How do you write an equation that contains points (-1, 2) and is parallel to x-2y=-3?

1 Answer
Alan P.
Sep 24, 2015

#1x-2y=-6#

Explanation:

All lines parallel to #x-2y= -3# will have the same slope as #x-2y=-3#

That is all lines parallel to #x-2y=-3# will have a slope #m=1/2# (see below if the reason for this isn't obvious)

Using the slope point form (with point #(hatx,haty)=(-1,2)# and slope #m=1/2#)
we have
#color(white)("XXXX")(y-2)=1/2(x-(-2))#
which we could simplify as
#color(white)("XXXX")2y-4= x+2#
or in standard form as
#color(white)("XXXX")1x-2y = -6#

further explanation
How do we know the slope of #x-2y=-3# has a slope of #1/2#?

We can rearrange #x-2y=-3# as
#color(white)("XXXX")-2y= -x-3#
or
#color(white)("XXXX")y =1/2x + 3/2#
which is ane equation in slope-intercept form
with a slope of #1/2#
(and a y-intercept of #3/2#, although this isn't relevant to this question)

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