### Instruction: A line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\]A second line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -9 \end{pmatrix} + u \begin{pmatrix} 4 \\ 2 \end{pmatrix}.\]Find the point where the lines intersect.### Response: Let's look at the first line. It is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\]So if we plug in $t=0$, we get the point $(1,1)$. If we plug in $t=1$, we get the point $(3,-2)$. So the first line goes through the points $(1,1)$ and $(3,-2)$.
Let's look at the second line. It is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -9 \end{pmatrix} + u \begin{pmatrix} 4 \\ 2 \end{pmatrix}.\]So if we plug in $u=0$, we get the point $(5,-9)$. If we plug in $u=1$, we get the point $(9,-7)$. So the second line goes through the points $(5,-9)$ and $(9,-7)$.
Let's find the equation of the first line. Since it goes through the points $(1,1)$ and $(3,-2)$, its slope is $\frac{-2-1}{3-1} = -3/2$. So its equation is $y-1 = -3/2 (x-1)$, which simplifies to $2y-2=-3x+3$ or $3x+2y=5$.
Now let's find the equation of the second line. Since it goes through the points $(5,-9)$ and $(9,-7)$, its slope is $\frac{-7-(-9)}{9-5} = 1/2$. So its equation is $y+9 = 1/2 (x-5)$, which simplifies to $2y+18=x-5$ or $x-2y=23$.
So the intersection of the two lines is the solution to the system of equations
\begin{align*}
3x+2y&=5,\\
x-2y&=23.\\
\end{align*}
Adding the two equations gives $4x=28$, so $x=7$. Substituting this back into the first equation gives $21+2y=5$, so $y=-8$.