Question

Write an equation that is perpendicular to 3x - 6y=1 and contains the points (-3,7)

Answer 1

`(-3,7), \ \ \ 3x - 6y = 1 \ subtract \ (-3x) \ from \ each \ side \\ \\ -6y = -3x + 1 \ divide \ each \term \ by \ (-6) \\ \\ y = \frac{3} {6}x - (1)/(6)\\ \\ y = \frac{1} {2}x - (1)/(6)\\ \\ The \ slope \ is :m _(1) = (1)/(2) \\ \\ If \ m_(1) \ and \ m _(2) \ are \ the \ gradients \ of \ two \ perpendicular \\ \\ lines \ we \ have \ m _(1)*m _(2) = -1 \\\\(1)/(2)*m_(2)=-1`


`(1)/(2)*m_(2)=-1\ \ | \ multiply\ both\ sides\ by\ 2 \\\\m_(2)=-2 \\\\ Now \ your \ equation \ of \ line \ passing \ through \ (-3,7) would \ be: \\ \\ y=m_(2)x+b \\ \\7=(-2} )\cdot (-3) + b \\ \\ 7=6+b\\ \\ b=7-6\\ \\b=1 \\ \\ y =-2x +1`