MCQ

MCQ 11 Mark

Two lines are said to be perpendicular if the product of their slope is equal to:

✓
-1
B
0
C
1
D
$\frac{1}{2}$

Answer

Correct option: A.

-1

-1

Solution:
When two lines are perpendicular, then the product of their slope is equal to -1. If two lines are perpendicular with slope $m_1$ and $m_2$, then $m_1.m_2$ = -1.

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MCQ 21 Mark

If slope of a line is 4 and y-intercept made by the line is 2 then the equation of line will be:

A
y = 4x - 2
✓
y = 4x + 2
C
y = 2x + 4
D
y = 2x - 4

Answer

Correct option: B.

y = 4x + 2

Let general equation of line be y = m × x + c.
Given, m = 4 and c = 2.
⇒ y = 4x + 2

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MCQ 31 Mark

Given the three straight lines with equations 5x + 4y = 0, x + 2y - 10 = 0 and 2x + y + 5 = 0, then these lines are:

A
None of these
B
The sides of a right angled triangle
✓
Concurrent
D
The sides of an equilateral triangle

Answer

Correct option: C.

Concurrent

Concurrent

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MCQ 41 Mark

Choose the correct answer. A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is:

A
(1, -1)
✓
(1, 1)
C
(0, 0)
D
(0, 1)

Answer

Correct option: B.

(1, 1)

(0, 0)

Solution:
Given equation are
$4x + 3y + 10 = 0 .....(i)$
$5x - 12y + 26 = 0 .....(ii)$
and $7x + 27y - 50 = 0 .....(iii)$
Let $(x_1, y_1)$ be any point equidistant from eq. (i), eq. (ii) and eq. (iii).
Distance of $(x_1, y_1)$ from eq. (i)
$=\Big|\frac{4\text{x}_1+3\text{y}_1+10}{\sqrt{16+9}}\Big|=\Big|\frac{4\text{x}_1+3\text{y}_1+10}{5}\Big|$
Distance of $(x_1, y_1)$ from eq. (iii)
$=\Big|\frac{5\text{x}_1-12\text{y}_1+26}{\sqrt{25+144}}\Big|=\Big|\frac{5\text{x}_1-12\text{y}_1+26}{13}\Big|$
Distance of $(x_1, y_1)$ from eq. (iii)
$=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{\sqrt{49+576}}\Big|=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{25}\Big|$
If the point $(x_1, y_1)$ is equidistant from the given lines, then
$\Big|\frac{4\text{x}_1+3\text{y}_1+10}{5}\Big|=\Big|\frac{5\text{x}_1-12\text{y}_!+26}{13}\Big|=\Big|\frac{7\text{x}_1+24\text{y}_1-50}{25}\Big|$
We see that putting $x_1 = 0$ and $y_1 = 0$, the above relation is satisfied i.e.,
$=\frac{10}{5}=\frac{26}{13}=\frac{50}{25}=2$
Hence, the correct option is (c).

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MCQ 51 Mark

Find the equation of line parallel to 4x + y = 2 and pass through (2, 5):

✓
4x + y - 13 = 0
B
4x + y + 13 = 0
C
4x - y - 13 = 0
D
4x - y + 13 = 0

Answer

Correct option: A.

4x + y - 13 = 0

Line 4x + y = 2 hasslope -4 Line parallel to it has
slope -4 and pass through (2, 5)
so equation will be y - 5 = (-4) (x - 2)
⇒ 4x + y - 13 = 0

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MCQ 61 Mark

If slope of a line is positive then its inclination is:

A
Right angle.
✓
Acute angle.
C
Obtuse angle.
D
Zero.

Answer

Correct option: B.

Acute angle.

Acute angle.

Solution:
If inclination is $\alpha$ slope is given by $\tan\alpha$ Given that slope of line is positive which means $\tan\alpha$ is positive. We know, $\tan\alpha$ is positive in $1^{st}$ quadrant i.e. $\alpha$ should be acute angle.

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MCQ 71 Mark

What is the distance of (5, 12) from origin?

A
6 units.
B
8 units.
C
10 units.
✓
13 units.

Answer

Correct option: D.

13 units.

13 units.

Solution:
We know, distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{({\text{x}}_{1}-{\text{x}}_{2})^{2}+({\text{y}}_{1}-{\text{y}}_{2})^{2}}$
So, distance between (5, 12) from origin (0, 0) is $\sqrt{({5-0})^{2}+({12-0})^{2}} = \sqrt{({5})^{2}+({12})^{2}} =13\text{ unit}.$

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MCQ 81 Mark

Three vertices of a parallelogram taken in order are (-1, -6), (2, -5) and (7, 2). The fourth vertex is:

A
(1, 4)
✓
(4, 1)
C
(1, 1)
D
(4, 4)

Answer

Correct option: B.

(4, 1)

(4, 1)

Solution:
Let A(-1, -6), B(2, -5) and C(7, 2) be the given vertex. Let D(h, k) be the fourth vertex.
The midpoints of AC and BD are (3, -2) and $\Big(\frac{2+\text{h}}{2},\frac{-5+\text{k}}{2}\Big)$ respectively.
We know that the diagonals of a parallelogram bisect each other.
$\therefore3=\frac{2+\text{h}}{2}$ and $-2=\frac{-5+\text{k}}{2}$
$\Rightarrow\text{h}=4$ and $\text{k}=1$

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MCQ 91 Mark

If slope of a line is $\frac{2}{3}$ then find the slope of line perpendicular to it:

✓
$\frac{-3}{2}$
B
$\frac{3}{2}$
C
$\frac{2}{3}$
D
$\frac{-2}{3}$

Answer

Correct option: A.

$\frac{-3}{2}$

$\frac{-3}{2}$

Solution:
If lines with slopes $m_1$ and $m_2$ are perpendicular then $m_1 \times m_2 = -1$.
If $\text{m}_{1} = \frac{2}{3}$ then $\text{m}_{2} = \frac{-3}{2} $

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MCQ 101 Mark

The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1:

A
Lies in the III quadrant.
B
Lies in the II quadrant.
C
Lies in the I quadrant.
✓
Cannot be found.

Answer

Correct option: D.

Cannot be found.

Cannot be found.

Solution:
The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1 is
$\Big(\frac{1\times3-1\times1}{1-1},\frac{1\times4-1\times2}{1-1}\Big)$
which is not defined.
Therefore, it is not possible to externally divide the line joining two points in the ratio 1 : 1

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MCQ 111 Mark

Find the equation of line parallel to y-axis and passing through (3, 4):

✓
x = 3
B
x = 4
C
y = 4
D
y = 3

Answer

Correct option: A.

x = 3

x = 3

Solution:
Let general equation of line be y = m (x - d)
$\Rightarrow\text{x} = \frac{\text{y}}{\text{m + d}}$
Since line is parallel to y-axis so, $\text{m}=\frac{1}{0}$ or $\frac{1}{\text{m}} =0$
⇒ x = d
⇒ x = 3 by substituting the point (3, 4).

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MCQ 121 Mark

Choose the correct answer. Equation of the line passing through (1, 2) and parallel to the line y = 3x - 1 is:

A
y + 2 = x + 1
B
y + 2 = 3 (x + 1)
✓
y - 2 = 3 (x - 1)
D
y - 2 = x - 1

Answer

Correct option: C.

y - 2 = 3 (x - 1)

y - 2 = 3 (x - 1)

Solution:
Given equation is y = 3x - 1
Slope = 3
Slope of the line passing through the given point (1, 2) and parallel to the given line = 3
So, the equation of the required line is
y - 2 = 3(x - 1)
Hence, the correct option is (c).

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MCQ 131 Mark

Choose the correct answer. The tangent of angle between the lines whose intercepts on the axes are a, -b and b, -a, respectively, is

A
$\frac{\text{a}^2-\text{b}^2}{\text{ab}}$
B
$\frac{\text{b}^2-\text{a}^2}{2}$
✓
$\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$
D
None of these.

Answer

Correct option: C.

$\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$

$\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$

Solution:
Intercepts of line are a and -b; i.e., line passes through the points (a, 0), (0, -b).
$\therefore$ Slope of line, $\text{m}_1=\frac{-\text{b}-0}{0-\text{a}}=\frac{\text{b}}{\text{a}}$
Intercepts of line are b, -a; i.e., line passes through the points (b, 0), (0, -a).
$\therefore$ Slope of line, $\text{m}_2=\frac{-\text{a}-0}{0-\text{b}}=\frac{\text{a}}{\text{b}}$
If $\theta$ is the angle between the lines, then
$\tan=\theta=\frac{\frac{\text{b}}{\text{a}}-\frac{\text{a}}{\text{b}}}{1+\frac{\text{a}}{\text{b}}\times\frac{\text{b}}{\text{a}}}=\frac{\frac{\text{b}^2-\text{a}^2}{\text{ab}}}{2}=\frac{\text{b}^2-\text{a}^2}{2\text{ab}}$

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MCQ 141 Mark

Find slope of line passing through origin and (3, 6):

✓
$2$
B
$3$
C
$\frac{1}{3}$
D
$\frac{1}{2}$

Answer

Correct option: A.

$2$

Solution:
We know, slope of line joining two points ($x_1, y_1$) and ($x_2, y_2$) is given by $\frac{(\text{y}_{2}-\text{y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$
So, slope of line joining (0, 0) and (3, 6) is $\frac{(6-0)}{(3-0)} = \frac{6}{3} = 2$

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MCQ 151 Mark

A triangle ABC is right angled at A has points A and B as (2, 3) and (0, -1) respectively. If BC = 5, then point C may be:

A
(-4, 2)
B
(4, -2)
✓
(0, 4)
D
(0, -4)

Answer

Correct option: C.

(0, 4)

(0, 4)

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MCQ 161 Mark

L is a variable line such that the algebraic sum of the distances of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero. The line L will always pass through:

✓
(1, 1)
B
(2, 1)
C
(1, 2)
D
none of these.

Answer

Correct option: A.

(1, 1)

(1,1)

Solution:
Let ax + by + c = 0 be the variable line. It is given that the algebraic sum of the distances of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero
$\therefore \ \frac{\text{a}+\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{2\text{a}+0+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{0+2\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}=0$
$\Rightarrow3\text{a}+3\text{b}+3\text{c}=0$
$\Rightarrow\text{a}+\text{b}+\text{c}=0$
Substituting c = -a - b in ax + by + c = 0, we get:
$\text{ax}+\text{by}-\text{a}-\text{b}=0$
$\Rightarrow\text{a}(\text{x}-1)+\text{b}(\text{y}-1)=0$
$\Rightarrow(\text{x}-1)+\frac{\text{a}}{\text{b}}(\text{y}-1)=0$
This line is of the form $\text{L}_1+\lambda\text{L}_2=0$ which passes through the intersection of $L_1 =0$ and $L_2 =0$,i.e. x - 1 = 0 and y - 1 = 0.
⇒ x = 1, y = 1

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MCQ 171 Mark

Equation of horizontal line below x-axis at 5 units from x-axis is:

A
x = 5
B
x = -5
C
y = 5
✓
y = -5

Answer

Correct option: D.

y = -5

y = -5

Solution:
Equation of x-axis is y = 0. Horizontal line is parallel to x-axis and below it by 5 units so, equation of line is y = -5.

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MCQ 181 Mark

Choose the correct answer. The equation of the line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0 is:

A
y - x + 1 = 0
✓
y - x - 1 = 0
C
y - x + 2 = 0
D
y - x - 2 = 0

Answer

Correct option: B.

y - x - 1 = 0

Slope of the given line +1 = 0 is -1.
So, slope of line perpendicular to above line is 1.
Line passes through the point (1, 2).
Therefore, equation of the required linens:
⇒ y - 2 = 1(x - 1)
⇒ y - x - 1 = 0.

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MCQ 191 Mark

Choose the correct answer. The equations of the lines passing through the point (1, 0) and at a distance $\frac{\sqrt{3}}{2}$ from the origin, are

A
$\sqrt{3}\text{x}+\text{y}-\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0$
B
$\sqrt{3}\text{x}+\text{y}+\sqrt{3}=0,\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0$
C
$\text{x}+\sqrt{3}\text{y}-\sqrt{3}=0,\text{x}-\sqrt{3}\text{y}-\sqrt{3}=0$
✓
None of these.

Answer

Correct option: D.

None of these.

Equation of any line passing through (1, 0) is
⇒ y - 0 = m(x - 1)
⇒ mx - y - m = 0
Distance of the line from origin is $\frac{\sqrt{3}}{2}$
$\therefore \frac{\sqrt{3}}{2} =\Big|\frac{\text{m}\times0-0-\text{m}}{\sqrt{1+\text{m}^2}}\Big|$
$\Rightarrow \frac{\sqrt{3}}{2}=\Big|\frac{-\text{m}}{\sqrt{1+\text{m}^2}}\Big|$
Squaring both sides, we get
$\frac{3}{4}=\frac{\text{m}^2}{1+\text{m}^2}$
$\Rightarrow 4\text{m}^2=3+3\text{m}^2$
$\Rightarrow 4\text{m}^2-3\text{m}^2=3$
$\Rightarrow \text{m}^2=3$
$\therefore \text{m}=\pm\sqrt{3}$
$\therefore$ Required equations are
$\pm\sqrt{3}\text{x}-\text{y}\mp\sqrt{3}=0$
i.e., $\sqrt{3}\text{x}-\text{y}-\sqrt{3}=0=0$ and $-\sqrt{3}\text{x}-\text{y}+\sqrt{3}=0=0$
$\Rightarrow \sqrt{3}\text{x}+\text{y}-\sqrt{3}=0$
Hence, the correct option is (a).

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MCQ 201 Mark

If the points A (1, 2), B (2, 4) and C (3, a) are collinear, what is the length BC?

A
2 unit
B
3 unit
✓
5 unit
D
5 unit

Answer

Correct option: C.

5 unit

5 unit

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MCQ 211 Mark

If the two lines with slope $m_1$ and $m_2$ are perpendicular then their slopes has relation:

A
$m_1+m_2=1$
B
$m_1 \times m_2=1$
✓
$m_1 \times m_2=-1$
D
$m_1+m_2=-1$

Answer

Correct option: C.

$m_1 \times m_2=-1$

$m_1 \times m_2=-1$

Solution:
If the two lines are perpendicular then if one line form angle $\alpha$ with positive x-axis then the other line form angle $90^\circ + \alpha$
If $\text{m}_{1} = \tan \alpha$ then $m_2$ will be $\tan (90^\circ + \alpha) = -\cot\alpha = \frac{-1}{\tan\alpha}$
$\Rightarrow m_1 \times m_2 = -1$.

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MCQ 221 Mark

If P (1, 2), Q (3, 5), R (7, 9) form a triangle then find the equation of median through P:

✓
5x - 4y + 3 = 0
B
5x + 4y + 3 = 0
C
5x - 4y - 3 = 0
D
5x + 4y - 3 = 0

Answer

Correct option: A.

5x - 4y + 3 = 0

Midpoint of QR line is $\Big(\frac{3+7}{2},\frac{5+9}{2}\Big) = (5, 7)$
Equation of line joining (1, 2) and (5, 7) is $\frac{\text{y - 2}}{7 - 2} = \frac{\text{x - 1}}{5-1}$
$\Rightarrow\frac{\text{y}-2}{5} = \frac{\text{x}-1}{4}$
⇒ 4y - 8 = 5x - 5
⇒ 5x - 4y + 3 = 0.

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MCQ 231 Mark

A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is:

A
$\frac{1}{3}$
B
$\frac{2}{3}$
C
$1$
✓
$\frac{4}{3}$

Answer

Correct option: D.

$\frac{4}{3}$

$\frac{4}{3}$

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MCQ 241 Mark

The acute angle between the medians drawn from the acute angles of a right angled isosceles triangle is:

A
$\cos^{-1}\big(\frac{2}{3}\big)$
✓
$\cos^{-1}\big(\frac{3}{4}\big)$
C
$\cos^{-1}\big(\frac{4}{5}\big)$
D
$\cos^{-1}\big(\frac{5}{6}\big)$

Answer

Correct option: B.

$\cos^{-1}\big(\frac{3}{4}\big)$

$\cos^{-1}\big(\frac{3}{4}\big)$

Solution:
Let the coordinates of the right-angled isosceles triangle be O(0, 0), A(a, 0) and B(0, a).

Here, BD and AE are the medians drawn from the acute angles B and A, respectively.
$\therefore$ Slope of $BD = m_1$
$=\frac{0-\text{a}}{\frac{\text{a}}{2}-0}$
$=-\frac{1}{2}$
Let $\theta$ be the angle between BD and AE.
$\tan\theta=\Big|\frac{-2+\frac{1}{2}}{1+1}\Big|$
$=\frac{3}{4}$
$\Rightarrow\cos\theta=\frac{4}{\sqrt{3^2+4^2}}$
$\Rightarrow\cos\theta=\frac{4}{5}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{4}{5}\Big)$
Hence, the acute angle between the medians is $\cos^{-1}\Big(\frac{4}{5}\Big).$

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MCQ 251 Mark

..... is the midpoint of (1, 2) and (5, 8):

A
(2, 5)
✓
(3, 5)
C
(5, 2)
D
(5, 3)

Answer

Correct option: B.

(3, 5)

(3, 5)

Solution:
We know, midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ is $\Big(\frac{{\text{x}}_{1}+{\text{x}}_{2}}{2}, \frac{{\text{y}}_{1}+{\text{y}}_{2}}{2}\Big)$
So, midpoint of (1, 2) and (5, 8) is $\Big(\frac{1+5}{2}, \frac{2+8}{2}\Big)$ is (3, 5)

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MCQ 261 Mark

If line joining (1, 2) and (5, 7) is parallel to line joining (3, 4) and (11, x):

A
10
B
11
C
12
✓
14

Answer

Correct option: D.

Solution:
We know, slope of line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{(\text{y}_{2}-\text{y}_{1})}{(\text{x}_{2}-\text{x}_{1})}$
Lines are parallel means slope is equal.
$\Rightarrow\frac{(\text{x}-4)}{(11-3)} =\frac {(7-2)}{(5-1)}$
$\text{x}-4 = 5\times\frac{8}{4} = 10$
⇒ x = 14

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MCQ 271 Mark

If (-4, 5) is one vertex and 7x - y + 8 = 0 is onediagonal of a square, then the equation of second diagonal is:

A
x + 3y = 21
B
2x - 3y = 7
✓
x + 7y = 31
D
2x + 3y = 21

Answer

Correct option: C.

x + 7y = 31

x + 7y = 31

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MCQ 281 Mark

Slope of a line is given by if inclination of line is $\alpha$:

A
$\sin\alpha$
B
$\cos\alpha$
✓
$\tan\alpha$
D
$\cot\alpha$

Answer

Correct option: C.

$\tan\alpha$

Slope of a line is given by $\tan\alpha$ if inclination of line is $\alpha$. Slope is denoted by tangent of the inclination angle.

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MCQ 291 Mark

Equation of vertical line to the right of y-axis at 5 units from y-axis is:

✓
x = 5
B
x = -5
C
y = 5
D
y = -5

Answer

Correct option: A.

x = 5

Equation of y-axis is x = 0. Vertical line is parallel to y-axis and to the right by 5 units so, equation of line is x = 5.

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MCQ 301 Mark

In what ratio does the line y - x + 2 = 0 cut the line joining (3, -1) and ( 8, 9)?

✓
2 : 3
B
3 : 2
C
3 : -2
D
1 : 2

Answer

Correct option: A.

2 : 3

2 : 3

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MCQ 311 Mark

The medians AD and BE of a triangle with vertices A(0, b), B(0, 0) and C(a, 0) are perpendicular to each other, if

A
$\text{a}=\frac{\text{b}}{2}$
B
$\text{b}=\frac{\text{a}}{2}$
C
$\text{ab}=1$
✓
$\text{a}=\pm\sqrt{2}\text{b}$

Answer

Correct option: D.

$\text{a}=\pm\sqrt{2}\text{b}$

The midpoints of BC and AC are $\text{D}\Big(\frac{\text{a}}{2},0\Big)$ and $\text{E}\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$
Slope of $\text{AD}=\frac{0-\text{b}}{\frac{\text{a}}{2}-0}$
Slope of $\text{BE}=\frac{-\frac{\text{b}}{2}}{\frac{\text{-a}}{2}}$
It is given that the medians are perpendicular to each other.
$\frac{0-\text{b}}{\frac{\text{a}}{2}-0}\times\frac{-\frac{\text{b}}{2}}{-\frac{\text{a}}{2}}=-1$
$\Rightarrow\text{a}=\pm\sqrt{2}\text{b}$

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MCQ 321 Mark

Choose the correct answer. Slope of a line which cuts off intercepts of equal lengths on the axes is:

✓
-1
B
-0
C
2
D
$\sqrt{3}$

Answer

Correct option: A.

-1

Intercept form of a line is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
$\Rightarrow \frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}=1(\because \text{a}=\text{b})$
⇒ x + y = a
⇒ y =- -x + a
$\therefore$ Slope is -1
Hence, the correct option is (a).

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MCQ 331 Mark

If the two lines are perpendicular then difference of their inclination angle is:

A
45°
B
60°
✓
90°
D
180°

Answer

Correct option: C.

90°

If the two lines are perpendicular then if one line form angle $\alpha$ with positive x-axis then the other line form angle $90^\circ+\alpha$

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MCQ 341 Mark

If the area of the triangle with vertices (x, 0), (1, 1) and (0, 2) is 4 square unit, then the value of x is:

A
-2
B
-4
✓
-6
D
8

Answer

Correct option: C.

-6

-6

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MCQ 351 Mark

What is the inclination of a line which is parallel to x-axis?

✓
0°
B
180°
C
45°
D
90°

Answer

Correct option: A.

0°

If a line is parallel to x-axis then angle formed by it with x-axis is zero. So, its inclination is zero.

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MCQ 361 Mark

The line segment joining the points (-3, -4) and (1, -2) is divided by y-axis in the ratio:

A
1 : 3
B
2 : 3
✓
3 : 1
D
3 : 2

Answer

Correct option: C.

3 : 1

Let the points (-3, -4) and (1, -2) be divided by y-axis at (0, t) in the ratio m : n.
$\therefore\Big(\frac{\text{m}-3\text{n}}{\text{m}+\text{n}},\frac{-2\text{m}-4\text{n}}{\text{m}+\text{n}}\Big)=(0,\text{t})$
$\Rightarrow0=\frac{\text{m}-3\text{n}}{\text{m}+\text{n}}$
$\Rightarrow\text{m}:\text{n}=3:1$

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MCQ 371 Mark

If a line with slope m makes x-intercept d. Then equation of the line is:

A
y = m(d - x)
✓
y = m(x - d)
C
y = m(x + d)
D
y = mx + d

Answer

Correct option: B.

y = m(x - d)

y = m(x - d)

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MCQ 381 Mark

The condition for the points (x, y), (-2, 2) and (3, 1) to be collinear is:

✓
x + 5y = 8
B
x + 5y = 6
C
5x + y = 8
D
5x + y = 6

Answer

Correct option: A.

x + 5y = 8

x (2 - 1) -1(1 - y) + 3 (y - 2) = 0 or x + 5y = 8

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MCQ 391 Mark

If equation of a line is y = 3x - 4 then find the slope of line:

✓
3
B
-3
C
4
D
-4

Answer

Correct option: A.

Comparing the above equation with general equation y = m × x + c,
m = 3 which is the slope of line.

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MCQ 401 Mark

The equation of a line that passes through the points (1, 5) and (2, 3) is:

✓
2x + y - 7 = 0
B
2x - y - 7 = 0
C
x + 2y - 7 = 0
D
2x + y + 7 = 0

Answer

Correct option: A.

2x + y - 7 = 0

2x + y - 7 = 0

Solution:
We know that the equation of a line passes through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is
$\frac{\left(y-y_1\right)}{\left(x_{-1}\right)}=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}$
$\left(x_1, y_1\right)=(1,5)$
$\left(x_2, y_2\right)=(2,3)$
Now, substitute the values in the formula, we get
$\frac{(y-5)}{(x-1)}=\frac{(3-5)}{(2-1)}$
$\frac{(y-5)}{(x-1)}=\frac{(-2)}{(1)}$
$y-5=-2(x-1)$
$y-5=-2 x+2$
$2 x+y-5-2=0$
$2 x+y-7=0$
$\therefore$ The equation of a line that passes through the points $(1,5)$ and $(2,3)$ is $2 x+y-7=0$.

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MCQ 411 Mark

The locus of a point, whose abscissa and ordinate are always equal is:

✓
x - y = 0
B
x + y = 1
C
x + y + 1 = 0
D
None of the above

Answer

Correct option: A.

x - y = 0

Let the abscissa and ordinate of a point “P” be (x, y)
Given condition: Abscissa = Ordinate
(i.e) x = y
The locus of a point is x - y = 0.

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MCQ 421 Mark

Find the equation of line parallel to 4x + y = 2 and pass through (2, 5):

✓
4x + y - 13 = 0
B
4x + y + 13 = 0
C
4x - y - 13 = 0
D
4x - y + 13 = 0

Answer

Correct option: A.

4x + y - 13 = 0

Line 4x + y = 2 has slope -4. Line parallel to it has slope -4 and pass through (2, 5)
so equation will be y - 5 = (-4) (x - 2)
⇒ 4x + y - 13 = 0

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MCQ 431 Mark

Choose the correct answer. A line cutting off intercept -3 from the y-axis and the tengent at angle to the xaxis is $\frac{3}{5}$, its equation is:

✓
5y - 3x + 15 = 0
B
3y - 5x + 15 = 0
C
5y - 3x - 15 = 0
D
None of these.

Answer

Correct option: A.

5y - 3x + 15 = 0

5y - 3x + 15 = 0

Solution:
Since the lines cut off intercepts -3 on $y$-axis then the line is passing through the point $(0,-3)$.
Given that: $\tan \theta=\frac{3}{5}$
$\Rightarrow$ Slope of the line $\mathrm{m}=\frac{3}{5}$
So, the equation of the line is
$y-y_1=m\left(x-x_1\right)$
$\Rightarrow y+3=\frac{3}{5}(x-0)$
$\Rightarrow 5 y+15=3 x$
$\Rightarrow 3 x-5 y-15=0$
$\Rightarrow 5 y-3 x+15=0$
Hence, the correct option is (a).

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MCQ 441 Mark

The number of real values of $\lambda$ for which the lines $\text{x} - 2\text{y} + 3 = 0, \lambda\text{x} + 3\text{y} + 1 = 0$ and $4\text{x} - \lambda\text{y} + 2 = 0$ are concurrent is:

✓
0
B
1
C
2
D
Infinite.

Answer

Correct option: A.

$\text{x} - 2\text{y} + 3 = 0 \ ...(\text{i})$
$\lambda\text{x} + 3\text{y} + 1 = 0 \ ...(\text{ii})$
$4\text{x} - \lambda\text{y} + 2 = 0 \ ...(\text{iii})$
It is given that (1), (2) and (3) are concurrent.
$\therefore\begin{vmatrix} 1&-2&3\\\lambda&3&1\\4&-\lambda&2\end{vmatrix}=0$
$\Rightarrow(6+\lambda)+2(2\lambda-4)+3(-\lambda^2-12)=0$
$\Rightarrow6+\lambda+4\lambda-8-3\lambda^2-36=0$
$\Rightarrow5\lambda-3\lambda^2-38=0$
$\Rightarrow3\lambda^2-5\lambda+38=0$
The discriminant of this equation is $25 - 4 \times 3 \times 38 = -431$
Hence, there is no real value of $\lambda$ for which the lines $\text{x} - 2\text{y} + 3 = 0, \lambda\text{x} + 3\text{y} + 1 = 0$ and $4\text{x} - \lambda\text{y} + 2 = 0$ are concurrent.

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MCQ 451 Mark

Find slope of line if inclination made by the line is 60°.

A
$\frac{1}{2}$
B
$\frac{1}{\sqrt{3}}$
✓
$\sqrt{3}$
D
$1$

Answer

Correct option: C.

$\sqrt{3}$

Slope of a line is given by $\tan\alpha$ if inclination of line is $\alpha$ If inclination is 60° the slope is $\tan 60^\circ = \sqrt{3}$

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MCQ 461 Mark

If the point (5, 2) bisects the intercept of a line between the axes, then its equation is:

A
5x + 2y = 20
✓
2x + 5y = 20
C
5x - 2y = 20
D
2x - 5y = 20

Answer

Correct option: B.

2x + 5y = 20

Let the equation of the line be $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
The coordinates of the intersection of this line with the coordinate axes are (a, 0) and (0, b).
The midpoint of (a, 0) and (0, b) is $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$
According to the question:
$\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)=(5,2)$
$\Rightarrow\frac{\text{a}}{2}=5,\frac{\text{b}}{2}=2$
$\Rightarrow\text{a}=10,\text{b}=4$
The equation of the required line is given below:
$\frac{\text{x}}{10}+\frac{\text{y}}{4}=1$
$\Rightarrow2\text{x}+5\text{y}=20$

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MCQ 471 Mark

Choose the correct answer. One vertex of the equilateral triangle with centroid at the origin and one side as x + y - 2 = 0 is:
[Hint: Let ABC be the equilateral triangle with vertex A (h, k) and let $\text{D}(\alpha,\beta)$ be the point on BC. Then $\frac{2\alpha+\text{h}}{3}=0=\frac{2\beta+\text{k}}{3}.$ Also $\alpha+\beta-2=0$ and $\frac{\text{k}-0}{\text{h}-0}\times(-1)=-1\Big].$

A
(-1, -1)
B
(2, 2)
✓
(-2, -2)
D
(2, -2)

Answer

Correct option: C.

(-2, -2)

Let ABC be the equilateral triangle with vertex A(h, k).
Also, centroid is G(0, 0).

Now, $\text{AG}\bot\text{BC}$
Slope of line BC or x + y - 2 = 0 is -1.
$\therefore$ Slope of $\text{AG},\frac{\text{k}}{\text{h}}=1$ or h = k.
Now distance of origin from $\text{BC}=\frac{|0+0-2}{\sqrt{1^2+1^2}|}=\sqrt{2}$
$\therefore$ Distance of A form $\text{BC}=3\sqrt{2}=\frac{|\text{h}+\text{k}-2|}{\sqrt{1^2+1^2}}$
$\therefore$ |h + k - 2| = 6
⇒ h + k - 8 = 0 or h + k + 4 = 0
⇒ h + k - 8 = 0 or h + k + 4 = 0
⇒ h = 4 or h = -2
$\therefore$ Vertex is (-2, -2).

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MCQ 481 Mark

What is the distance between (1, 3) and (5, 6):

A
3 units
B
4 units
✓
5 units
D
25 units

Answer

Correct option: C.

5 units

5 units

Solution:
We know, distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(\text{x}_{1}−\text{x}_{2})^2+(\text{y}_{1}−\text{y}_{2})^2.}$
So, distance between (1, 3) and (5, 6) is $\sqrt{{(1-5)}^2+{(3-6)}^2}$
= $(4)^2+ (3)^2 = 5$ units

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MCQ 491 Mark

Choose the correct answer. If the line $\frac{\text{x}}{\text{a}} + \frac{\text{y}}{\text{b}} =1$ passes through the points (2, -3) and (4, -5), then (a, b) is:

A
(1, 1)
B
(-1, 1)
C
(1, -1)
✓
(-1, -1)

Answer

Correct option: D.

(-1, -1)

Equation of line passing through the points (2, -3) and (4, -5) is
$\text{y}+3=\frac{-5+3}{4-2}(\text{x}-2)$
$\Rightarrow \text{y}+3=\frac{-2}{2}(\text{x}-2)$
$\Rightarrow \text{y}+3=-(\text{x}-2)$
$\Rightarrow \text{y}+3=-\text{x}+2$
$\Rightarrow \text{x}+\text{y}=-1$
$\Rightarrow \frac{\text{x}}{-1}+\frac{\text{y}}{-1}=1$ (intercept from)
$\therefore \text{a}=-1,\text{b}=-1$
Hence, the correct option is (d).

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MCQ 501 Mark

The points $(-a, -b), (0 , 0), (a, b)$ and $(a^2, ab)$ are:

A
Vertices of a square
B
Vertices of a parallelogram
✓
Collinear
D
Vertices of a rectangle

Answer

Correct option: C.

Collinear

Collinear

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