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

12th Grade > Mathematics

STRAIGHT LINES MCQs

Straight Lines

Total Questions : 60 | Page 3 of 6 pages

Question 21. 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 22. The lines joining the points of intersection of line x + y = 1 and curve

x^{2} + y^{2} - 2 y + λ = 0

to the origin are perpendicular, then the value of

λ

will be

12
−12
1√2
0

Discuss Question

Answer: Option D. -> 0
:
D
Making the equation of curve homogeneous with the help of line x + y =1,we get

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

\Rightarrow x^{2} (1 + λ) + y^{2} (- 1 + λ) - 2 y x = 0

Therefore the lines be perpendicular, if A +B = 0.

\Rightarrow 1 + λ - 1 + λ = 0 \Rightarrow λ = 0

Question 23. The orthocenter of the triangle formed by (0, 0), (8, 0) and (4, 6) is

4,83
3,4
4,3
−3,4

Discuss Question

Answer: Option A. -> 4,83
:
A
Sol: Let A ≡ (0, 0), B ≡ (8, 0) and C ≡ (4, 6).

The Orthocenter Of The Triangle Formed By (0, 0), (8, 0) And...

Slope of BC =

\frac{6 - 0}{4 - 0} = \frac{3}{2}

Equation of the line through A(0, 0) and perpendicular to BC is
(y – 0) =

\frac{2}{3}

(x – 0) i.e. 2x – 3y = 0 …… (1)
Slope of CA =

\frac{6 - 0}{4 - 0} = \frac{3}{2}

Equation of the line through B(8, 0) and perpendicular to CA is
(y – 0) =

- \frac{2}{3}

(x – 8) i.e., 2x + 3y = 16 …… (2)
Solving (1) and (2), the orthocenter is

4, \frac{8}{3}

Question 24. O is the origin and A is the point (3,4). If a point P moves so that the line segment OP is always parallel to the line segment OA, then the equation to the locus of P is

4x - 3y = 0
4x + 3y = 0
3x + 4y = 0
3x - 4y = 0

Discuss Question

Answer: Option A. -> 4x - 3y = 0
:
A
Since OA and OP will be parallel only when O, A and P are collinear.
Therefore,

∣ ∣∣ ∣ \begin{matrix} 0 & 0 & 1 3 & 4 & 1 x & y & 1 \end{matrix} ∣ ∣∣ ∣

= 0

\Rightarrow

4x - 3y = 0.

Question 25. 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 26. A point P moves so that its distance from the point (a, 0) is always equal to its distance from the line x + a = 0. The
locus of the point is

y2=4ax
x2=4ay
y2+4ax=0
x2+4ay=0

Discuss Question

Answer: Option A. -> y2=4ax
:
A

(x - a)^{2} + y^{2} = (x + a)^{2} \Rightarrow y^{2} = 4 a x

Note: This is also the definition of parabola

y^{2}

= 4ax.

Question 27. If P = (1, 0), Q = (-1, 0) and R = (2, 0) are three given points, then the locus of a point S satisfying the relation

S Q^{2} + S R^{2} = 2 S P^{2}

A straight line parallel to x-axis
A circle through origin
A circle with centre at the origin
A straight line parallel to y-axis

Discuss Question

Answer: Option D. -> A straight line parallel to y-axis
:
D
Let S(x, y), then

(x + 1)^{2} + y^{2} + (x - 2)^{2} + y^{2} = 2 [(x - 1)^{2} + y^{2}]

\Rightarrow

2x +1 + 4 - 4x = - 4x + 2

\Rightarrow

x = -

\frac{3}{2}

Hence it is a straight line parallel to y-axis.

Question 28. The distance between the parallel lines

8 x + 6 y + 5 = 0

and

4 x + 3 y - 25 = 0

Discuss Question

Answer: Option C. -> 112
:
C
Distance between parallel lines

a x + b y + c_{1} = 0

and

a x + b y + c_{2} = 0

\frac{| c_{1} - c_{2} |}{\sqrt{a^{2} + b^{2}}}

Given lines are

8 x + 6 y + 5 = 0

and

4 x + 3 y - 25 = 0

8 x + 6 y - 50 = 0

∴

Required distance =

\frac{5 + 50}{\sqrt{8^{2} + 6^{2}}} = \frac{55}{10} = \frac{11}{2}

Question 29. If (

α, β

(¯ x, ¯ y) a n d (u, v)

are respectively coordinates of the circumcentre, centroid and orthocentre of a triangle.

3¯x =2α+u and 3¯y =2β+v
3¯x =2α−u and 3¯y =2β−v
3¯x =2α−u and 3¯y =2β+v
3¯x =2α+u and 3¯y =2β−v

Discuss Question

Answer: Option C. -> 3¯x =2α−u and 3¯y =2β+v
:
C
We know that, the centroid of a triangle divides the segment joining the orthocentre and circumcentre internally in the ratio 2 : 1. Therefore,

¯ x = \frac{2 α + u}{2 + 1} a n d ¯ y = \frac{2 β + v}{2 + 1}

\Rightarrow 3 ¯ x = 2 α + u a n d ¯ y = 2 β + v

Question 30. 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