What is the standard form of a line that goes through #(5, -4)# and is perpendicular to #y = 5/4x -5#?

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Answer 1 · 2017-04-20T12:56:54

#5y+4x=0#

Since the line is perpendicular to another line with slope #5/4#, its slope will be the negative reciprocal of the other line's slope.

Hence the slope of the line is #-4/5#. We also know it passes through #(5, -4)#.

Using

#y=mx+c#

we know

#"m (slope) =" -4/5#

therefore

#y = -4/5x+c#

Substituting #(5, -4)# gives you

#-4=-4/5 (5)+c#

#c=0#

Therefore

#y=-4/5x #

#5y=-4x#

#5y+4x=0#