Prove that locus of orthocentre is $x+y=0$ The base of a triangle is axis of $x$ and its other $2$ sides are given by the equations :$y = (1+α)\frac{x}{α} + (1+α)$ and $y = (1+β) \frac{x}{β} + (1+β)$.Prove that the locus of its orthocentre is the line $x+y = 0 $
Finding orthocentre by finding two perpendiculars and then their point of intersection is proving to be tedious. Could someone suggest an alternate approach.
 A: Personally I don't find it too tedious, though (or because) I have skipped the "finding two perpendiculars" step. There could be geometric reasoning behind the simple locus, but I have failed to see it so far.

Representing the slanted sides in intercept form,
$$\frac{x}{-k} + \frac y{1+k} = 1$$
And so the two vertices on the $x$-axis are $A(-\alpha,0)$ and $B(-\beta, 0)$.
Representing them in slope-intercept form,
$$y = \left(\frac{1+k}k\right)x + (1+k)$$
And so the two slanted slopes are $\frac{1+\alpha}\alpha$ and $\frac{1+\beta}\beta$. The slopes of the lines perpendicular to them are $\frac{-\alpha}{1+\alpha}$ and $\frac{-\beta}{1+\beta}$ respectively.

For every point on the altitude through $A(-\alpha, 0)$, its coordinates can be represented by
$$(-\alpha,0) + s\, (1+\beta,-\beta) = (s-\alpha+s\beta, -s\beta)$$
where $s\in\mathbb R$. Similarly, the points on the altitude through $B(-\beta,0)$ are
$$(-\beta,0) + t\, (1+\alpha,-\alpha) = (t-\beta+t\alpha, -t\alpha)$$
where $t\in\mathbb R$. Equating these two to find their intersection point, i.e. the orthocentre $H$, 
$$\begin{align*}
s-\alpha+s\beta&= t-\beta+t\alpha\tag1\\
-s\beta &= -t\alpha\tag2
\end{align*}$$
From $(2)$,
$$t = s\cdot\frac\beta\alpha\tag3$$
Substituting $(3)$ into $(1)$, and using $s\beta = t\alpha$,
$$\begin{align*}
s -\alpha + s\beta &= s\cdot\frac\beta\alpha - \beta + s\beta\\
s &= \frac{\alpha -\beta}{1-\beta/\alpha}\\
&= \frac{\alpha(\alpha-\beta)}{\alpha-\beta}\\
&= \alpha\\
H&= (\alpha \beta, -\alpha\beta)
\end{align*}
$$
