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 $\ell_4$.
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 $(x,y)=(-7,5)$.
To find the point of intersection of two lines, we need to find the coordinates of a point that lies on both lines.
The equation of a line can be written in slope-intercept form as $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept. In order to find the equation of each line, we need to find the slope and the y-intercept.
Line $\ell_1$ is horizontal, which means the slope is 0. Since the line passes through $(2,-8)$, the equation of the line is $y = 0x - 8$, or simply $y = -8$.
Line $\ell_2$ is vertical, which means the slope is undefined. The equation of the line is $x = -7$.
To find the equation of Line $\ell_3$, we need to find the slope. The formula for the slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is:
\[m = \frac{y_2 - y_1}{x_2 - x_1}\]
Using the coordinates $(-1,5)$ and $(6,-2)$, we substitute the values into the formula:
\[m = \frac{-2 - 5}{6 - (-1)} = \frac{-7}{7} = -1\]
The equation of Line $\ell_3$ is $y = -x + b$. To find the y-intercept $b$, we substitute the coordinates $(-1,5)$ into the equation:
\[5 = -(-1) + b \Rightarrow 5 = 1 + b \Rightarrow b = 4\]
Therefore, the equation of Line $\ell_3$ is $y = -x + 4$.
To find the equation of Line $\ell_4$, we first need to find the slope. Using the formula for slope with the coordinates $(13,-5)$ and $(7,-6)$:
\[m = \frac{-6 - (-5)}{7 - 13} = \frac{-1}{-6} = \frac{1}{6}\]
Substituting the coordinates $(13,-5)$ into the equation $y = mx + b$, we have:
\[-5 = \frac{1}{6}(13) + b \Rightarrow -5 = \frac{13}{6} + b \Rightarrow b = -\frac{5}{6} - \frac{13}{6} = -\frac{18}{6} = -3\]
Therefore, the equation of Line $\ell_4$ is $y = \frac{1}{6}x - 3$.
To find the point of intersection of $\ell_3$ and $\ell_4$, we need to solve the system of equations formed by the two line equations:
\[\begin{aligned}
y &= -x + 4 \\
y &= \frac{1}{6}x - 3
\end{aligned}\]
Setting the equations equal to each other, we have:
\[-x + 4 = \frac{1}{6}x - 3\]
Adding $x$ to both sides and adding 3 to both sides:
\[4 + 3 = \frac{1}{6}x + x\]
\[7 = \frac{1}{6}x + \frac{6}{6}x\]
Combining like terms:
\[7 = \frac{7}{6}x\]
Dividing both sides by $\frac{7}{6}$:
\[x = 6\]
Substituting $x=6$ into either equation to find $y$:
\[y = -6 +4 = -2\]
Therefore, the point of intersection of $\ell_3$ and $\ell_4$ is $(6,-2)$.
To find the point of intersection of two lines, you need to find the coordinates $(x, y)$ that satisfy both equations simultaneously. Let's start with lines $\ell_3$ and $\ell_4$.
Line $\ell_3$ passes through $(-1, 5)$ and $(6, -2)$. We can find the slope of this line using the formula:
\[m = \frac{{y_2 - y_1}}{{x_2 - x_1}}\]
where $(x_1, y_1)$ and $(x_2, y_2)$ are the coordinates of the two given points. So,
\[m_{\ell_3} = \frac{{-2 - 5}}{{6 - (-1)}} = \frac{{-7}}{{7}} = -1\]
Now we can find the equation of line $\ell_3$ in point-slope form using the slope we just found and one of the given points. Let's use $(-1, 5)$:
\[y - 5 = -1(x - (-1))\]
\[y - 5 = -1(x + 1)\]
\[y - 5 = -x - 1\]
\[y = -x + 4 \quad \quad (1)\]
Line $\ell_4$ passes through $(13, -5)$ and $(7, -6)$. We can find the slope of this line using the same formula:
\[m_{\ell_4} = \frac{{-6 - (-5)}}{{7 - 13}} = \frac{{-1}}{{-6}} = \frac{1}{6}\]
Now we can find the equation of line $\ell_4$ in point-slope form using the slope we just found and one of the given points. Let's use $(7, -6)$:
\[y - (-6) = \frac{1}{6}(x - 7)\]
\[y + 6 = \frac{1}{6}(x - 7)\]
\[y + 6 = \frac{1}{6}x - \frac{7}{6}\]
\[y = \frac{1}{6}x - \frac{19}{6} \quad \quad (2)\]
Now we have two equations for the lines $\ell_3$ and $\ell_4$. To find their point of intersection, we can solve the system of equations formed by equating the two equations:
\[-x + 4 = \frac{1}{6}x - \frac{19}{6}\]
Let's solve this equation for $x$:
\[6(-x + 4) = 6\left(\frac{1}{6}x - \frac{19}{6}\right)\]
\[-6x + 24 = x - 19\]
\[24 + 19 = x + 6x\]
\[43 = 7x\]
\[x = \frac{43}{7}\]
Substituting this value of $x$ back into Equation (1), we can find $y$:
\[y = -\frac{43}{7} + 4\]
\[y = -\frac{43}{7} + \frac{28}{7}\]
\[y = -\frac{43}{7} + \frac{28}{7}\]
\[y = -\frac{15}{7}\]
So the point of intersection of lines $\ell_3$ and $\ell_4$ is $\left(\frac{43}{7}, -\frac{15}{7}\right)$.