How do you write an equation of a line Perpendicular to #3x+4y=12#, through (7, 1)?

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Answer 1 · 2016-11-14T10:45:29

#4x-3y-25=0#

Product of the slopes of two lines perpendicular to each other is #-1#. Let the slope of desired line be #m#.

Writing the equation of given line #3x+4y=12# in slope intercept form, we get

#4y=-3x+12# or #y=-3/4x+3#

Hence slope of this line is #-3/4#.

As such we have #(-3)/4 xxm=-1#

or #m=-1xx4/(-3)=4/3#

Hence, slope of desired line is #4/3# and as it passes through #(7,1)#

we can have equation of desired line using point slope form of equation i.e.

#(y-1)=4/3(x-7)#

or #3((y-1)=4(x-7)#

or #3y-3=4x-28#

or #4x-3y-25=0#
graph{(3x+4y-12)(4x-3y-25)=0 [-6.21, 13.79, -3.32, 6.68]}