Straight Lines(11th And 12th > Mathematics ) Questions and answers for exam Preparation

11th And 12th > Mathematics

STRAIGHT LINES MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

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 2.

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

$4, \frac{8}{3}$
$3, 4$
$4, 3$
$- 3, 4$

Discuss Question

Answer: Option A. ->

4, \frac{8}{3}

:
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 3.

Without changing the direction of coordinate axes, origin is transferred to (h, k), so that the linear (one degree)

terms in the equation $x^{2} + y^{2} - 4 x + 6 y - 7 = 0$ are eliminated. Then the point (h, k) is

(3, 2)
(- 3, 2)
(2, - 3)
(1.7)

Discuss Question

Answer: Option C. -> (2, - 3)
:
C

Putting x = x' + h, y = y' + k, the given equation transforms to

$x^{' 2} + y^{' 2} + x^{'} (2 h - 4) + y^{'} (2 k + 6) + h^{2} + k^{2} - 7 = 0$

To eliminate linear terms, we should have

2h - 4 = 0, 2k + 6 = 0 $\Rightarrow$ h = 2, k = -3

i.e., (h, k) = (2, -3).

Question 4.

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 5.

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 6.

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 7.

Let A (h, k), B (1, 1) and C (2, 1) be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is 1, then the set of values which `k' can take is given by