Question

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

Answer 1

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

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