Find the equation of the line passing through the point $(2, 3)$ and making intercept of length $2$ units between the lines $y + 2 x = 3$ and $y + 2 x = 5$ .

3 x - 4 y = 18

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x = 2

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3 x + 4 y = 18

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x + 2 = 0

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Solution

The correct option is C $3 x + 4 y = 18$
$L_{1} : y + 2 x - 3 = 0$
$L_{2} : y + 2 x - 5 = 0$

$L_{1}$ & $L_{2}$ are parallel lines

distance between parallel lines

$= ∣ ∣ ∣ ∣ \frac{- 3 - (- 5)}{\sqrt{2^{2} + 1^{2}}} ∣ ∣ ∣ ∣$

$= \frac{2}{\sqrt{5}}$

In $△ A P B$

$\Rightarrow$ ${A P}^{2} = {B P}^{2} + {A B}^{2} A P = 2$ (given)
$\Rightarrow$ $2^{2} = {(\frac{2}{\sqrt{5}})}^{2} + {(A B)}^{2}$
$\Rightarrow$ ${A B}^{2} = 4 - \frac{4}{5} = \frac{16}{5}$
$\Rightarrow$ $A B = \frac{4}{\sqrt{5}}$
$\Rightarrow$ $tan α = \frac{\frac{2}{\sqrt{5}}}{\frac{4}{\sqrt{5}}} = \frac{1}{2}$

Slope of line: $L_{1} y = - 2 x + 3$

$m = - 2$

Let slope of line $P A = m$

$tan α = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}$

$\frac{1}{2} = \frac{m - (- 2)}{1 - 2 m}$

$2 m + 4 = 1 - 2 m$

$4 m = - 3$

$m = - \frac{3}{4}$

Slope of line $P A = - \frac{3}{4}$

Equation of line which passes through $(2, 3)$ and has slope $= - \frac{3}{4}$

$y - y_{1} = m (x - x_{1})$

$y - 3 = - \frac{3}{4} (x - 2)$

$4 y - 12 = - 3 x + 6$

$3 x + 4 y = 18$

Ans: C

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