Question

Find the equation of the line that is perpendicular to the line x+3y+10=0 and goes through the point (5,-1)

Answer 1

Answer:

Perpendicular line's gradient = -1/gradient

x+3y+10=0

3y = -x - 10

y = -⅓x - 10/3

gradient = -⅓

y = -⅓x + c

At 5,-1

-1 = -⅓(5) + c

-1 = -5/3 + c

-1 + 5/3 = c = ⅔

y = -⅓x + ⅔