### Instruction: The equation of the line joining the complex numbers $-2 + 3i$ and $1 + i$ can be expressed in the form
\[az + b \overline{z} = 10\]for some complex numbers $a$ and $b$.  Find the product $ab$.### Response: I need to find two complex numbers $a$ and $b$ that satisfy the given equation for both $z = -2 + 3i$ and $z = 1 + i$.
Let me substitute $z = -2 + 3i$ into the equation and see what I get.
I get $a(-2 + 3i) + b(-2 - 3i) = 10$.
Expanding the brackets, I get $-2a + 3ai - 2b - 3bi = 10$.
I can separate the real and imaginary parts and write this as $(-2a - 2b) + (3a - 3b)i = 10 + 0i$.
For this to be true, I need the real parts and the imaginary parts to match up, so I get a system of two equations: $-2a - 2b = 10$ and $3a - 3b = 0$.
I can try to solve this system by elimination.
If I multiply the first equation by $3$ and the second one by $2$, I get $-6a - 6b = 30$ and $6a - 6b = 0$.
Adding these equations, I get $-12b = 30$, so $b = -\frac{5}{2}$.
Plugging this into the first equation, I get $-2a + 5 = 10$, so $a = -\frac{5}{2}$ as well.
Therefore, I have found $a$ and $b$, and their product is $ab = -\frac{5}{2} \cdot -\frac{5}{2} = \frac{25}{4}$.