### 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 of intersection, I need to find values of t and u that make the x and y coordinates of the two lines equal.
I can set up a system of equations from the x and y components of the parameterizations:

\[1 + 2t = 5 + 4u\]
\[1 - 3t = -9 + 2u\]
I can rearrange the equations to get t and u on the same side:

\[2t - 4u = 4\]
\[-3t - 2u = -10\]
I can multiply the first equation by 3 and the second by 2 to eliminate the t terms:

\[6t - 12u = 12\]
\[-6t - 4u = -20\]
I can add the two equations to get:

\[-16u = -8\]
I can solve for u by dividing by -16:

\[u = \frac{1}{2}\]
I can plug in u = 1/2 into either of the original equations to get t. I'll use the first one:

\[1 + 2t = 5 + 4(\frac{1}{2})\]
I can simplify and solve for t:

\[2t = 7 - 1\]
\[t = 3\]
Now that I have t and u, I can plug them into either of the parameterizations to get the point of intersection.
I'll use the first one:

\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + 3 \begin{pmatrix} 2 \\ -3 \end{pmatrix}\]