If $2x-py-3=0$ and $6x+3y+2=0$ are perpendicular what is $p$?

Question

If $2x-py-3=0$ and $6x+3y+2=0$ are perpendicular what is $p$?

Answer 1

First method: rewrite the two equations as $y = \frac{2x}{p} - \frac{3}{p}$ (so $p\neq 0)$ and $y = -2x -\frac{2}{3}$. The lines are perpendicular if the products of the two coefficients for $x$ are $-1$: $\frac{2}{p} \cdot -2 = -1$.

If you are familiar with the inproduct and the direction vectors, there is a second method: take this inproduct of $(2,-p)$ and $(6,3)$. This should be equal to zero: $2\cdot 6 - p\cdot 3 = 0$.

Answer 2

Slope of the first line $=\dfrac{2}{p}$ Slope of the second line$=-2$

To be perpendicular, $$ \dfrac{2}{p}*(-2)=-1$$ Hence,$$p=4$$