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

`y=(-3)/(7)x-(32)/(7)`

Explanation:

Given that,

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

The equation can be written as :

`-3y=68-7x\\\\y=(68)/(-3)+(7x)/(3)\\\\y=(7x)/(3)+((-68)/(3))`

The slope is :7/3

Line is perpendicular so use m = –3/7

`-8=(-(3)/(7))* 8+b\\\\-8+(3)/(7)* 8=b\\\\b=(-32)/(7)`

So, required equation is :

`y=(-3)/(7)x-(32)/(7)`