Question

Find the equation of the line perpendicular to the line y=−3x+5 that passes through point (2, 6).

Answer 1

`m = ( - 1)/( - 3) = (1)/(3) \\ y - mx = b \\ 6 - (2)/(3) = b \\ (16)/(3) = b \\ 5 (1)/(3) = b \\ y = (x)/(3) + 5 (1)/(3)`

Answer 2

`y=(x)/(3)+(16)/(3)`

Explanation:

Hello, I think I can help you with this

Step 1

let line 1

y=-3x+5

this equation is in the form y= mx+b, where m is the slope,Hence

-3x=mx

-3=m

m(1)=-3

Step 2

two lines are perpendicular if the product of their slopes is equal to -1

`m_(1)*m_(2) =-1\\m_(1)=-3\\-3*m_(2) =-1\\\\m_(2)=(-1)/(-3)\\m_(2)=(1)/(3)\\\\`

Step 3

find the equation of the line

`y-y_(0)=m(x- x_(0))`

Let

`P(2,6)\\slope=(1)/(3) \\ put\ the\ values\ into\ the\ equation\\y-y_(0)=m(x- x_(0))\\y-6=(1)/(3)(x-2)\\y-6=(x)/(3)-(2)/(3)\\y=(x)/(3)-(2)/(3)+6\\\ y=(x)/(3)+(16)/(3)`

Have a nice day.