[tex]\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\\text{Convert the equation to the slope-intercept form}\ y=mx+b.\\\\l:2x-3y=5\qquad\text{subtract 2x from both sides}\\\\-3y=-2x+5\qquad\text{divide both sides by (-3)}\\\\l:y=\dfrac{2}{3}x-\dfrac{5}{3}\to m_1=\dfrac{2}{3}\\\\m_2=\dfrac{2}{3}\\\\\text{Therefore we have the equation of a line}\ m:y=-\dfrac{3}{2}x+b.[/tex]
[tex]\text{We know the line passes through the point (3, -10)}.\\\text{Substitute the coordinates of the point to the equation of a line:}\\\\-10=\dfrac{2}{3}(3)+b\\\\-10=2+b\qquad\text{substract 2 from both sides}\\\\-12=b\to b=-12\\\\y=\dfrac{2}{3}x-12\qquad\text{multiply both sides by 3}\\\\3y=2x-12\qquad\text{subtract 2x from both sides}\\\\-2x+3y=-12\qquad\text{change the signs}\\\\2x-3y=12\\\\Answer:\ \boxed{m:y=\dfrac{2}{3}x-12}\to\boxed{m:2x-3y=12}[/tex]
