Straight Lines(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

STRAIGHT LINES MCQs

Straight Lines

Total Questions : 60 | Page 5 of 6 pages

Question 41. If the line

(3 x + 14 y + 7) + k (5 x + 7 y + 6) = 0

is parallel to the y-axis, then the value of

k

13
−35
−2
2

Discuss Question

Answer: Option C. -> −2
:
C
Given line is

(3 x + 14 y + 7) + k (5 x + 7 y + 6) = 0

\Rightarrow (3 + 5 k) x + (14 + 7 k) y + (7 + 6 k) = 0

If it is parallel to y- axis, thencoefficient of y

= 0

\Rightarrow 14 + 7 k = 0

\Rightarrow k = - 2

Question 42. If the coordinates of the points A, B, C, be (4,4), (3,-2) and (3,-16) respectively, then the area of the triangle ABC is

Discuss Question

Answer: Option D. -> 7
:
D

△

\frac{1}{2}

[4(- 2 + 16) + 3(-16 - 4) + 3(4 + 2)]
=

\frac{1}{2}

[56 - 60 + 18] = 7.

Question 43. If the lines

(p - q) x^{2} + 2 (p + q) x y + (q - p) y^{2} = 0

are mutually perpendicular, then

p = q
q = 0
p = 0
p and q may have any value

Discuss Question

Answer: Option D. -> p and q may have any value
:
D
Pair of straight lines represented by a second degree equation with coefficient of

x^{2}

as a and coefficient of

y^{2}

as bare perpendicular if a+b = 0. Herere a + b = 0 for every p and q.

Question 44. The angle between the pair of straight lines

x^{2} + 4 y^{2} - 7 x y = 0,

tan−113
tan−13
tan−1√335
tan−15√33

Discuss Question

Answer: Option C. -> tan−1√335
:
C

tan θ = \frac{2 \sqrt{h^{2} - a b}}{a + b}

θ = t a n^{- 1} \frac{2 \sqrt{\frac{49}{4} - 4}}{5} = t a n^{- 1} \frac{\sqrt{33}}{5}

Question 45. The equations of two equal sides of an isosceles triangle are 7x – y + 3 = 0 and x + y –3 = 0 and the third side passes through the point (1, -10). The equation of the third side is

x – 3y – 31 = 0 but not 3x + y + 7 = 0
3x + y + 7 = 0 but not x – 3y – 31 = 0
3x + y + 7 = 0 or x – 3y – 31 = 0
Neither 3x + y + 7 nor x – 3y – 31 = 0

Discuss Question

Answer: Option C. -> 3x + y + 7 = 0 or x – 3y – 31 = 0
:
C
Any line through (1, - 10) is given by y + 10 = m(x - 1)
Since it makes equal angle

^{'} α^{'}

with the given lines 7x – y + 3 = 0 and x + y – 3 = 0, therefore

t a n α = \frac{m - 7}{1 + 7 m} = \frac{m + 1}{1 + m (- 1)} \Rightarrow m = \frac{1}{3} o r - 3

Hence the two possible equations of third side are 3x + y + 7 = 0 and x - 3y - 31 = 0.

Question 46. Let PS be the median of the triangle with vertices P(2, 2), Q(6, -1) and R(7, 3). The equation of the line passing through (1, -1) and parallel to PS is

2x – 9y – 7 = 0
2x – 9y – 11 = 0
2x + 9y – 7 = 0
2x – 9y + 7 = 0

Discuss Question

Answer: Option D. -> 2x – 9y + 7 = 0
:
D
S = midpoint of QR =

(\frac{6 + 7}{2}, \frac{- 1 + 3}{2}) = (\frac{13}{2}, 1) ∴ slope of P S = \frac{2 - 1}{2 - \frac{13}{2}} = - \frac{2}{9} ∴ The required equation is y + 1 = \frac{- 2}{9} (x - 1) i . e ., 2 x + 9 y + 7 = 0

Question 47. If the co-ordinates of the middle point of the portion of a line intercepted between coordinate axes is (3, 2), then the equation of the line will be