Precalculus Question

Question

Find an equation for the line, in the indicated form, with the given properties.

Perpendicular to the line x-6y=3; containing the point (5,3)

Answer 1

x-6y = 3

x-6y-3 = 0

x-3 = 6y

(x-3)/6 = y

(1/6)x - (3/6) = y

(1/6)x - 1/2 = y

THe given line has slope 1/6

the perpendicular line has slope -6 and eats (5,3)

B = y - mx = 3 - (-6)(5) = 3 - -30 = 3 + 30 = 33

y = -6x + 33