How do you write an equation in standard form for a line perpendicular to x+3y=6 and passing through (-3,5)?

1 Answer
George C.
May 25, 2015

Fun fact:

Given an equation in #x# and #y# of a line, then you can construct the equation of a perpendicular line by swapping #x# and #y# and reversing the sign of one of them.

This is equivalent to reflecting the original line in the #45^o# line with equation #y=x# then reflecting in one of the axes.

So in your example, we can replace the original

#x+3y=6#

with

#y-3x=6#

to get a perpendicular line.

Then adding #3x# to both sides we get:

#y=3x+6#

which is in standard slope intercept form, with slope #3# and intercept #6#

The line we want to construct is parallel to this, so will have the same slope, #3#, but probably a different intercept.

Given the slope #3# and the point #(-3, 5)#, we can write the equation of the desired line in standard point slope form as:

#y-5 = 3(x-(-3)) = 3(x+3)#

To convert this to slope intercept form, add #5# to both sides to get:

#y = 3(x+3)+5 = 3x+9+5 = 3x+14#

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