Question

Find the equation of the line passing through the point of intersection of $2x+y=5$ and $x+3y+8=0$ and parallel to the line $3x+4y=7$.

Answer 1

Given : $2x+y=5$ and $x+3y+8=0$

$2x+y=5$

$\Rightarrow y=5-2x$ ...$\left(1\right)$

$\Rightarrow x+3(5-2x)+8=0$

$\Rightarrow -5x+23=0$

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

$y=5-\frac{46}{5}=-\frac{21}{5}$

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

Since, the line is parallel to

$3x+4y=7$

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

Slope of the line is

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

Therefore, the equation of the required line is

$y+\frac{21}{5}=-\frac{3}{4}(x-\frac{23}{5})$

$\Rightarrow y+\frac{3}{4}x+\frac{21}{5}-\frac{69}{20}=0$

$\Rightarrow 4y+3x+4\left(\frac{84-69}{20}\right)=0$

$\Rightarrow 3x+4y+3=0$

Hence, the equation of line is

$3x+4y+3=0$