Answer:
[tex]slope \ of \ y = -2x + 11, m_{1} = -2\\\\Let \ the \ slope\ of \ the \ new \ line \ be \ m_{2} \\\\ Since \ the \ line \ is \ perpendicular \ to \ the\ given \ line, m_{1} \cdot m_{2} = -1\\\\-2 \cdot m _{2} = -1\\ \\m _{2} = \frac{1}{2} \\\\Equation \ of \ the \ line \ passing \ through (-6, 8) \ slope = \frac{1}{2} \\\\(y - y_{1}) = m_{2}(x-x_{2} )\\\\(y-8) =\frac{1}{2} (x-(-6))\\\\y -8 =\frac{1}{2} (x+6)\\\\y = \frac{1}{2}x +3 +8\\\\y =\frac{1}{2}x + 11[/tex]
option 4
