### Instruction: A line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} -1 \\ 5 \end{pmatrix}.\]A second line is parameterized by
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \end{pmatrix} + u \begin{pmatrix} -1 \\ 4 \end{pmatrix}.\]Find the point where the lines intersect.### Response: I need to find the values of $t$ and $u$ that make the coordinates of the two points equal.
That means I need to solve the system of equations
\[\begin{pmatrix} 2 \\ 3 \end{pmatrix} + t \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 0 \\ 7 \end{pmatrix} + u \begin{pmatrix} -1 \\ 4 \end{pmatrix}.\]
I can write this as
\[\begin{cases} 2 - t = -u \\ 3 + 5t = 7 + 4u \end{cases}.\]
I can use substitution or elimination to solve this.
I'll try elimination.
If I multiply the first equation by 5, I get
\[\begin{cases} 10 - 5t = -5u \\ 3 + 5t = 7 + 4u \end{cases}.\]
Then I can add the two equations and get
\[13 = 7 - u.\]
So $u = -6$.
Then I can plug this into the first equation and get
\[2 - t = -(-6).\]
So $t = -4$.
Now that I have $t$ and $u$, I can plug them into either parameterization and get the point of intersection.
I'll use the first one.
I get
\[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} + (-4) \begin{pmatrix} -1 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 \\ -17 \end{pmatrix}.\]