Answer:
[tex]\large\boxed{y=-\dfrac{3}{5}+4}[/tex]
Step-by-step explanation:
[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}\\===========================\\\\\text{We have}\ m_1=\dfrac{5}{3},\ \text{therefore}\ m_2=-\dfrac{1}{\frac{5}{3}}=-\dfrac{3}{5}.\\\\l:y=-\dfrac{3}{5}x+b\\\\\text{Put the coordinates of the point (-5, 7) to the equation of a line }\ l:\\\\7=-\dfrac{3}{5}(-5)+b\\\\7=3+b\qquad\text{subtract 3 from both sides}\\\\4=b\to b=4\\\\\text{Fimally we have the equation:}\\\\l:y=-\dfrac{3}{5}+4[/tex]
