Question

Find equation of line passes from point of intersection of perpendicular to the line $x-7y+5=0$ and having $x$ intercept $3$.

Answer 1

The given equation of line is $x-7y+5=0$
$y=\frac{1}{7}x+\frac{5}{7}$, which is of the form $y=mx+c$
$\therefore$ slope of the given line $=\frac{1}{7}$
Thus the slope of the line perpendicular to the line having a slope of $\frac{1}{7}$ is $m=-\frac{1}{\left\{\frac{1}{7}\right\}}=-7$
Therefore equation of the line with slope $-7$ and x-intercept $3$ is given by
$y=m(x-d)$
$\Rightarrow y=-7(x-3)$
$\Rightarrow y=-7x+21$
$\Rightarrow 7x+y=21$