Question

Find the equation of the line passing through the point of intersection of $2x-7y+11=0$ and $x+3y-8=0$ and is parallel to (i) x-axis (ii) y-axis.

Answer 1

The required line is

$2x-7y+11+\lambda (x+3y-8)=0$

or,x$(2+\lambda )+y(-7+3\lambda )+11-8\lambda =0$

(i) When the line is parallel to x-axis.It slope is 0

$\therefore -\frac{(2+\lambda )}{3\lambda -7}=0$

$\therefore$ Equation of line is

$2x-7y+11-2(x+3y-8)=0$

$-13y+27=0$ $13y-27=0$

(ii) When the line is parallel to y-axis then,

$\frac{-1}{slope}=0$

$\lambda =\frac{7}{3}$

$\therefore$ Equation of line is

$2x-7y+11+\frac{7}{3}(x+3y-8)=0$

$\Rightarrow \frac{6x-21y+33+7x+21y-56}{3}=0$

$\Rightarrow 6x-21y+33+7x+21y-56=0$

$\Rightarrow 13x-23=0$