### 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: I want to find the values of t and u that make the two expressions for x and y equal.
That means I have to solve a system of two equations with two unknowns.
The first equation is $x = 1 + 2t = 5 + 4u$, or equivalently, $2t - 4u = 4$.
The second equation is $y = 1 - 3t = -9 + 2u$, or equivalently, $-3t - 2u = -10$.
I can use the elimination method to solve this system.
I can multiply the first equation by 3 and the second equation by 2, and then add them to eliminate t.
That gives me $6t - 12u = 12$ and $-6t - 4u = -20$, so $-16u = -8$, or $u = 1/2$.
Now I can plug in u = 1/2 into either equation to find t. I'll use the first one: $2t - 4(1/2) = 4$, or $t = 3$.
Finally, I can plug in t = 3 and u = 1/2 into either expression for x and y to find the point of intersection.