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

(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

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