D. x + y + 1 = 0
Slope = tan135° = −1, so y + 3 = −(x − 2), i.e. x + y + 1 = 0.
15. The line 2x + y = 5 divides the segment joining (1, 1) and (3, 4) in the ratio:
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Ratio = −L(P₁)/L(P₂), where L(x, y) = 2x + y − 5.
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Answer: A. 2 : 5 internally
Ratio = −(2·1 + 1 − 5)/(2·3 + 4 − 5) = −(−2)/5 = 2/5, positive, so 2 : 5 internally. Check: the point (11/7, 13/7) satisfies 2x + y = 5.
16. The acute angle between the pair of lines x² − 4xy + y² = 0 is:
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tanθ = 2√(h² − ab)/|a + b|.
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Answer: B. 60°
Here a = 1, h = −2, b = 1. tanθ = 2√(h² − ab)/|a + b| = 2√(4 − 1)/2 = √3, so θ = 60°.
17. The pair of straight lines represented by x² − 5xy + 6y² = 0 is:
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Factorise the quadratic expression in x and y.
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Answer: A. x − 2y = 0 and x − 3y = 0
x² − 5xy + 6y² factorises as (x − 2y)(x − 3y).
18. The equation kx² + 6xy + 9y² = 0 represents a pair of coincident lines when k equals:
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Condition for coincident lines: h² = ab.
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Answer: C. 1
Coincident lines require h² = ab: 3² = 9k, so k = 1. Indeed x² + 6xy + 9y² = (x + 3y)².
19. The pair of lines 2x² + 7xy + ky² = 0 is perpendicular when k equals:
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Perpendicular pair: coefficient of x² + coefficient of y² = 0.
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Answer: B. −2
For ax² + 2hxy + by² = 0 to give perpendicular lines, a + b = 0, so 2 + k = 0 and k = −2.
20. If m₁ and m₂ are the slopes of the lines represented by x² + 4xy − 3y² = 0, then m₁ + m₂ equals:
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Substitute y = mx and use the sum of roots of the quadratic in m.
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Answer: D. 4/3
Put y = mx: 1 + 4m − 3m² = 0, i.e. 3m² − 4m − 1 = 0. Sum of roots = 4/3. (In general m₁ + m₂ = −2h/b.)
4.2 Circles, tangents and normals
20 questions
21. The centre and radius of the circle x² + y² − 4x + 6y − 12 = 0 are:
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Centre (−g, −f), radius √(g² + f² − c).
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Answer: B. (2, −3) and 5
Centre (−g, −f) = (2, −3); r = √(4 + 9 + 12) = √25 = 5.
22. The equation of the circle with centre (1, −2) passing through (4, 2) is:
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Radius = distance from the centre to the given point.
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Answer: D. x² + y² − 2x + 4y − 20 = 0
r² = (4 − 1)² + (2 + 2)² = 25. (x − 1)² + (y + 2)² = 25 expands to x² + y² − 2x + 4y − 20 = 0.
23. The equation of the circle drawn on the segment joining (1, 2) and (3, −4) as diameter is:
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Diameter form: (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.
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Answer: A. x² + y² − 4x + 2y − 5 = 0
(x − 1)(x − 3) + (y − 2)(y + 4) = 0 gives x² − 4x + 3 + y² + 2y − 8 = 0, i.e. x² + y² − 4x + 2y − 5 = 0.
24. The curve x = 3 + 5cosθ, y = −1 + 5sinθ represents the circle:
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Eliminate θ using cos²θ + sin²θ = 1.
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Answer: A. (x − 3)² + (y + 1)² = 25
(x − 3)/5 = cosθ and (y + 1)/5 = sinθ; squaring and adding gives (x − 3)² + (y + 1)² = 25.
25. The length of the tangent from the point (5, 3) to the circle x² + y² − 2x + 4y − 4 = 0 is:
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Length of tangent = √S₁, where S₁ is the circle expression at the point.
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Answer: C. 4√2
S₁ = 25 + 9 − 10 + 12 − 4 = 32, so the length = √32 = 4√2.
26. The equation of the tangent to the circle x² + y² = 25 at the point (3, 4) is:
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Use xx₁ + yy₁ = a².
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Answer: A. 3x + 4y = 25
Tangent at (x₁, y₁) to x² + y² = a² is xx₁ + yy₁ = a², giving 3x + 4y = 25.
27. The line y = x + c touches the circle x² + y² = 8 when c equals:
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For y = mx + c to touch x² + y² = a², c² = a²(1 + m²).
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Answer: D. ±4
Tangency condition c² = a²(1 + m²) = 8(1 + 1) = 16, so c = ±4.
28. The line 3x + 4y = k touches the circle x² + y² − 2x − 4y − 4 = 0 for:
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Distance of the centre from the line must equal the radius.
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Answer: B. k = 26 or k = −4
Centre (1, 2), radius √(1 + 4 + 4) = 3. Need |3 + 8 − k|/5 = 3, so |11 − k| = 15 and k = 26 or −4.
29. The equation of the normal to the circle x² + y² − 2x + 2y − 7 = 0 at the point (1, 2) is:
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Every normal to a circle passes through its centre.
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Answer: C. x = 1
The centre is (1, −1). The normal passes through the point (1, 2) and the centre (1, −1), so it is x = 1. (y = 2 is the tangent there.)
30. The equation x² + y² + 4x − 6y + k = 0 represents a circle of radius 5 when k equals:
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Use r² = g² + f² − c.
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Answer: C. −12
r² = g² + f² − c = 4 + 9 − k = 25, so k = −12.
31. The circle of radius 3 that touches both coordinate axes and has its centre in the first quadrant is:
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Touching both axes means the centre is (r, r).
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Answer: A. x² + y² − 6x − 6y + 9 = 0
The centre is (3, 3) and r = 3: (x − 3)² + (y − 3)² = 9, i.e. x² + y² − 6x − 6y + 9 = 0.
32. The circle with centre (3, 4) that touches the x-axis is:
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The radius equals the distance of the centre from the x-axis.
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Answer: B. x² + y² − 6x − 8y + 9 = 0
Touching the x-axis means r = |y-coordinate of centre| = 4. (x − 3)² + (y − 4)² = 16 gives c = 9 + 16 − 16 = 9.
33. With respect to the circle x² + y² − 4x + 2y − 11 = 0, the point (1, 2) lies:
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Check the sign of S₁ at the point.
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Answer: A. inside the circle
S₁ = 1 + 4 − 4 + 4 − 11 = −6 < 0, so the point is inside. (The centre is (2, −1), not (1, 2).)
34. The number of common tangents to the circles x² + y² = 4 and x² + y² − 10x + 16 = 0 is:
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Compare the distance between centres with r₁ + r₂ and |r₁ − r₂|.
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Answer: D. 3
C₁ = (0, 0), r₁ = 2; C₂ = (5, 0), r₂ = √(25 − 16) = 3. C₁C₂ = 5 = r₁ + r₂, so the circles touch externally and have 3 common tangents.
35. The circle passing through the origin and making intercepts 4 and 6 on the positive x-axis and positive y-axis respectively is:
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A circle through the origin is x² + y² + 2gx + 2fy = 0; use the two intercept points.
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Answer: C. x² + y² − 4x − 6y = 0
The circle passes through (0, 0), (4, 0) and (0, 6); the segment joining (4, 0) and (0, 6) is a diameter, giving x(x − 4) + y(y − 6) = 0.
36. The length of the chord cut off by the line x = 3 on the circle x² + y² = 25 is:
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Put x = 3 in the circle and find the two y-values.
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Answer: C. 8
At x = 3, y² = 16, so y = ±4 and the chord length is 8.
37. The tangents to the circle x² + y² = 4 which are parallel to the line 3x + 4y = 0 are:
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A parallel line is 3x + 4y = k; set its distance from the centre equal to the radius.
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Answer: B. 3x + 4y = ±10
Take 3x + 4y = k. The distance from the origin must equal 2: |k|/5 = 2, so k = ±10.
38. The circle x² + y² − 6x + 8y = 0 passes through the origin. The other end of the diameter through the origin is:
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The centre is the midpoint of every diameter.
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Answer: D. (6, −8)
The centre is (3, −4). The centre is the midpoint of the diameter, so the other end is (2·3 − 0, 2·(−4) − 0) = (6, −8).
39. The circle concentric with x² + y² − 4x − 6y − 3 = 0 and having twice its radius is:
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Keep g and f the same; find the new c from r² = g² + f² − c.
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Answer: B. x² + y² − 4x − 6y − 51 = 0
Centre (2, 3), r = √(4 + 9 + 3) = 4. The new radius is 8, so c = g² + f² − r² = 13 − 64 = −51.
40. The line x cosα + y sinα = p touches the circle x² + y² = a² if:
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Perpendicular distance from the centre = radius.
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Answer: B. p² = a²
The distance of the centre (0, 0) from the line is |p|/√(cos²α + sin²α) = |p|. Tangency requires |p| = a, i.e. p² = a².
4.3 Parabola, ellipse and hyperbola
26 questions
41. If the line x + y − 1 = 0 touches the parabola y² = kx, then the value of k is
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Substitute the line into the parabola and set the discriminant to zero.
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Answer: D. −4
Put x = 1 − y: y² = k(1 − y), so y² + ky − k = 0. Tangency needs a zero discriminant: k² + 4k = 0, so k = 0 or k = −4. k = 0 is not a parabola, so k = −4.
42. The focus of the parabola y² = 12x is:
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Compare with y² = 4ax.
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Answer: B. (3, 0)
Compare with y² = 4ax: 4a = 12, so a = 3 and the focus is (a, 0) = (3, 0).
43. The directrix of the parabola x² = −8y is:
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For x² = −4ay the directrix is y = a.
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Answer: C. y = 2
x² = −4ay with a = 2 opens downward, focus (0, −2), so the directrix is y = 2.
44. The length of the latus rectum of the parabola y² = −20x is:
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Latus rectum of a parabola = 4a.
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Answer: C. 20
Latus rectum = 4a = 20 (the sign only tells the direction of opening).
45. The focus of the parabola y² − 4y − 8x + 12 = 0 is:
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Complete the square in y to get (y − k)² = 4a(x − h).
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Answer: C. (3, 2)
(y − 2)² = 8x − 8 = 8(x − 1). Vertex (1, 2), 4a = 8 so a = 2; focus = (1 + 2, 2) = (3, 2).
46. The focal distance of the point on the parabola y² = 8x whose x-coordinate is 6 is:
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Focal distance on y² = 4ax is x + a.
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Answer: B. 8
For y² = 4ax, focal distance = x + a = 6 + 2 = 8.
47. The equation of the parabola with vertex at the origin and focus at (0, −3) is:
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Decide the axis and direction from the focus, then use 4a.
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Answer: A. x² = −12y
The focus lies on the negative y-axis with a = 3, so the parabola is x² = −4ay = −12y.
48. The line y = 2x + c is a tangent to the parabola y² = 16x when c equals:
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Tangency condition for y² = 4ax: c = a/m.
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Answer: D. 2
For y² = 4ax, y = mx + c is a tangent when c = a/m. Here a = 4, m = 2, so c = 2.
49. Which conic section has eccentricity exactly equal to 1?
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Recall the range of e for each conic.
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Answer: B. Parabola
A parabola has e = 1; an ellipse has 0 < e < 1, a hyperbola e > 1, and a circle e = 0.
50. The eccentricity of the ellipse x²/25 + y²/16 = 1 is:
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For an ellipse, e² = 1 − b²/a².
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Answer: A. 3/5
e² = 1 − b²/a² = 1 − 16/25 = 9/25, so e = 3/5.
51. The foci of the ellipse x²/25 + y²/9 = 1 are:
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ae = √(a² − b²).
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Answer: D. (±4, 0)
c² = a² − b² = 25 − 9 = 16, so c = 4 and the foci are (±4, 0) since the major axis is along x.
52. The length of the latus rectum of the ellipse x²/16 + y²/9 = 1 is:
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Latus rectum of an ellipse = 2b²/a.
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Answer: C. 9/2
Latus rectum = 2b²/a = 2·9/4 = 9/2.
53. For the ellipse 9x² + 4y² = 36, the major axis is:
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Write in standard form and see which denominator is larger.
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Answer: B. of length 6 along the y-axis
Dividing by 36: x²/4 + y²/9 = 1. The larger denominator 9 is under y², so the major axis is along the y-axis with length 2·3 = 6.
54. The ellipse with foci (±3, 0) and eccentricity 1/2 is:
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Use ae = 3 to find a, then b² = a²(1 − e²).
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Answer: C. x²/36 + y²/27 = 1
ae = 3 with e = 1/2 gives a = 6. b² = a² − a²e² = 36 − 9 = 27.
55. The directrices of the ellipse x²/9 + y²/5 = 1 are:
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Directrices of an ellipse: x = ±a/e.
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Answer: D. x = ±9/2
a = 3, ae = √(9 − 5) = 2, so e = 2/3. Directrices: x = ±a/e = ±3/(2/3) = ±9/2.
56. For any point on the ellipse x²/49 + y²/24 = 1, the sum of its distances from the two foci is:
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Sum of focal distances = length of major axis.
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Answer: C. 14
The sum of the focal distances of any point on an ellipse equals the major axis 2a = 2·7 = 14.
57. If the latus rectum of an ellipse is equal to half of its major axis, the eccentricity is:
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Set 2b²/a = a and use e² = 1 − b²/a².
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Answer: A. 1/√2
2b²/a = a gives b² = a²/2. Then e² = 1 − b²/a² = 1/2, so e = 1/√2.
58. The eccentricity of the hyperbola x²/16 − y²/9 = 1 is:
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For a hyperbola, e² = 1 + b²/a².
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Answer: D. 5/4
e² = 1 + b²/a² = 1 + 9/16 = 25/16, so e = 5/4.
59. The foci of the hyperbola y²/9 − x²/16 = 1 are:
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For a hyperbola c² = a² + b²; the foci lie on the transverse axis.
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Answer: B. (0, ±5)
The transverse axis is along y. c² = 9 + 16 = 25, so the foci are (0, ±5).
60. The eccentricity of a rectangular (equilateral) hyperbola is:
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Put a = b in e² = 1 + b²/a².
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Answer: C. √2
For a rectangular hyperbola a = b, so e² = 1 + 1 = 2 and e = √2.
61. The length of the latus rectum of the hyperbola x²/9 − y²/16 = 1 is:
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Latus rectum of a hyperbola = 2b²/a.
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Answer: B. 32/3
Latus rectum = 2b²/a = 2·16/3 = 32/3.
62. The equation x²/(9 − k) + y²/(5 − k) = 1 represents a hyperbola when:
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The denominators must have opposite signs.
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Answer: B. 5 < k < 9
A hyperbola needs the two denominators to have opposite signs: 9 − k > 0 and 5 − k < 0, i.e. 5 < k < 9. (For k < 5 it is an ellipse.)
63. The hyperbola whose vertices are (±3, 0) and foci are (±5, 0) is:
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For a hyperbola b² = c² − a².
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Answer: A. x²/9 − y²/16 = 1
a = 3, c = 5, so b² = c² − a² = 25 − 9 = 16.
64. The asymptotes of the hyperbola x²/9 − y²/4 = 1 are:
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Asymptotes: y = ±(b/a)x.
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Answer: B. y = ±(2/3)x
Asymptotes of x²/a² − y²/b² = 1 are y = ±(b/a)x = ±(2/3)x.
65. The eccentricity of a hyperbola is 2. The eccentricity of its conjugate hyperbola is:
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Use 1/e₁² + 1/e₂² = 1.
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Answer: C. 2/√3
For conjugate hyperbolas 1/e₁² + 1/e₂² = 1. So 1/4 + 1/e₂² = 1, giving e₂² = 4/3 and e₂ = 2/√3.
66. The centre of the hyperbola 9x² − 16y² − 18x − 64y − 199 = 0 is:
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Complete the squares in x and in y.
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Answer: B. (1, −2)
9(x − 1)² − 9 − 16(y + 2)² + 64 = 199 gives 9(x − 1)² − 16(y + 2)² = 144, i.e. (x − 1)²/16 − (y + 2)²/9 = 1. Centre (1, −2).
4.4 Coordinates in space and the plane
15 questions
67. The distance between the points (1, 2, 3) and (4, 6, 15) is:
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d = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²].
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Answer: C. 13
d = √(3² + 4² + 12²) = √(9 + 16 + 144) = √169 = 13.
68. The point dividing the join of A(2, −1, 3) and B(5, 2, −3) internally in the ratio 1 : 2 is:
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Section formula: (mx₂ + nx₁)/(m + n), etc.
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Answer: C. (3, 0, 1)
P = (1·B + 2·A)/3 = ((5 + 4)/3, (2 − 2)/3, (−3 + 6)/3) = (3, 0, 1). (4, 1, −1) is the 2 : 1 point.
69. The direction cosines of a line with direction ratios 2, −1, 2 are:
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Divide each direction ratio by √(a² + b² + c²).
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Answer: B. 2/3, −1/3, 2/3
√(2² + 1² + 2²) = 3, so the direction cosines are 2/3, −1/3, 2/3.
70. If a line makes angles α, β, γ with the coordinate axes, then sin²α + sin²β + sin²γ equals:
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Use cos²α + cos²β + cos²γ = 1.
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Answer: B. 2
sin²α + sin²β + sin²γ = 3 − (cos²α + cos²β + cos²γ) = 3 − 1 = 2.
71. A line makes an angle of 45° with the x-axis and 60° with the y-axis. The acute angle it makes with the z-axis is:
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Use cos²α + cos²β + cos²γ = 1.
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Answer: B. 60°
cos²γ = 1 − cos²45° − cos²60° = 1 − 1/2 − 1/4 = 1/4, so cosγ = ±1/2 and the acute angle is 60°.
72. The angle between two lines with direction ratios 1, 2, 2 and 2, 2, 1 is:
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cosθ = (a₁a₂ + b₁b₂ + c₁c₂)/(√(a₁² + b₁² + c₁²)·√(a₂² + b₂² + c₂²)).
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Answer: A. cos⁻¹(8/9)
cosθ = (1·2 + 2·2 + 2·1)/(√9·√9) = 8/9.
73. Lines with direction ratios 1, k, 3 and 2, 1, −2 are perpendicular when k equals:
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Use a₁a₂ + b₁b₂ + c₁c₂ = 0.
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Answer: B. 4
Perpendicular: 1·2 + k·1 + 3·(−2) = 0, so k − 4 = 0 and k = 4.
74. The direction cosines of the line joining (1, 2, 3) and (3, 5, 9) are:
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Direction ratios = differences of coordinates; then normalise.
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Answer: C. 2/7, 3/7, 6/7
Direction ratios: 3 − 1, 5 − 2, 9 − 3 = 2, 3, 6; √(4 + 9 + 36) = 7, so the direction cosines are 2/7, 3/7, 6/7.
75. The equation of the plane making intercepts 2, 3 and 6 on the x-, y- and z-axes respectively is:
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Intercept form: x/a + y/b + z/c = 1.
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Answer: A. 3x + 2y + z = 6
x/2 + y/3 + z/6 = 1; multiplying by 6 gives 3x + 2y + z = 6.
76. The perpendicular distance of the point (1, 1, 1) from the plane 2x + y + 2z + 4 = 0 is:
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d = |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²).
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Answer: C. 3
d = |2 + 1 + 2 + 4|/√(4 + 1 + 4) = 9/3 = 3.
77. The angle between the planes 2x − y + z = 6 and x + y + 2z = 7 is:
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The angle between two planes equals the angle between their normals.
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Answer: A. 60°
Angle between the normals: cosθ = (2 − 1 + 2)/(√6·√6) = 3/6 = 1/2, so θ = 60°.
78. The plane through (1, −1, 2) whose normal has direction ratios 2, 3, −1 is:
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Point-normal form: a(x − x₁) + b(y − y₁) + c(z − z₁) = 0.
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Answer: B. 2x + 3y − z + 3 = 0
2(x − 1) + 3(y + 1) − (z − 2) = 0 gives 2x + 3y − z + 3 = 0. Check: 2 − 3 − 2 + 3 = 0.
79. The planes 2x + ky + 3z = 5 and 3x − 2y + 4z = 7 are perpendicular when k equals:
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Use a₁a₂ + b₁b₂ + c₁c₂ = 0 for the normals.
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Answer: C. 9
Normals must be perpendicular: 2·3 + k(−2) + 3·4 = 0, so 18 − 2k = 0 and k = 9.
80. The centroid of the triangle with vertices (1, 2, 3), (3, 0, 1) and (2, 4, −1) is:
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Average the coordinates of the three vertices.
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Answer: A. (2, 2, 1)
Centroid = ((1 + 3 + 2)/3, (2 + 0 + 4)/3, (3 + 1 − 1)/3) = (2, 2, 1).
81. The distance of the point (3, −4, 5) from the x-axis is:
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Distance from the x-axis = √(y² + z²).
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Answer: D. √41
The foot of the perpendicular on the x-axis is (3, 0, 0), so the distance = √(4² + 5²) = √41.