Answer:
[tex]y=\dfrac{3}{4}x+7\dfrac{1}{4}[/tex]
Step-by-step explanation:
[tex]k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\ \iff\ m_1m_2=-1[/tex]
Let [tex]k:4x+3y=6\to 3y=-4x+6\ \ \ \ |:3\\\\y=-\dfrac{4}{3}x+2\\\\m_1=-\dfrac{4}{3}[/tex]
[tex]l:y=m_2x+b\\\\l\ \perp\ k\ \iff\ -\dfrac{4}{3}m_2=-1\qquad|\cdot\left(-\dfrac{3}{4}\right)\\\\m_2=\dfrac{3}{4}\\\\l:y=\dfrac{3}{4}x+b[/tex]
The line l passes through the point (-3, 5).
Substitute the coordinates of the point to the equation of the function l:
[tex]5=\dfrac{3}{4}(-3)+b\\\\5=-\dfrac{9}{4}+b\\\\5=-2\dfrac{1}{4}+b\qquad|+2\dfrac{1}{4}\\\\b=7\dfrac{1}{4}[/tex]
Finally [tex]l:y=\dfrac{3}{4}x+7\dfrac{1}{4}[/tex]
