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Find the coordinates of the orthocenter of the triangle, the equations of whose sides are $x + y = 1$, $2x + 3y = 6$, $4x − y + 4 = 0$.

I know how to find it by first finding all the vertices (the orthocenter is the point of intersection of all altitudes). How can we find it without finding the coordinates of its vertices?

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    @JackD'Aurizio he seems to be asking whether we can find the orthocenter without finding the vertices, using only the equations – 2017-01-27
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    @jackd'aurizio pretty much nothing. Why? It was not said by me, the Op specifies he doesn't want to find vertices directly – 2017-01-27
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    @N.S.JOHN is saying correct – 2017-01-27
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    @jackd'aurizio sir I think you are mistaking me for the OP – 2017-01-27

2 Answers 2

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(See figure below)

Here is a solution bypassing the obtention of vertices' coordinates, by using "pencils of lines".

Being given two lines $L_1$ and $L_2$ with resp. equations $$\begin{cases}u_1x+v_1y+w_1=0\\u_2x+v_2y+w_2=0\end{cases}$$

Any line passing through point $L_1 \cap L_2$ has general equation

$$m(u_1x+v_1y+w_1=0)+(1-m)(u_2x+v_2y+w_2=0)=0, \ \ m \in \mathbb{R}.$$

The set of all these lines is called the "pencil of lines" defined by $L_1$ and $L_2$.

Thus, the pencil of lines defined by $x+y-1=0$ and $2x+3y-6=0$ is:

$$a(x+y-1)+(1-a)(2x+3y-6)=0$$

$$\tag{1} \iff \ (2-a)x+(3-2a)y+(5a-6)=0$$

Among these lines, one is the height. This height is characterized by the fact that its normal vector is orthogonal to the normal vector of the third line:

$$\binom{2-a}{3-2a} \perp \binom{4}{-1} \ \ \iff \ \ (2-a)4+(3-2a)(-1)=0 \ \ \iff \ \ a=\frac52$$

(recall: a line with equation $ux+vy+w=0$ has normal vector $\binom{u}{v}.$)

Plugging this value of $a$ in (1) gives the equation of the height:

$$\tag{2}x+4y-13=0.$$

Working in the same way with a group of 2 other sides:

$$b(2x+3y-6)+(1-b)(4x-y+4)=0$$

$$\tag{3} \iff \ \ (4-2b)x+(4b-1)y+(4-10b)=0$$

$$\binom{4-2b}{4b-1} \perp \binom{1}{1} \ \ \iff \ \ (4-2b)1+(4b-1)1=0 \ \ \iff \ \ b=-\frac32$$

Plugging this value of $b$ in (3) gives the equation of the second height:

$$\tag{4}7x-7y+19=0.$$

The solution of the system formed by (2) and (4) are the coordinates of the orthocentre

$$H\left(\frac{3}{7},\frac{22}{7}\right).$$

enter image description here

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    (+1) This is exactly what I had in my mind. I still wonder *why* one might be tempted to choose such detour, but anyway it does what it is supposed to. – 2017-01-27
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    Although the letter $B$ is equivalent to find a vertex, you explicitly didn't use it, well done anyway. – 2017-01-28
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    Just a remark, in my place, the "pencil of lines" is more popularly called as the "family of lines" instead. – 2018-05-05
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Call $O=(x_0,y_0)$ the orthocenter. A general line that goes through $O$ is like:

$$y=ax+(y_0-ax_0)$$

Once it is perpendicular to all three lines then all heights must have the form:

$$y=x+y_0-x_0\\ y=\frac{3}{2}x+y_0-\frac{3}{2}x_0\\ y=-\frac{1}{4}x+y_0-\frac{1}{4}x_0$$

See that all three lines pass through $O$ but in order to be a height all of them must pass through some vertex. It means that there is not a way to find $O$ without at least two vertices.

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    (-1) Usually $O$ denotes the circumcenter, and the orthocenter is denoted through $H$. Additionally, these lines do not prove we strictly need to find the coordinates of two vertices to locate $H$. For instance, we may find the equation of an altitude since we know it is orthogonal to a side, and by imposing it is concurrent with the other two sides (through a determinant $=0$). If we do that twice and intersect two altitudes, we locate $H$ without using the coordinates of two vertices. – 2017-01-27
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    @JackD'Aurizio: Just to be fair. Before downvoting you should post your solution and see if that works. – 2017-01-27
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    I just sketched the proof. I am not saying it is more efficient than the trivial approach by locating two (or three) vertices, I am just saying your conclusion is clearly wrong: *there is* a way. – 2017-01-27
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    @JackD'Aurizio: I'm really interesting to see if what you said is not equivalent to find a vertex. – 2017-01-27
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    @JackD'Aurizio: anyway, at least you said why you are downvoting, what is good, once many people don't do that. Thanks! – 2017-01-27
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    *It means that there is not a way to find O without at least two vertices.* : as stated, **this is wrong**. At most, there is no *simple* way to find O without the coordinates of $\geq 2$ vertices. I agree that my approach and yours are equivalent, I am just debating your conclusion. – 2017-01-27
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    @Jack D'Aurizio If you have some time, have a look at my solution. – 2017-01-27