Find orthocentre of a triangle given equations of its three sides, without finding triangle's vertices 
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?
 A: (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 

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


A: 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.  
