Answer:
[tex]\large\boxed{y-7=\dfrac{1}{4}(x+5)}[/tex]
Step-by-step explanation:
[tex]\text{The point-slope form of an equation of a line:}\\\\y-y_1=m(x-x_1)\\\\m-slope\\\\\text{Let a line}\ k\ \text{has a slope}\ m_1\ \text{and a line}\ l\ \text{has a slope}\ m_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\\================================[/tex]
[tex]\text{We have the equation:}\ y-3=-4(x+2)\to m_1=-4.\\\\\text{Therefore, the slope of the line perpendicular to the given line is}\\\\m_2=-\dfrac{1}{-4}=\dfrac{1}{4}.\\\\\text{Put it and the coordinates of the point (-5, 7) to the equation of a line:}\\\\y-7=\dfrac{1}{4}(x-(-5))\\\\y-7=\dfrac{1}{4}(x+5)[/tex]
