Question

Find the equation of the line that passes through the point (7,5) and is perpendicular to the line 2x - 3y=6

Answer 1

$(7,5);\ \ \ \ 2x - 3y=6 \ \ / subtract \ 2x \ from \ each \ side \\ \\-3y = -2x + 6\ \ / divide \ each \term \ by \ (-3) \\ \\ y = \frac{2} {3}x -2\\ \\ The \ slope \ is :m _{1} = \frac{ 2}{3} \\ \\ If \ m_{1} \ and \ m _{2} \ are \ the \ gradients \ of \ two \ perpendicular \\ \\ lines \ we \ have \\\\\ m _{1}*m _{2} = -1$

$\frac{2}{3}\cdot m_{2}=-1\ \ / \cdot (\frac{3}{2})\\\\m_{2}=-\frac{3}{2}\\\\Now \ your \ equation \ of \ line \ passing \ through \ (7,5) would \ be: \\ \\ y=m_{2}x+b \\ \\5=-\frac{3}{2}\cdot 7 + b \\ \\ 5= -\frac{21}{2}+b\\ \\b=5+\frac{21}{2} \\ \\b= \frac{10}{2}+\frac{21}{2}\\ \\b= \frac{31}{2}\\\\b=15.5 \\ \\ y = -\frac{3}{2}x +15.5$