Answer:
[tex]\large\boxed{y=\dfrac{1}{4}x+\dfrac{15}{2}}[/tex]
Step-by-step explanation:
[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=-4x-1\to m_1=-4\\\\\text{Therefore}\ m_2=-\dfrac{1}{-4}=\dfrac{1}{4}.\\\\\text{The equation of a line perpendicular to}\ y=-4x-1:\\\\y=\dfrac{1}{4}x+b\\\\\text{Put the coordinates of the point (-2, 7) to the equation:}\\\\7=\dfrac{1}{4}(-2)+b\\\\7=-\dfrac{1}{2}+b\qquad\text{add}\ \dfrac{1}{2}\ \text{to both sides}\\\\7\dfrac{1}{2}=b\to b=7\dfrac{1}{2}=\dfrac{7\cdot2+1}{2}=\dfrac{15}{2}\\\\\text{Finally:}\\\\y=\dfrac{1}{4}x+\dfrac{15}{2}[/tex]
