analytic geometrytriangles
The line equations:
$$x+2y-1=0$$
$$5x+4y-17=0$$
$$x-4y+11=0$$
form a triangle,
How can I find the line equations for the heights of that triangle, without searching for the coordinates of the vertices?
I know it has to do with the intersection of the lines, but I don't know how to find the symmetrical line of that angle.
Thanks in advance
one line segment from $(x_1,y_1)$ to $(x_2,y_2)$
another line segment from $(x_3,y_3)$ to $(x_4,y_4)$
the set of points on the first line segment is $$A = \{ (x_1 + (x_2-x_1)u,y_1 + (y_2-y_1)u) \in \mathbb R^2 \mid u \in [0,1] \}$$
the set of points on the second line segment is $$B = \{ (x_3 + (x_4-x_3)t,y_3 + (y_4-y_3)t) \in \mathbb R^2 \mid t \in [0,1] \}$$
we want to find $$A \cap B$$ which means finding $u \in [0,1]$, $t \in [0,1]$ such that $$(x_1 + (x_2-x_1)u,y_1 + (y_2-y_1)u)=(x_3 + (x_4-x_3)t,y_3 + (y_4-y_3)t)$$
split into components
$$x_1 + (x_2-x_1)u=x_3 + (x_4-x_3)t$$
$$y_1 + (y_2-y_1)u=y_3 + (y_4-y_3)t$$
solve for $u$ by eliminating $t$
$$\frac{(x_1-x_3) + (x_2-x_1)u}{(x_4-x_3)}=\frac{(y_1-y_3) + (y_2-y_1)u}{(y_4-y_3)}$$
$$\frac{(y_4-y_3)(x_1-x_3) - (x_4-x_3)(y_1-y_3)}{(x_4-x_3)(y_2-y_1) - (y_4-y_3)(x_2-x_1)} = u$$
now you can find the intersection of two lines by calculating this $u$ and checking its between 0 and 1, then calculating t (which is easy once you know u) and checking it's also between 0 and 1.
First, observe that just like @Nicholas said, the equation $\,x^2-y^2+2y=1\,$ defines two lines:
\begin{align} x^2-y^2+2y=1 \iff x^2 = (y-1)^2 \implies \begin{cases} l_1: & y = x + 1 \\ l_2: & y = -x + 1 \end{cases} \end{align}
The slope of the first one is $\,\dfrac{\pi}{4} = 45º,\,$ and for the second one the slope is $\,\dfrac{3\pi}{4} = 135º,\,$ i.e. they are perpendicular.
Equations of bisecting lines can be found by adding and subtracting equations of original lines:
\begin{align} \begin{cases}b_1:&x=0\\b_2:&y=1\end{cases} \end{align}
Points of intersection of these lines with each other and with the line $\,l_0: \ \; y = 3-x\,$ are
\begin{align}
A & := l_0 \cap b_1 &\iff& &&\begin{cases}y = 3-x\\ x = 0\end{cases}
&\implies&& A &= \big(\,0,\,3\,\big) \\
B & := b_1 \cap b_2 &\iff& &&\begin{cases}x = 0\\y=1\end{cases}
&\implies&& B &= \big(\,0,\,1\,\big) \\
C & := l_0 \cap b_1 &\iff& &&\begin{cases}y = 3-x\\ y=1\end{cases}
&\implies&& C &= \big(\,2,\,1\,\big) \\
\end{align}
Observe that $\;b_1\perp b_2\;$ so that $\;\triangle\, ABC\;$ is right triangle.
Therefore the area $\,S\,$ of $\;\triangle\, ABC\;$ is just a half of product of legs $\, AB\,$ and $\,BC\,$:
\begin{align}
S_{\triangle ABC} = \dfrac{1}{2}\,\big\|\left(0,\,1\right) - \left(0,\,3\right)\big\|\, \big\|\left(2,\,1\right) - \left(0,\,1\right)\big\| = 2
\end{align}
Best Answer
There isn't any need to "search" for the coordinates of the vertices, because the coordinates of a vertex are nothing but the solution of the equations of the two lines that intersect to form that vertex. I will therefore solve the three equations to get the vertex coordinates since I have been provided with them.
Suppose the vertices are $A, B$ and $C$, and the equations correspond to lines $AB, BC$ and $AC$. Thus we have three equations for three lines: $$AB: x + 2y - 1 = 0$$ $$BC: 5x + 4y -17 = 0$$ $$AC: x - 4y + 11 = 0$$
By solving the equations for AB & AC, AB & BC, and BC & AC respectively, we get the following coordinates: $A(5, -2)$, $B(1, 3)$, $C(-3, 2)$. Also, by expressing the equations in slope-intercept form, we get $m_{AB} = \frac{-1}2$, $m_{BC} = \frac{-5}4$ and $m_{AC} = \frac{1}4$.
Again suppose, the three heights are $AD, BE$ and $CF$. That means, line $AD$ is perpendicular to $BC$, $BE$ is perpendicular to $AC$ and $CF$ is perpendicular to $AB$. We now recall that the slope of a perpendicular line is the negative inverse of the slope of the base (the proof is fairly easy so I will leave that out). Using this knowledge, we get the following slopes: $m_{AD} = -\frac{1}{m_{BC}} = \frac{4}5$, $m_{BE} = -\frac{1}{m_{AC}} = -4$ and $m_{CF} = -\frac{1}{m_{AB}} = 2$.
Since we have the coordinates of the vertices and the slopes of the perpendiculars from the vertices (which are of course lines passing through the vertices), we can easily construct equations for the three perpendiculars using $m = \frac{y-y_1}{x-x_1}$ which we get from the definition of slope. Therefore, the three required equations for three perpendiculars are: $$AD: 4x - 5y + 30 = 0$$ $$BE: -4x - y + 7 = 0$$ $$CF: 2x - y + 8 = 0$$