Answer:
[tex](-10,9)[/tex]
Recall that if you have the two points, [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], the midpoint between them is [tex](\frac{x_1 +x_2}{2}, \frac{y_1 +y_2}{2})\\[/tex].
We let [tex]x_c[/tex] be the [tex]x[/tex]-coordinate of point [tex]C[/tex] and [tex]y_c[/tex] for the [tex]y[/tex]-coordinate of point [tex]C[/tex].
The problem tells us that point [tex]B[/tex] is the midpoint of [tex]\overline{AC}[/tex]. This means that the point [tex](-3,5)[/tex] is the same point as [tex](\frac{4 +x_c}{2}, \frac{1 +y_c}{2})\\[/tex].
Solving for [tex]x_c[/tex]:
[tex]\frac{4 +x_c}{2} = -3 \\ \frac{4 +x_c}{2} \cdot 2 = -3 \cdot 2 \\ 4 +x_c = -6 \\ 4 +x_c -4 = -6 -4 \\ x_c = -10[/tex]
Solving for [tex]y_c[/tex]:
[tex]\frac{1 +y_c}{2} = 5 \\ \frac{1 +y_c}{2} \cdot 2 = 5 \cdot 2 \\ 1 +y_c = 10 \\ 1 +y_c -1 = 10 -1 \\ y_c = 9[/tex]
The coordinates of point [tex]C[/tex] is [tex](-10,9)[/tex].