Question

Find the equation of a line passing through $(3,-2)$ and perpendicular to the line $x-3y+5=0$

Answer 1

Any equation passing through $3,-2$ and perpendicular to given line is

$[y-y_1=\frac{1}{m}(x-x_1\left)\right]...\left(1\right)$

where $(x_1-y_1$ is $(3,-2)$ and m is slope of line

$\frac{-1}{m}$ is taken as lines are perpendicular

Finding slope of line $x-3y+5=0$

$3y=x+5$

$y=\frac{x}{3}+\frac{5}{3}$

$\Rightarrow m=\frac{1}{3}$

Substituting the value of m and $(x_1-y_1)$ in

$y-(-2)=-\frac{1}{\frac{1}{3}}(x-3)$

$y+2=-3(x-3)=-3x+9$

$3x+y=7$