1 Marks Question

Question 11 Mark

Write the equation of line perpendicular to line $-3 x+$ $2 y+4=0$.

Answer

Given equation of line
$-3 x+2 y+4=0$
$\Rightarrow \quad 3 x-2 y-4=0$
Equation of line perpendicular to it is $2 x+3 y+\lambda=0$

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Question 21 Mark

If one of the line in given lines is parallel to $x$-axis, then calculate the angle between the lines, which is made by the second lines with $x$-axis.

Answer

Let the equation of first line be
\[
y=\text { Constant }
\]
Equation of second line
\[
y=m x+c
\]
If second line makes an angle $\theta$ with $x$-axis then
\[
m=\tan \theta \Rightarrow \theta=\tan ^{-1}(m)
\]
$\therefore$ Angle between both lines $=\tan ^{-1}(m)$

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Question 31 Mark

Write the equation of line passing through point ( -4 , -3 ) and parallel to $x$-axis.

Answer

According to question,
Slope of line $m=0$
$x_1=-4, y_1=-3$
So, equation of straight line
$\begin{array}{ll} & y-y_1=m\left(x-x_1\right) \\ \Rightarrow & y+3=0(x+4)\end{array}$
$\Rightarrow \quad y+3=0 \quad \therefore y=-3$

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Question 41 Mark

Find the equation of line perpendicular to line $3 x-5 y$ $+7=0$ and passing through point $(1,5)$.

Answer

Equation of line perpendicular to the given line
$5 x+3 y+\lambda=0$
This line passes through point $(1,5)$,
$5 \times 1+3 \times 5+\lambda=0$
$\Rightarrow \quad \lambda=-20$
So, equation of required line $5 x+3 y-20=0$

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Question 51 Mark

Find the distance between parallel lines $2 x+3 y+4$ $=0$ and $2 x+3 y-2=0$.

Answer

Distance between lines
\[
\begin{array}{l}
=\frac{\left|c_2-c_1\right|}{\sqrt{a^2+b^2}}=\frac{|-2-4|}{\sqrt{(2)^2+(3)^2}} \\
=\frac{6}{\sqrt{4+9}}=\frac{6}{\sqrt{13}} \text { units }
\end{array}
\]

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Question 61 Mark

Find the length of the perpendicular drawn from point $(4,5)$ on line $y=m x+c$.

Answer

Here equation of line is $m x-y+c=0$
$\begin{array}{l}\therefore \text { Length of perpendicular }\end{array}=\frac{|m \times 4-5+c|}{\sqrt{m^2+(-1)^2}}$
$\therefore \quad=\frac{|4 m-5+c|}{\sqrt{m^2+1}}$

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Question 71 Mark

Find the length of perpendicular drawn from point (2, -3 ) on straight line $3 x-4 y+5=0$.

Answer

\[
\begin{aligned}
d & =\frac{|3 \times 2-4(-3)+5|}{\sqrt{(3)^2+(-4)^2}} \\
& =\frac{|6+12+5|}{5}=\frac{|23|}{5}=\frac{23}{5} \text { units }
\end{aligned}
\]

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Question 81 Mark

Find the distance between the parallel lines $3 x-4 y+$ $2=0$ and $6 x-8 y+3=0$.

Answer

On multiplying 2 in equation (1)
$\begin{array}{ll}
& 6 x-8 y+4=0 .......(1)\\
\text { and } & 6 x-8 y+3=0........(2)
\end{array}$
$\therefore$ Distance between lines
$\begin{array}{l}
=\frac{|4-3|}{\sqrt{(6)^2+(-8)^2}}=\frac{1}{\sqrt{36+64}} \\=\frac{1}{\sqrt{100}}=\frac{1}{10} \text {unit}\end{array}$

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