Let $z = x +iy$
$z' = x - iy$
$(3+7i)z + (10-2i)z' + 100 = 0$
$(3+7i)(x + iy) + (10-2i)(x-iy) + 100 = 0$
on solving we get,
$13x-9y + i(5x-7y) + 100 = 0$
$(13x-9y+100 ) + i(5x-7y) = 0$
the real part and imaginary part must be equal on both side
$13x-9y+100 = 0$
$5x-7y = 0 $
on solving we get value of $x$ and $y$
So, on putting value of $x$ and $y$ , we get $z$
$z$ is a $point$
