Answer:
[tex]y=\frac{x}{3}+\frac{16}{3}[/tex]
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
[tex]m_{1}*m_{2} =-1\\m_{1}=-3\\-3*m_{2} =-1\\\\m_{2}=\frac{-1}{-3}\\m_{2}=\frac{1}{3}\\\\[/tex]
Step 3
find the equation of the line
[tex]y-y_{0}=m(x- x_{0})[/tex]
Let
[tex]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}[/tex]
Have a nice day.
