complex number geometry

Question

Give any two complex numbers lie along the angle bisector of the line
L1 : z= (1+ 3r) + i(1+4r)
L2 : z =(1+3s) + i(1-4s)
find any two points on the angle bisector

Answer 1

Samagra,

 

These lines can be written is a different way:

 

L1: z = (1 + i) + r(3 + 4i)

L2: z = (1 + i) + s(3 − 4i)

 

If you plot these two lines in the xy-plane (using the y-axis for the imaginary axis), you will see that L1 passes through the points (1,1) and (4,5) and the line L2 passes through (1,1) and (4,−3).

 

The angle bisector could be horizontal or vertical. It is not clear what is needed here.

 

If you take the angle bisector to be horizontal, it has equation:

 

ABh: z = (1 + i) + t.

 

This is a horizontal line passing through (1,1). You could take any two points on this line as the answer. For example,

 

(1,1) = 1 + i or (2,1) = 2 + i.

 

If the angle bisector is taken to be vertical, it has equation:

 

ABv: z = (1 + i) + ui = 1 + i(u + 1).

 

This is the vertical line passing through (1,1). Two points on THIS line would be (1,1) = 1 + i and (1,2) = 1 + 2i.

 

Again, I cannot be sure WHICH angle bisector this question demands, so I have included both of them.

 

If I have not answered your question sufficiently, please post a reply and tell me why.