# Does the orthocenter really exist for all triangles?

Imagine I draw a triangle on a cartesian plane so that it has vertices at $P=(0,0), Q=(b,c), R=(a,0)$ where $b<a$ and $b,a>0$. As far as basic logic goes we should be able to arrange all triangles in this way. The equation of the the three lines that outline the triangle are:

line PR is $y=0$

line PQ is $y=\frac{c}{b}x$

Point slope form of RQ is $y-0=\frac{c-0}{b-a}(x-a)$ --> $y=\frac{c}{b-a}x+a*(\frac{c}{b-a})$

The altitude is a perpendicular line extending from one side to the opposite vertex. Let PR(Q) denote the altitude extending from line PR to vertex Q.

PR(Q) is $x=b$

PQ(R) is $y-0=-\frac{b}{c}(x-a)$ --> $y=-\frac{b}{c}x+\frac{ba}{c}$

QR(P) is $y-0=-\frac{b-a}{c}(x-0)$ --> $y=-\frac{b-a}{c}x$

By definition they all must intersect at $x=b$ and yield the same y value. Thus for every value of a,b,c

$\frac{b^2}{c}+\frac{ba}{c}=\frac{b-a}{c}$ must be true.

$b(b+a)=(b-a)$ --> $(b-1)(b+a)=0$ --> either b=1 or a=-b which does not apply to every triangle.

I have seen many proofs validating the concurrency of the altitudes and the existence of the orthocenter and I have no doubt that it is true. However, that would imply that I have made a mistake in my work and I have been unable to spot it. So, what did I miss?

• There is some confusion in the parameters. What is $c$? Also, if $Q$ and $R$ has the same $x$-coordinate, then the triangle is right, and not a generic one. – Anatoly Nov 10 '14 at 7:59
• You've fixed your notation, and your equations are now valid ... up until you claim that this must be true: $$\frac{b^2}{c}+\frac{ba}{c}=\frac{b-a}{c}$$ What you should have is $$-\frac{b^2}{c}+\frac{ba}{c} = -\frac{b-a}{c}b$$ which is an identity. – Blue Nov 10 '14 at 18:10
• By the way: It's not "by definition" that the equations $PQ(R)$ and $QR(P)$ have the same $y$-value when $x=b$. It's "by the Theorem of Altitude Concurrency" (for lack of a better name). – Blue Nov 10 '14 at 18:13

If $P=(0,0)$, $Q=(u,v)$, $R=(w,0)$ (where we may assume $u\neq w$ and $v\neq 0$), then \begin{align} PR &:\quad y = 0 \\[4pt] PQ &:\quad y = \frac{v}{u}x \\[4pt] QR &:\quad y = -\frac{v}{w-u}x+\frac{vw}{w-u} \end{align}
and then \begin{align} PR(Q) &:\quad x = u \\[6pt] PQ(R) &:\quad y = -\frac{u}{v} x + \frac{uw}{v} = \frac{u(w-x)}{v} \\[6pt] QR(P) &:\quad y = \frac{w-u}{v} x \end{align}
For $x=u$, equations $PQ(R)$ and $QR(P)$ yield the same $y$-coordinate (namely, $u(w-u)/v$) for a point lying on all three altitudes. Concurrency is confirmed.
Moreover, because $QR(P)$ and $PQ(R)$ are "the same equation" when $x=u$, there's no isolating $w$ to show dependence on $u$ and $v$.