Q:

What is the slope of a line that is perpendicular to a line whose equation is 3y=βˆ’4x+2 ?

Accepted Solution

A:
we know thatif two lines are perpendicular, then the product of their slopes is equal to minus oneso[tex]m1*m2=-1[/tex]Step 1Find the slope of the given linewe have[tex]3y=-4x+2[/tex]Solve for yDivide by [tex]3[/tex] both sides[tex]3y/3=(-4x+2)/3[/tex][tex]y=-\frac{4}{3}x+ \frac{2}{3}[/tex]the slope of the given line is [tex]m1=-\frac{4}{3}[/tex]Step 2Find the slope Β of a line that is perpendicular to the given linewe have [tex]m1=-\frac{4}{3}[/tex][tex]m1*m2=-1[/tex][tex]m2=-1/m1[/tex]substitute the value of m1[tex]m2=-1/(-4/3)=\frac{3}{4}[/tex]thereforethe answer isthe slope is [tex]\frac{3}{4}[/tex]