How do you write an equation in standard form for a line passing through (–4, 2) and is parallel to y = –x + 6?

1 Answer
May 1, 2015

All lines parallel to #y=ax+b# have to have the same coefficient at #x# in the form #y=px+q#. That is, #p=a#.

In our case all lines parallel to #y=-x+6# have to have a form #y=-x+q#, where #q# can be any real number.

The condition that our line should pass through #(-4,2)# results in the equation, from which we can determine an unknown variable #q#:
#2=-(-4)+q#
#q=-2#

Therefore, the equation of a line we are looking for is
#y=-x-2#

It passes through #(-4,2)# since, if we substitute #-4# for #x#, the value of #y# will be #2#.
It also parallel to #y=-x+6# because the coefficients at #x# are the same in both cases.

Graphs of these two equations are:
#y=-x+6#
graph{-x+6 [-10, 10, -5, 5]}
#y=-x-2# - notice the line is passing point #(-4,2)# and is parallel to #y=-x+6# above
graph{-x-2 [-10, 10, -5, 5]}