How do you write an equation in standard form for a line passing through (–2, 8) with a slope of 2?

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Answer 1 · 2015-05-01T16:51:29

Standard form for a line is #ax+by=c#

Start in point slope form: #y-y_1 = m(x-x_1)# where #(x_1, y_1)# is a point on the line and #m# is the slope.

#y-8=2(x-(-2))#

Now simplify and remove the parentheses (by distributing the #2#):

#y-8=2(x +2)#

#y-8=2x +4#

To get slope-intercept form, we would get the #y# alone by adding #8#.

But we've been asked for standard from, so we need to get the #x# and #y# together and everything else on the other side:

Subtract #2x# (or add #-2x# -- it's the same) and add #8# to get:

#y-2x=4+8=10#

We need the #x# term first, (remember the #-2# is stuck to the #x#) so:

#-2x+y=12#

If you don't like the #-# sign in front of the #x#, you can multiply both sides by #-1# (although this will put a #-# sign in front of the #12#)

#-1(-2x+y) = (-1)(12)#

#2x-y=-12#

Answer 2 · 2015-05-01T16:51:50

The most commonly accepted version of "standard form" for a linear equation in two unknowns is:
#Ax+By=C# for some constants #A, B, C#

The equation of a line through #(-2,8)# with a slope of #2# is
#(y-8)/(x-(-2)) = 2#

#y-8 = 2x+4#

#-2x+y = 12#