Hint :-
- The product of slopes of two perpendicular lines is -1 .
- The slope intercept form of the line is [tex] y = mx + c [/tex] .
- The point slope form of the line is [tex] m(x-x_1) = (y-y_1) [/tex]
Solution :-
The given equation to us is ,
[tex]\implies x = \dfrac{-3}{4}y - 5 [/tex]
Convert it to slope intercept form we have ,
[tex]\implies 4x =-3y -20 \\\\\implies 3y = -4x -20 \\\\\implies y =\dfrac{-4}{3}x -\dfrac{20}{3}[/tex]
On comparing to the slope intercept form of the line we have ,
- [tex]m = \dfrac{-4}{3}[/tex]
As we know that the product of slopes of two perpendicular lines is-1 , henceforth ,
- [tex] m_{perp}= \dfrac{3}{4}[/tex]
Now using slope point form rewrite the equation of the perpendicular line ,
[tex]\implies y-y_1 = m(x-x_1) \\\\\implies y -6 = \dfrac{3}{4}(x+12)\\\\\implies 4y -24 = 3x + 36 \\\\\implies 3x -4y +36+24=0\\\\\implies \underline{\boxed{ \gray{3x -4y +60=0}}}[/tex]
