If I have $2$ lines$AB$ and $PQ$ perpendicular to each other drawn on a graph- then will the slope of $AB$ times the slope of $PQ$ be equal to $-1$?
i.e $ [(y2-y1)/(x2-x1)] \times [(b2-b1)/(a2-a1)] = -1 ? $
If I have $2$ lines$AB$ and $PQ$ perpendicular to each other drawn on a graph- then will the slope of $AB$ times the slope of $PQ$ be equal to $-1$?
i.e $ [(y2-y1)/(x2-x1)] \times [(b2-b1)/(a2-a1)] = -1 ? $
Yes, that is so. A small trigonometrical proof will be like this:
$\tan( \theta + \frac{\pi}{2} ) = - \cot {\theta} = - \frac{1}{\tan{\theta}}$