Two lines are perpendicular if:
[tex]m1\cdot m2=-1[/tex]For the line:
[tex]y=\frac{2}{3}x[/tex]We can see that its slope is 2/3, so:
Let:
m1 = 2/3
Therefore:
[tex]\begin{gathered} \frac{2}{3}\cdot m2=-1 \\ \text{Solving for m2:} \\ m2=-\frac{1}{\frac{2}{3}}=-\frac{3}{2} \end{gathered}[/tex]Now, we have the slope and a point P(6,6), let's use the slope point equation:
[tex]\begin{gathered} y-y1=m(x-x1) \\ y-6=-\frac{3}{2}(x-6) \\ y-6=-\frac{3}{2}x+9 \\ \text{Solving for y:} \\ y(x)=-\frac{3}{2}x+9+6 \\ y(x)=-\frac{3}{2}x+15 \end{gathered}[/tex]
