### 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: To find the point where the lines intersect, I need to find the values of t and u that make the x and y coordinates of the points equal.
That is, I need to solve the system of equations
\[\begin{align*}
1 + 2t &= 5 + 4u \\
1 - 3t &= -9 + 2u
\end{align*}\]
I can eliminate u by multiplying the first equation by 2 and the second by 4, and then subtracting them.
\[\begin{align*}
2(1 + 2t) &= 2(5 + 4u) \\
4(1 - 3t) &= 4(-9 + 2u) \\
\\
2 + 4t &= 10 + 8u \\
4 - 12t &= -36 + 8u \\
\\
-2 - 8t &= -26 \\
8t &= 24 \\
t &= 3
\end{align*}\]
Now that I have t, I can plug it into either equation to find u. I'll use the first one.
\[\begin{align*}
1 + 2(3) &= 5 + 4u \\
7 &= 5 + 4u \\
4u &= 2 \\
u &= \frac{1}{2}
\end{align*}\]
Finally, I can use either t or u to find the x and y coordinates of the intersection point.
I'll use t and the first parameterization.