What is represented by the set of complex numbers z satisfying the equation $(3+7i)z+(10-2i)\bar{z}+100=0$? I have an objective type question :- The set complex numbers $z$ satisfying the equation:-
$$(3+7i)z+(10-2i)\bar{z}+100=0$$
represents:- 
A) A straight line
B) A pair of intersecting straight lines
C) A point 
D) A pair of distinct parallel straight lines
My approach:-
I have put $z=x+iy$ and solved equation to get :-
$$13x-9y+i(5x-7y)+100=0$$
i.e.
$$y=\frac{13}{9}x+\frac{100}{9}\tag{1.}$$&
$$y=\frac{5}{7}x\tag{2.}$$
So eq. $1$ & $ 2$ are two straight lines which are intersecting each other.So my answer is B)
Now i want to ask that ,is my approach and answer correct? if yes ,then is there any simpler approach? and if no then how would i get the answer?

Once you get to the step where you have the equations of the two lines, there is still more work to do.  Or there would be if you had to actually find the point of intersection.  But the answer is C, not B.  Everything you've done up to that point is correct and I'm not sure if there's a better way to do it.
So why is the answer C and not B?  You started with the given equation:
$$(3+7i)z + (10-2i)\overline{z} + 100 = 0$$
Then you let $z = x+iy$:
$$(3+7i)(x+iy) + (10-2i)(x-iy) + 100 = 0$$
Then you did some algebra and ended up with:
$$13x - 9y + i(5x-7y) + 100 = 0 \tag{*}$$
Then you equated real and imaginary parts to obtain two equations, and the key here is that both equations must be true at the same time in order for equation (*) above to hold.  And this only happens at the point of intersection.
