1. ## graphing

If the graphs of $x + 3y – 5 = 0$ and $px + 2y +1 = 0$ are to meet at right angles, then the value of $p$ is:

-6, 6, $2/3$, - $2/3$ or - $3/2$ ?

2. Originally Posted by foreverbrokenpromises
If the graphs of $x + 3y – 5 = 0$ and $px + 2y +1 = 0$ are to meet at right angles, then the value of $p$ is:

-6, 6, $2/3$, - $2/3$ or - $3/2$ ?
your first eqn is $x+3y-5=0$ ?

its slope $= \frac{-1}{3}$

slope of second line $=\frac{-p}{2}$

if the two lines are perpendicular then their slopes will be negative reciprocal of each other.

$\frac{-1}{3}=\frac{2}{p}$

$p=-6$