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I need to prove that the altitudes of a triangle intersect at a given point using co-ordinate geometry.

I am thinking of assuming that point to be $(x,y)$ and then using slope equations to prove that the point exists and I can think of another way too by taking two equations (altitudes) forming a family of line which must be equal to the third equation for satisfying the condition given in the question, But thinking is all I am able to do, I am unable to put it on the paper. A hint towards solution would be great.

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2 Answers 2

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If we assume that two vertices of the triangle lie on the $x$-axis (which can be obtained by rotation and translation) then the vertices are $(0,0)$, $(x_1,0)$ and $(x_2,y_2)$. Let us try to find the equations of the altitudes.

The first one is easy $$x=x_2 \qquad (1).$$

What about second one? We want to find a line with normal vector $(x_2-x_1,y_2)$ which goes trough the point $(0,0)$. (We know the normal vector since the line has to be to perpendicular on the segment connecting the points $(x_1,0)$ and $(x_2,y_2)$.) Using standard form of the equation of line we get $$(x_2-x_1)x+y_2y=0 \qquad (2).$$

The third of them: We are looking for a line with normal vector $(x_2,y_2)$ which goes through the point $(x_1,0)$. This gives the equation $$x_2x+y_2y=x_2x_1 \qquad (3).$$

The only one thing is missing - to show that the system of linear equations (1), (2) and (3) has a solution.


I think that the solution would not be much more complicated even without the assumption that $y_1=0$ (i.e. without using the rotation). But I would say that moving one of the vertices to $(0,0)$ simplifies things a little. (EDIT: But I agree with Srivatsan's comment that working out the details of the actual computation could be messy.)

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This table lists the coordinates of the various triangle centers: en.wikipedia.org/wiki/…. I am expecting that the expression would look slightly messy because it involves $\sec A, \sec B, \sec C$. Perhaps this problem might become easier if we used complex numbers instead. –  Srivatsan Dec 25 '11 at 19:38
    
I have a slicker proof that uses vector algebra and I'll refrain from adding it here as OP seems to not want it. –  user21436 Dec 25 '11 at 19:45
    
@KannappanSampath Even if it is (perhaps) not that much interesting for OP, it could be useful for other users... –  Martin Sleziak Dec 25 '11 at 19:50
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Thanks! Your approach to place the triangle on one of the axis is great. I am taking the co-ordinates as $(b,c), (0,0)$ and $(a,0)$. Now, the equations I am getting are $x = b, y = \frac{a-b}{c}x $ and $y = \frac{-c}{b}x + \frac{ac}{b}$. And now I think we can apply the condition for concurrency $$\left[\begin{matrix}a_1&a_2&a_3\\b_1&b_2&b_3 \\ c_1 &c_2&c_3\end{matrix}\right] =0$$ comparing it with standard equation $a_ix + b_iy + c_i=0$ –  Ishaan Singh Dec 25 '11 at 20:02
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@Ishaan Singh: The line $y = \frac{-c}{b}x + \frac{ac}{b}$ from your comment is perpendicular to the vector $(c,b)$ and you want to get a line perpendicular to the vector $(b,c)$. (I hope I did not miss something.) –  Martin Sleziak Dec 25 '11 at 20:12

I'll prove this proposition by Vector algebra, (not to solve OP's problem, but essentially for other users who might find it useful):

Let $\Delta ABC$ be a triangle whose altitudes $AD$, $BE$ intersect at $O$. In order to prove that the altitudes are concurrent, we'll have to prove that $CO$ is perpendicular to $AB$.

Taking $O$ as the origin, let the position vectors of $A$, $B$, $C$ be $\vec{a}$, $\vec{b}$, $\vec{c}$ respectively. Then $\vec{OA}=\vec{a}$, $\vec{OB}=\vec{b}$ and $\vec{OC}=\vec{c}$.

Now, as $AD \perp BC$, we have $\vec{OA} \perp \vec{BC}$. This means $\vec{OA} \cdot \vec{BC}=0$. This means, $$\vec{a} \cdot (\vec{c}-\vec{b})=0$$

Similarly $\vec{OB} \perp \vec{CA}$ and that gives you, $$ \vec{b} \cdot (\vec{a} -\vec{c})=0$$ Adding these you'll have, $$ (\vec{a}-\vec{b}) \cdot \vec{c} =0$$ This reads off immediately that $\vec{OC} \perp \vec{BA}$. This proves the proposition.

Our Triangle

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