If the graphs of $\displaystyle x + 3y – 5 = 0$ and $\displaystyle px + 2y +1 = 0$ are to meet at right angles, then the value of $\displaystyle p $ is:
-6, 6, $\displaystyle 2/3$, -$\displaystyle 2/3$ or -$\displaystyle 3/2$ ?
your first eqn is $\displaystyle x+3y-5=0$ ?
its slope $\displaystyle = \frac{-1}{3}$
slope of second line $\displaystyle =\frac{-p}{2}$
if the two lines are perpendicular then their slopes will be negative reciprocal of each other.
$\displaystyle \frac{-1}{3}=\frac{2}{p}$
$\displaystyle p=-6$