How do you write an equation of a line that is parallel to #y+3x=7# and passes through point #(7,2)#?

1 Answer
Feb 9, 2015

I would use first your equation to find the SLOPE of your line. Basically the slope is a number that tells you what the inclination of your line is.
So, to find the parallel to your line you need a line with the same inclination...the same slope:
Your line: #y+3x=7# can be written (isolating the #y# on the left) as:
#y=-3x+7# this allows you to "read" immediately the slope of your line, the coefficient of #x#, which in your case is #-3#.

Now the difficult bit...
The slope represents the inclination of your line and basically tells you how #y# changes when #x# changes!

For example, a big slope means that at every fixed change in #x# the value of #y# changes a lot and your line is VERY steep!!!
Have a look at this picture:

enter image source here

slope #5# is steeper than slope #2#!

To find your slope you simply take the change in #y# divided by the change in #x#:

In your case:

#slope=(Deltay)/(Deltax)=(y_2-y_1)/(x_2-x_1)# but you want a slope which is equal to the one of your original line: #-3#
Together with the coordinates of your point you can write:
#slope=(y_2-y_1)/(x_2-x_1)#
#-3=(y-2)/(x-7)#
#-3x+21=y-2# and finally your line:
#y+3x=23#
Graphically:

enter image source here

hope it helps