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A triangle $ABC$ is given where vertex $A$ is $(1,1)$ and the orthocentre is $(2,4)$. Also sides $AB$ and $BC$ are members of the family of the lines $ax+by + c = 0$, where $a,b,c$ are in Arithmetic Progression..Find the other two vertices of the triangle.

Using information about Arithmetic Progression and vertex $A$, I have deduced that $ax-a=0$. But now I am able to proceed further and use coordinates of orthocentre. Could someone help me with this?

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The arithmetic progression condition amounts to $a+c=2b$.

Now, consider $AB$. Because $A$ is on this line, we have $a+b+c=0$, and as you deduced, we get $ax-a=0\implies x=1$. Thus, the $x$-coordinate of point $B$ must be 1 as well i.e. $B=(1,m)$.

Then, consider point $C$. Because $AB$ is on the line $x=1$, and the orthocenter is at $(2,4)$, $C$ must lie on the line passing through $(2,4)$ that is perpendicular to $x=1$ i.e. the line $y=4$. Thus, $C=(n,4)$.

Now, since $BC$ lies on a line of the form $ax+by+c=0$, we plug in $B$ to get $a+bm+c=0$ Recalling that $a+c=2b$, the first equation becomes $b(m+2)=0$, and from this we must have $m=-2$ or $b=0$. But $b=0$ forces $a+c=0$, and thus gives a degenerate triangle, so $m=-2$.

So $B=(1,-2)$, $A=(1,1)$, and $C=(n,4)$. The slope of the line through $B$ and $C$ is $\frac{6}{n-1}.$ The line with perpendicular to $BC$ passing through $A$ then, is $$y-1=\frac{1-n}{6}(x-1).$$ This must pass through the orthocenter, $(2,4)$, so we find $$3=\frac{1-n}{6}\cdot 1\implies n=-17.$$

Thus, the triangle is $A=(1,1), B=(1,-2), C=(-17,4)$.

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  • $\begingroup$ How did you say that $x$ -coordinate of point B must be $1$ as well? $\endgroup$ – MathGeek Aug 21 '16 at 20:40
  • $\begingroup$ Because the line through $AB$ has equation $ax-a=0$, as you said, so it is the line $x=1$. $\endgroup$ – Samir Khan Aug 21 '16 at 21:01

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