What is the equation of the line which passes through the point of intersection of the lines y=x and x+y=6 and which is perpendicular to the line with equation 3x+6y=12?

1 Answer
Aviv S.
Jan 27, 2018

The line is #y=2x-3#.

Explanation:

First, find the intersection point of #y=x# and #x+y=6# using a system of equations:

#y+x=6#
#=>y=6-x#

#y=x#
#=>6-x=x#
#=>6=2x#
#=>x=3#
and since #y=x#:
#=>y=3#

The lines' intersection point is #(3,3)#.

Now we need to find a line that goes through the point #(3,3)# and is perpendicular to the line #3x+6y=12#.

To find the slope of the line #3x+6y=12#, convert it to slope-intercept form:

#3x+6y=12#

#6y=-3x+12#

#y=-1/2x+2#

So the slope is #-1/2#. The slopes of perpendicular lines are opposite reciprocals, so that means the slope of the line we're trying to find is #-(-2/1)# or #2#.

We can now use point-slope form to make an equation for our line from the point and slope that we found before:

#y-y_1=m(x-x_1)#

#=>y-3=2(x-3)#

#=>y-3=2x-6#

#=>y=2x-3#

The line is #y=2x-3#.

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