What is the equation of the line passing through (2, –3) and parallel to the line #y = –6x – 1# in standard form?

1 Answer
Daniel L.
May 18, 2015

The answer is #6x+y-9=0#

You start by noting that the function you are looking for can be written as #y=-6x+c# where #c in RR# because two parallel lines have the same "x" coeficients.
Next you have to calculate #c# using the fact that the line passes through #(2,-3)#
After solving the equation #-3=-6*2+c#
#-3=-12+c#
#c=9#
So the line has the equation #y=-6x+9#
To change it to the standard form you just have to move #-6x+9# to the left side to leave #0# on the right side, so you finally get:
#6x+y-9=0#

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