How do you find the equation of a line perpendicular to the line y= 2/3 (fraction) x -3 , while also passing through the point (4,8)

Answer 1

When you are given the slope and a point, use the point-slope form:

(x - x₁)·m = (y - y₁)

The slope of line perpendicular to y = (2/3)x - 3 is -3/2 (the negative reciprocal). The point (x₁,y₁) = (4,8). Plug and go:

(x - x₁)·m = (y - y₁)

(x - 4)(-3/2) = y - 8

Solve for y to put it into y = mx + b form.