Answer:
y = [tex]\frac{3}{2}[/tex]x + 15
Step-by-step explanation:
We are to find the equation of a line ([tex]l_{2}[/tex]) which is perpendicular to line ([tex]l_{1}[/tex]) whose equation is; y = [tex]\frac{-2}{3}[/tex]x + 5 and that;
[tex]l_{2}[/tex] goes through (-8,3)
The slope of [tex]l_{2}[/tex] is [tex]-\frac{2}{3}[/tex]
The product of slope of [tex]l_{1}[/tex] and that of [tex]l_{2}[/tex] = -1 (The product of two perpendicular slopes is -1)
Therefore, [tex]-\frac{2}{3}[/tex] times slope of [tex]l_{2}[/tex] = -1
Slope of [tex]l_{2}[/tex] = -1 ÷ [tex]-\frac{2}{3}[/tex] = [tex]\frac{3}{2}[/tex]
Taking another point (x,y) on [tex]l_{2}[/tex];
Slope = Change in y ÷ change in x = [tex]\frac{y - 3}{x - -8}[/tex] = [tex]\frac{3}{2}[/tex]
y - 3 = [tex]\frac{3}{2}[/tex]x + 12
y = [tex]\frac{3}{2}[/tex]x + 12 + 3
Finally,
y = [tex]\frac{3}{2}[/tex]x + 15
