Drawing an altitude from origin to the opposite side of a triangle If I know the vertexes of a triangle and one of them is origin O(0, 0). Then how can I draw altitude from origin to the opposite side of the triangle.
 A: The altitude will pass through the origin and be perpendicular to the opposite side. If you know the three vertices, you can find the slope of the side opposite the origin with the usual slope formula between two points. Then the altitude will have a slope equal to the negative reciprocal of the opposite side, assuming this slope is defined. So the altitude is described completely by its slope and the fact that it must pass through the origin.
A: If the vertices are $(a,b)$, $(p,q)$, and $(0,0)$, then the altitude from the opposite side to $(0,0)$ will be along the line that is perpendicular to the line joining $(a,b)$ and $(p,q)$.
If $a=p$, then the side is vertical, and the altitude goes from $(0,0)$ to $(a,0)$ (horizontal).
If $a\neq p$, then the side is in the line with slope
$$\frac{q-b}{p-a},$$
so you want the line through $(0,0)$ with slope
$$m = \frac{a-p}{q-b}.$$
This is the line $y=\left(\frac{a-p}{q-b}\right)x$. 
The intersection of this line with the line that contains the opposite side, which is
$$ y - b = \left(\frac{q-b}{p-a}\right)(x-a)$$
can be found by plugging in $ y = \left(\frac{a-p}{q-b}\right)x$ and solving for $x$.
