Question

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

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

Answer 1

The correct option is B $3x+4y+3=0$

To find point of intersection

solve equation of both lines

$2x+y=5$ …………$\left(1\right)$

$x+3y+8=0$ …………..$\left(2\right)$

Point of inter-section$=(\frac{23}{5},\frac{-21}{5})$

The required line is parallel to $3x+4y=7$

Line parallel to $ax+by+c_1=0$ is $ax+by+\lambda =0$ required line is $3x+4y+\hat{i}=0$

We know that it passes through $(\frac{23}{5},\frac{-21}{5})$

$=3\left(\frac{23}{5}\right)-4\left(\frac{21}{5}\right)+\lambda =0$

$=-(\frac{69}{5}-\frac{84}{5})=\lambda$

$\lambda =(-3)$

$\Rightarrow \lambda$

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