Question

A line passes through the point of intersection of $2x+y=5$ and $x+3y=-8$ and parallel to the line $3x+4y=7$ is

• A. $3x+4y+3=0$
• B. $3x+4y=0$
• C. $4x–3y+3=0$
• D. $4x–3y=3$

Answer 1

The correct option is A

$3x+4y+3=0$

Step1. Finding intersection point:

Given equations of lines are

$2\mathrm{x}+\mathrm{y}=5....\left(\mathrm{i}\right)$,

$\mathrm{x}+3\mathrm{y}=-8....\left(\mathrm{ii}\right)$

On solving equation $\left(\mathrm{i}\right)\&\left(\mathrm{ii}\right)$ we get,

$\Rightarrow x=\frac{23}{5}$,

$\Rightarrow y=-\frac{21}{5}$

So the point of intersection of lines is $(\frac{23}{5},-\frac{21}{5})$

Step2. Finding equation of the line.

Also the line is parallel to

$3x+4y=7$

$\Rightarrow y=\frac{-3}{4}x+\frac{7}{4}$

Slope of the line is $-\frac{3}{4}$

Slope of parallel line are same

Therefore, equation of line passing passing through the point $(\frac{23}{5},-\frac{21}{5})$ with slope $-\frac{3}{4}$is given by

$$\begin{gathered} y+\frac{21}{5}=\frac{-3}{4}(x-\frac{23}{5})\mathbf{}\mathbf{}\mathit{y}\mathbf{-}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{m}\mathbf{(}\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{)} \\ \Rightarrow 20y+84=-15x+69 \\ \Rightarrow 15x+20y+15=0 \\ \Rightarrow 3x+4y+3=0 \end{gathered}$$

Hence, the correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}$.