Question

What is the equation of the line that is perpendicular to line y= -2x+11 and crosses

point (-6, 8)?
y = 1 /2x
y = 1/2x -21
y = 1/2x – 10
y =1/2x + 11

Answer 1

`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) = (1)/(2) \\\\Equation \ of \ the \ line \ passing \ through (-6, 8) \ slope = (1)/(2) \\\\(y - y_(1)) = m_(2)(x-x_(2) )\\\\(y-8) =(1)/(2) (x-(-6))\\\\y -8 =(1)/(2) (x+6)\\\\y = (1)/(2)x +3 +8\\\\y =(1)/(2)x + 11`

option 4