Answer:
[tex]\large\boxed{y=-\dfrac{8}{5}x+\dfrac{31}{5}}[/tex]
Step-by-step explanation:
[tex]\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept[/tex]
[tex]\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\--------------------------[/tex]
[tex]\text{We have:}\\\\y=\dfrac{5}{8}x-4\to m_1=\dfrac{5}{8}\\\\\text{The slope of a perpendicular line:}\ m_2=-\dfrac{1}{\frac{5}{8}}=-\dfrac{8}{5}\\\\\text{The equation:}\\\\y=-\dfrac{8}{5}x+b\\\\\text{Put the coordinates of the point (2, 3) to the equation:}\\\\3=-\dfrac{8}{5}(2)+b\qquad\text{solve for}\ b\\\\3=-\dfrac{16}{5}+b\qquad\text{add}\ \dfrac{16}{5}\ \text{to both sides}\\\\\dfrac{15}{5}+\dfrac{16}{5}=b\to b=\dfrac{31}{3}\\\\\text{Finally:}\\\\y=-\dfrac{8}{5}x+\dfrac{31}{5}[/tex]
