Question

Find the equation of the line passing through the point $(-6,10)$ and perpendicular to the straight line $7x+8y=5$.

Answer 1

Given point $(-6,10)$ and the line $7x+8y=5$

Slope of the line is $m_1=\frac{-7}{8}$

And slope of the perpendicular line is $m_2=-\frac{1}{m_1}=\frac{8}{7}$

So, the equation of line in slope-intercept form is $y=\frac{8}{7}x+c$ .....(1)

And point $(-6,10)$ is passing through the above line

$\Rightarrow 10=\frac{8}{7}\times (-6)+c$

$\Rightarrow c=\frac{118}{7}$

Therefore line is $7y-8x=118$ .... (Substitute ${}^{\prime }{c}^{\prime }$ in (1) and simplify)