Question

Write an equation of the line Perpendicular to the given line that contains PP(6,6)y=(2/3)x

Answer 1

Two lines are perpendicular if:

`m1\cdot m2=-1`

For the line:

`y=(2)/(3)x`

We can see that its slope is 2/3, so:

Let:

m1 = 2/3

Therefore:

`\begin{gathered} (2)/(3)\cdot m2=-1 \\ \text{Solving for m2:} \\ m2=-(1)/((2)/(3))=-(3)/(2) \end{gathered}`

Now, we have the slope and a point P(6,6), let's use the slope point equation:

`\begin{gathered} y-y1=m(x-x1) \\ y-6=-(3)/(2)(x-6) \\ y-6=-(3)/(2)x+9 \\ \text{Solving for y:} \\ y(x)=-(3)/(2)x+9+6 \\ y(x)=-(3)/(2)x+15 \end{gathered}`