How do you find a line that passes through (3,5) perpendicular to a line whose slope is -1/9?

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Answer 1 · 2017-02-17T12:47:19

#y=9x-22#

The standardised equation of a strait line is #y=mx+c# where #m# is the gradient.

Any line perpendicular to this will have the gradient of:

#" " (-1)xx1/m#

So in this case as #m=-1/9# the gradient of the perpendicular is:

#(-1)xx-9/1 = +9#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus the equation of the line we are after is #y=9x+c#

This line passes through the point #(x,y)->(3,5)#

so by substitution we have:

#y=9x+c" "->" "5=9(3)+c#

#" "5=27+c#

subtract 27 from both sides

#" "5-27=27-27+c#

#" "c=-22#

Thus we have:

#y=9x-22#