Geometry Question

Question

Write the equation of a line that passes through the point (5, 27) and is perpendicular to the line 22x 1 6y 5 24.

Answer 1

You need to start by putting the original equation in y= mx+b form

-2x +6y = -4

6y= 2x-4

y=(1/3)x -2/3

The slope of the perpendicular will be the negative reciprocal of the slope of the original line or -3

The equation of the perpendicular is y= -3x +b

You have a point on the line at (5,27) so 27 = (-3)(5) + b

27= -15 +b

b= 42

so equation is y= -3x+42