How do you write an equation of a line passing through (-3, 4), perpendicular to 3y=x-2?

1 Answer
May 3, 2018

3x+y+5=0 is the required equation of the straight line. graph{(3x+y+5)(x-3y-2)=0 [-8.44, 2.66, -4.17, 1.38]}

Explanation:

Any line perpendicular to ax+by+c=0 is bx-ay+k=0 where k is constant.

Given equation is

rarr3y=x-2

rarrx-3y=2

Any line perpendicular to x-3y=2 will be 3x+y+k=0

As 3x+y+k=0 passes through (-3,4), we have,

rarr3*(-3)+4+k=0

rarr-9+4+k=0

rarrk=5

So, the required equation of the straight line is 3x+y+5=0