What is the equation of the line perpendicular to #y=-3/2x # that passes through # (2,-4) #?

1 Answer
Dec 20, 2015

#y=2/3x-16/3#

Explanation:

The slope-intercept form of a line is written in the form:

#y=mx+b#

where:
#y=#y-coordinate
#m=#slope
#x=#x-coordinate
#b=#y-intercept

Start by finding the slope that is perpendicular to #-3/2x#. Recall that when a line is perpendicular to another line, it is #90^@# to it.

We can find the slope of the line perpendicular to #-3/2x# by finding the negative reciprocal. Recall that the reciprocal of any number is #1/"number"#. In this case, it is #1/"slope"#. To find the negative reciprocal we can do:

#-(1/"slope")#
#=-(1/(-3/2x))#
#=-(1-:-3/2x)#
#=-(1*-2/3x)#
#=-(-2/3x)#
#=2/3xrArr# negative reciprocal, perpendicular to #-3/2x#

So far, our equation is: #y=2/3x+b#

Since we do not know the value of #b# yet, this is going to be what we are trying to solve for. We can do this by substituting the point, #(2,-4)#, into the equation:

#y=mx+b#
#-4=2/3(2)+b#
#-4=4/3+b#
#-16/3=b#

Now that you know all your values, rewrite the equation in slope-intercept form:

#y=2/3x-16/3#