Question

Find an equation for the line with the given properties. Perpendicular to the line 7x - 3y = 68; containing the point (8, -8)

Answer 1

Answer:

$y=\dfrac{-3}{7}x-\dfrac{32}{7}$

Step-by-step explanation:

Given that,

A line 7x - 3y = 68 and containing the point (8, -8).

The equation can be written as :

$-3y=68-7x\\\\y=\dfrac{68}{-3}+\dfrac{7x}{3}\\\\y=\dfrac{7x}{3}+(\dfrac{-68}{3})$

The slope is :7/3

Line is perpendicular so use m = –3/7

$-8=(-\dfrac{3}{7})\times 8+b\\\\-8+\dfrac{3}{7}\times 8=b\\\\b=\dfrac{-32}{7}$

So, required equation is :

$y=\dfrac{-3}{7}x-\dfrac{32}{7}$