Answer
4x+3y=17
Work Step by Step
We have to find equation of the line passing through the point (5,-1) and is perpendicular to the line 3x-4y=12.
Let's say slope of 3x-4y=12 is m1 and slope of the line passing through the point (5,-1) and perpendicular to the line 3x-4y=12 is m2.
First we need to find slope of the line 3x-4y=12
4y=3x-12
y=(3x-12)/4
y=(3/4)x - 3
we get m1=3/4
Condition for the lines to be perpendicular is,
m1*m2 = -1
(3/4)*m2 = -1
m2 = -4/3
The line passes through (5,-1) so equation of the line is,
y-(-1) = (-4/3)*(x-5)
y+1 = (-4x+20)/3
3y+3 = -4x+20
4x+3y=17
So the equation of the line passing through the point (5,-1) and perpendicular to the line 3x-4y=12 is 4x+3y=17\leq
