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One segment is of length A and lays on the line L_a which is

y = k_ax + c_a

The other segment is of length B and lays on the line L_b which is

y = k_bx + c_b

The red lines are parallel to L_a and L_b respectively, with the distances D_a and D_b.

One point where two red lines intersect is p and it defines two triangles. The area of the triangles are D_a × A/2 and D_b × B/2.

These areas should be the same, so:

D_a × A/2 = D_b × B/2
D_a × A = D_b × B
D_b = A/B × D_a

So D_b is proportionate to D_a and thus all p will be located on the two green "solution" lines, S₁ and S₂.

Note that it is irrelevant where on L_a and L_b the segments A and B are located. The areas of the triangles will be the same.

I have not made the equations for S₁ and S₂ yet, will do that later.

A way to formulate the solution would be like this:

Call the first segment A and the other segment B. See them as vectors, V_a and V_b. Take the intersection between the lines going through A and B and call that point s.

The solution is then the points which lay on the two lines going through s and of which one has the same slope as V_a + V_b and the other as V_a - V_b.