For this case we have to by definition, if two lines are perpendicular then the product of its slopes is -1. That is to say:
[tex]m_ {1} * m_ {2} = - 1[/tex]
We have the following equation:
[tex]3y = -4x + 2\\y = - \frac {4} {3} x + \frac {2} {3}[/tex]
Thus, the equation is of the pending-intersection form[tex]y = mx + b[/tex]
Where:
m: It's the slope
b: It is the cut-off point with the y axis
We have to:
[tex]m_ {1} = - \frac {4} {3}[/tex]
Thus, we find [tex]m_ {2}[/tex]:
[tex]m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {- \frac {4} {3}}\\m_ {2} = \frac {3} {4}[/tex]
Thus, the slope of the line perpendicular to the given line is:
[tex]m_ {2} = \frac {3} {4}[/tex]
ANswer:
[tex]m_ {2} = \frac {3} {4}[/tex]
