Find the equation of the line through the intersection of $5x-3y=1$and $2x-3y-23=0$and perpendicular to the line $5x-3y-1=0$.

Solution: Let $m1$ be the slope of required line.
$m2=\frac{5}{3}$(slope of given line)
As, $m1\cdot \left(m2\right)=-1$
so, $m1=-\frac{3}{5}$.On Solving given two equations,
$5x-3y-1=0,2x+3y-23=0$
We get,$\left(x,y\right)=\left(\frac{24}{7},\frac{113}{7}\right)$.
Equation of required line,$\left(y-\frac{113}{7}\right)=\left(-\frac{3}{5}\right)\left(x-\frac{24}{7}\right)$
$63x+105y-781=0$