Line $\ell_1$ is horizontal and passes through $(2,-8)$. Line $\ell_2$ is vertical and passes through $(-7,5)$. Line $\ell_3$ passes through $(-1,5)$ and $(6,-2)$. Line $\ell_4$ passes through $(13,-5)$ and $(7,-6)$. Find the point of intersection of $\ell_3$ and $\ell_4$.

Answer 1

The point of intersection is the solution to the system of equations \begin{align*}
y&=k \\
y-5&=m(x+1) \\
y+5&=n(x-13).
\end{align*}Eliminating $y$ gives \begin{align*}
m(x+1)-5&=n(x-13)+5 \\
mx+m-10&=nx-13n+5.
\end{align*}To make the $x^2$ terms cancel, we solve for $n$ and get \begin{align*}
n&=\frac{mx+m-10}{x-13} \\
&=\frac{m(x-13)+13m+m-10}{x-13} \\
&=\frac{mx-13m+m-10}{x-13} \\
&=\frac{mx-12m-10}{x-13}.
\end{align*}Substituting into the first equation gives $k=\frac{mx-12m-10}{x-13}$, or \[(mx-12m-10)(x-13)-(x-13)k=0.\]Expanding gives $13k-12mx-25x+13m+100=0$, or \[13k=12mx+25x-13m-100.\]Substituting into the second equation gives \begin{align*}
y-5&=m(x+1) \\
y&=m(x+1)+5 \\
k&=m(k+1)+5.
\end{align*}This implies that $k=5$, so \[13(5)=12mx+25x-13m-100,\]or $12mx+25x-13m-100=65$. Either way, the line $y=\boxed{5}$ intersects the line $12mx+25x-13m-100=0$ at $(x,y)=(-7,5)$.