Let  $3x-2y-7=0$ be a line. The points $(3,-5),(7,5)$ lie in

Given $f(x,y)=3x-2y-7=0$

$\begin{array}{l}\frac{f(3,-5)}{f(7,5)}=\frac{3\left(3\right)-2(-5)-7}{3\left(7\right)-2\left(5\right)-7}\\ =\frac{9+10-7}{21-10-7}\\ =\frac{12}{4}\\ =3>0\end{array}$

$\therefore$ Same side of the line