An equation of a line perpendicular to the line represented by the equation #y=-1/2x - 5# and passes through #(6,-4)# is what?

1 Answer
Bill Jorgensen
May 21, 2018

#y-6 = 2(x+4)#

Explanation:

First find the slope #m# for the new line, rule for a perpendicular slope is:

your equation is in slope intercept form #y=mx+b#:

#y=-1/2x - 5#

#m'=-1/m#

#m=-1/2# so your new slope #m'=-1/(-1/2) = 2#

Now just use the slope point formula to solve your new line:

#y-y_1 = m(x-x_1)#

your point #(x_1,y_1) = (6, -4)#

#y-6 = 2(x-(-4))#

#y-6 = 2(x+4)#

you can do the algebra and convert this to standard form #Ax + By = C# or slope intercept for #y=mx+b# but the problem just asks for "an equation" so I assume this one would do fine.

here is the original graph:

graph{y=-1/2x - 5 [-18.92, 21.08, -14.56, 5.44]}

and here is the new one:

graph{y-6 = 2(x+4) [-41.75, 38.25, -17.36, 22.64]}

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