Straight Lines(11th Grade > Mathematics ) Questions and answers for exam Preparation

11th Grade > Mathematics

STRAIGHT LINES MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11.

If the vertices of a triangle be (a, 1), (b, 3) and (4, c), then the centroid of the triangle will lie on x-axis, if

a + c = -4
a + b = -4
c = -4
b + c = -4

Discuss Question

Answer: Option C. -> c = -4
:
C

The point lies on axis of x, if y = 0.

Therefore, $\frac{1 + 3 + c}{3}$ = 0 $\Rightarrow$ c = -4.

Question 12.

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 13.

If the vertices of a triangle be (a, b - c), (b, c - a) and (c, a - b), then the centroid of the triangle lies

At origin
On x-axis
On y-axis
(a+b+c,0)

Discuss Question

Answer: Option B. -> On x-axis
:
B

x = $\frac{a + b + c}{3}$ , y = $\frac{b - c + c - a + a - b}{3}$ = 0

Hence, centroid lies on x - axis.

Question 14.

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

$y^{2} = 4 a x$
$x^{2} = 4 a y$
$y^{2} + 4 a x = 0$
$x^{2} + 4 a y = 0$

Discuss Question

Answer: Option A. ->

y^{2} = 4 a x

:
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 15.

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

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

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 16.

If two vertices of a triangle are (6,4), (2,6) and its centroid is (4, 6), then the third vertex is

(4,8)
(8,4)
(6,4)
(0,0)

Discuss Question

Answer: Option A. -> (4,8)
:
A

Given:
Centroid = $(4, 6)$
Vertices $(6, 4) & (2, 6)$
Let the Co-ordinates of $C$ be $(x_{3}, y_{3})$
$x_{1} = 6, x_{2} = 32, y_{1} = 4 & y_{2} = 6$

Centroid $(4, 6) = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

$\Rightarrow 4 = \frac{6 + 2 + x_{3}}{3}$ and $6 = \frac{4 + 6 + y_{3}}{3}$
$\Rightarrow$ $x_{3} = 4$ and $y_{3} = 8$

$∴$ Third vertex is $(4, 8)$ .

Question 17.

If A(4, -3), B(3, -2) and C(2, 8) are the vertices of a triangle, then its centroid will be

(-3,3)
(3,3)
(3,1)
(1,3)

Discuss Question

Answer: Option C. -> (3,1)
:
C

Let the centroid of the triangle be (x, y).

The centroid of a triangle is given by $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$

$x = \frac{4 + 3 + 2}{3} = 3$

$y = \frac{- 3 - 2 + 8}{3} = 1$

Question 18.

The equation of the internal bisector of $∠$ BAC of $Δ$ ABC with vertices A(5, 2), B(2, 3) and C(6, 5), is

2x + y + 12 = 0
x + 2y – 12 = 0
2x + y – 12 = 0
x + 2y +12 = 0

Discuss Question

Answer: Option C. -> 2x + y – 12 = 0
:
C
Let AD be the internal bisector of angle BAC cutting BC at D.

$N o w, A B = \sqrt{(5 - 2)^{2} + (2 - 3)^{2}} = \sqrt{10} a n d A C = \sqrt{(5 - 6)^{2} + (2 - 5)^{2}} = \sqrt{10}$

since AD is the internal bisector of angle BAC,

$∴ \frac{B D}{D C} = \frac{A B}{A C} = \frac{\sqrt{10}}{\sqrt{10}} = \frac{1}{1}$

$∴$ Coordinates of D are $(\frac{2 + 6}{2}, \frac{3 + 5}{2})$ i.e. (4, 4)

So, the equation of AD is

$y - 2 = \frac{2 - 4}{5 - 4}$ (x – 5) or 2x + y – 12 = 0

Question 19.

If the orthocenter and circumcentre of a triangle are (0,0) and (3,6) respectively then the centroid of the triangle is

(1,2)
(2,4)
$(\frac{2}{3}, \frac{4}{3})$
$(\frac{1}{3}, \frac{2}{3})$

Discuss Question

Answer: Option B. -> (2,4)
:
B
In any triangle centroid divides the line joining orthocenter and circumcentre internally in the ratio 2 : 1.
Applying section formula to find the point which divides the line joining (0,0) in the ratio 2:1 , we get the coordinated of centroid equal to (2,4).

Question 20.

If h denote the A.M, k denote G.M of the intercepts made on axes by the lines passing through (1, 1) then (h, k) lies on