Question

What is the orthocenter of a triangle with vertices at `O(0,0 )`, `P(a,b)`, and Q(c,d)#?

Answer 1

`(x,y) = { ac + bd }/{ad - bc} (d-b,a-c)`

Explanation:

I've generalized this old question rather than asking a new one. I did this before for a circumcenter question and nothing bad happened, so I continue the series.

As before I put one vertex at the origin to try to keep the algebra tractable. An arbitrary triangle is easily translated and the result easily translated back.

The orthocenter is the intersection of the altitudes of a triangle. Its existence is based on the theorem that the altitudes of a triangle intersect at a point. We say the three altitudes are concurrent.

Let's prove the altitudes of triangle OPQ are concurrent.

The direction vector of side OP is `P-O=P=(a,b),` which is just a fancy way of saying the slope is `b/a` (but the direction vector also works when `a=0`). We get the direction vector of the perpendicular by swapping coordinates and negating one, here `(b,-a).` Perpendicularly is confirmed by the zero dot product:

`(a,b)cdot(b,-a)=ab-ba=0 quad sqrt`

The parametric equation of the altitude from OP to Q is thus:

`(x,y) = Q + t(b,-a)= (c,d)+t(b,-a) quad` for real `t`

The altitude from OQ to P is similarly

`(x,y) = (a,b)+u(d,-c) quad` for real `u`

The direction vector of PQ is `Q-P=(c-a,d-b)`. The perpendicular through the origin, i.e. the altitude from PQ, is thus

`(x,y) = v(d-b,a-c) quad` for real `v`

Let's look at the meet of the altitudes from OP and PQ:

`(c,d)+t(b,-a) = v(d-b,a-c)`

That's two equations in two unknowns, `t` and `v`.

`c+bt = v(d-b)`

`d-at = v(a-c)`

We'll multiply the first by `a` and the second by `b`.

`ac+abt = av(d-b)`

`bd-abt = bv(a-c)`

Adding,

`ac + bd = v(a(d-b) + b(a-c)) = v(ad - ab + ab -bc)`

`v = { ac + bd }/{ad - bc}`

Way cool with the dot product in the numerator and cross product in the denominator.

The meet is the presumed orthocenter `(x,y)`:

`(x,y) = v(d-b,a-c) = { ac + bd }/{ad - bc} (d-b,a-c)`

Let's find the meet of the altitudes from OQ and PQ next. By symmetry we can just swap `a` with `c` and `b` with `d`. We'll call the result `(x',y').`

`(x',y') = { ca + db }/{cb - da} (b-d,c-a) = { ac + bd }/{ad - bc} (d-b,a-c)`

We have these two intersections are the same, `(x',y')= (x,y),` so we've proved the altitudes are concurrent. `quad sqrt`

We've justified the naming of the common intersection the orthocenter , and we've found its coordinates.

`(x,y) = { ac + bd }/{ad - bc} (d-b,a-c)`