How do you write an equation of a line passing through (2, -8), perpendicular to #4x+5y=7#?

1 Answer

Answer:

#y=5/4x-21/2#

Explanation:

First, you would want to get only #y# on one side of the given equation (for converting the equation to point slope form), first by subtracting #4x# from both sides:

#5y=7-4x# or #5y=-4x+7#

Then divide by 5:

#y=7/5-4/5x# or #y=-4/5x+7/5#

Product of slopes of two perpendicular lines is #-1#, so you know that the slope of the line perpendicular to given line would be

#(-1)/(-4/5)=1xx5/4=5/4#

Now, make your new equation in point-slope format:

#y=mx+b#

#y=5/4x+b#

Now, to figure out the #y#-intercept (b), you can substitute #2# and #-8# into x and y, respectively i.e.

#(-8)=5/4xx2+b#

Multiply #5/4# and #2#:

#(-8)=10/4+b# or #(-8)=5/2+b#

Adding #-5/2# to both sides:

#(-8)-5/2=b#

Make -8 and 5/2 have a common denominator:

#-16/2-5/2=b#

#-21/2=b#

Then you can put that into the point-slope formula and you're done!

#y=5/4x-21/2#
graph{(y-5/4x+21/2)(5y-7+4x)=0 [-8.62, 11.38, -8.44, 1.56]}