Orthocentre of a triangle defined by three lines Problem: If the orthocentre of the triangle formed by the lines $2x+3y-1=0$,$x+2y-1=0$,$ax+by-1=0$ is at the origin, then $(a,b)$ is given by?
I would solve this by finding poins of intersection and the standard text-book methods. But, I was discouraged by the algebraic labour that I have to bear. So, I turn to you for a general, quick, intuitive formula for the orthocentre of a triangle defined by three lines
I do not require the answer to the original problem but more so, a formula or method, and obviously not the one I already mentioned, i.e. determining points of intersection and then finding the orthocenter.

Since the three lines intersect precisely at the orthocenter = the origin, we have that:
$$\text{intersection of lines I, III}\;:\;\begin{cases}2x+3y=1\\
ax+by=1\end{cases}\implies \begin{cases}\;2ax+3ay=a\\\!\!-2ax-2by=-2\end{cases}\implies$$
$$(3a-2b)y=a-2\implies y=\frac{a-2}{3a-2b}$$
And since we're given $\;y=0\implies a=2\;$ , and from here...continue.
You don't need all the equations for the intersections, but you need, as far as I can tell, at least one set of equations for one intersection point (which is known: the orthocenter)
