How do you write an equation of a line passing through (6, 3), perpendicular to #y = 2x + 2#?

1 Answer
Jun 20, 2016

Answer:

#x+2y=12#

Explanation:

#y=color(green)(2)x+2# is an equation in slope-intercept form with a slope of #color(green)(m=2)#

All lines perpendicular to #y=color(green)(2)x+2# will have a slope
#color(white)("XXX")-1/color(green)(m)=color(red)(-1/2)#

If a line has a slope of #color(red)(""(-1/2))# and passes through #color(blue)(""(6,3))#
then we can write its equation in slope-point form as
#color(white)("XXX")y-color(blue)(3)=color(red)(""(-1/2))(x-color(blue)(6))#

While this is a valid answer, we would normally rearrange it into a simpler form:
#color(white)("XXX")2y-6=6-x#
or
#color(white)("XXX")x+2y=12#