a and b can't be real numbers, if they were we would have
a(x+ iy) + b(x - iy) = 38, (where z = x + iy),
and on equating reals and imaginaries,
(a + b)x = 38 and (a - b)y = 0, so a = b.
That doesn't make much sense in the context of the question.
So, let
\(\displaystyle a = a_{1} + a_{2} i \quad\text{and }\quad b = b_{1}+b_{2}i \)
then, after substituting and equating reals and imaginaries,
\(\displaystyle a_{1}x-a_{2}y+b_{1}x+b_{2}y = 38,\\ \text{and} \\ a_{1}y+a_{2}x-b_{1}y+b_{2}x=0. \)
Now substituting the two points (-5, 4) and (7, 2), yields 4 equations in the 4 unknowns.
which produce the solutions
\(\displaystyle a = 1-6i \quad \text{and}\quad b = 1+6i.\)
\(\displaystyle (1-6i)z+(1+6i)\overline{z}=38\)
is satisfied by both z = -5 + 4i and z = 7 + 2i,
and the general equation
\(\displaystyle (1-6i)(x+iy)+(1+6i)(x-iy)=38 \\ \text{simplifies to} \\ x+6y=19, \\ \text{which is the equation of the line between the two points.}\)
.
...correctly (mostly).
