Question

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

Answer 1

$m = \frac{ - 1}{ - 3} = \frac{1}{3} \\ y - mx = b \\ 6 - \frac{2}{3} = b \\ \frac{16}{3} = b \\ 5 \frac{1}{3} = b \\ y = \frac{x}{3} + 5 \frac{1}{3}$

Answer 2

Answer:

$y=\frac{x}{3}+\frac{16}{3}$

Step-by-step 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

$m_{1}*m_{2} =-1\\m_{1}=-3\\-3*m_{2} =-1\\\\m_{2}=\frac{-1}{-3}\\m_{2}=\frac{1}{3}$

Step 3

find the equation of the line

$y-y_{0}=m(x- x_{0})$

Let

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

Have a nice day.