Let z_1=2+3i and z_2=3+4i.

Question

Let $z_1=2+3i$ and $z_2=3+4i$. The set $S=\{z\in\mathbb C: |z-z_1|^2-|z-z_2|^2=|z_1-z_2|^2\}$ represents a

Accepted Answer

With $z=x+iy$: $|z_1-z_2|^2=|{-1-i}|^2=2$.

$$\big[(x-2)^2+(y-3)^2\big]-\big[(x-3)^2+(y-4)^2\big]=2.$$

Expanding: $(-4x+4-6y+9)-(-6x+9-8y+16)=2 \implies 2x+2y-12=2 \implies x+y=7.$

This is a straight line with $x$-intercept 7 and $y$-intercept 7, so the sum of intercepts is $14$.

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