geometrytriangles
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?
To solve the problem as stated, it's not necessary to write equations for the altitudes, or even to find point $C$. It just asks for equations representing the three sides of the triangle.
Side $AB$ is the simplest case, since points $A$ and $B$ are given. You have already computed its slope $m_{AB}=4$, and can use either $A$ or $B$ as the point on the line.
For side $AC$, notice that it must be perpendicular to the line $BN$. We don't need to know where these lines intersect to conclude that $m_{AC}m_{BN} = -1$. And $m_{BN}$ is simple to compute from the two given points $B$ and $N$. Knowing the slope and a point $A$ on the line, you can write the equation for $AC$.
Similarly, side $BC$ is perpendicular to the line $AN$, so you can compute $m_{AN}$ and from that determine $m_{BC}$, then write the equation for $BC$.
The orthocenter of $ABC$ is the incenter of the orthic triangle.
If you check that $(10,15)$ is the incenter of the orthic triangle (or simply notice it is the only point lying inside the orthic triangle) it follows that $(5,10),(15,30),(50,-5)$ are the vertices of the original triangle, also since (by computing dot products) the given options form a orthocentric system.
Best Answer
(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