Question

Write the equation of a line perpendicular to the given line and passing through the given point: y=-2/3×+2;(6,3)

Answer 1

The slope of the given line y=-(2/3)x+2
is m1=-2/3
The slope of any line perpendicular to it is m2=-1/m1=3/2

Since it passes through point (6,3) and the slope m2=3/2 is known, we use the point-slope form to define the new line:
y-y1=m(x-x1) where (x1,y1)=(6,3).
Substituting x1=6, y1=3 into the point slope form,
y-y1=m2(x-x1)
y-3=(3/2)(x-6)
simplify
y=(3/2)x-9+3, =>

y=(3/2)x-6
is the line perpendicular to y=-(2/3)x+2 passing through (6,3)

Check: y=(3/2)6-6=3 ok, and
m1m2=-2/3*(3/2)=-1 ok

Answer 2

y=-2/3x +2

slope = -2/3

perpendicular slope = 3/2 (opposite and reciprocal)

passing thru (6,3)
y = mx+b
3 = 3/2 (6) + b
b = 3 -9
b =-6

equation
y = 3/2x - 6