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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.

Cheeku
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  • Find where the first two lines meet. The perpendicular line to the third one (what is its form in terms of a and b?) must go through (0,0) and this point of intersection. That should give a and b. – Paul Apr 12 '14 at 15:42
  • Let's call the first line $AB$, the second line $AC$ and the third line $BC$. A line that passes the orthocenter $O$ and perpendicular to $AB$ will intersect $AC$ at $C$, because that's the altitude drawn from $C$. Same goes for the line passing $O$ and perpendicular to $AC$. Once you have points $B$ and $C$, you can derive the equation for $BC$ – Dylan Apr 12 '14 at 17:00
  • Actually, this problem is even easier because the 3rd constant is known so you'll only need to find the slope of the third line. (it's equivalent to $y = -\frac{a}{b}x + \frac{1}{b}$).

    Find the point where the first 2 lines intersect. The line passing through that point and the orthocenter will be perpendicular to the third line. You don't need to find these lines, only their slopes. 2 perpendicular lines with slopes $m_1$ and $m_2$ will have the relation $m_2 = -1/m_2$

     – Dylan Apr 12 '14 at 17:03

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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)

DonAntonio
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