Find the equation of the line through point (4,−1) and perpendicular to y=x+2 .

Answer 1

Follow the solution provided by William W.

Answer 2

The line y₁=×+2 has a slope = 1. m₁₌₁

The perpendicular line to y₁=×+2 has a slope as a negative recíprocal of m₁ m_p=-1

The perpendicular line as y_p=mpx+b througt (4,-1) is

-1=-1(4)+b. b=4-1 b=3

Now y_p= -×+3 is a perpendicular line to y_p=×+2

The product of m_{1 times} m_{p= -1}