Let z_1 = 2 + 3i and z_2 = 3 + 4i. The set S | vedclass

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(A) Let $z = x + iy$. Then $|z - z_1|^2 = (x - 2)^2 + (y - 3)^2$ and $|z - z_2|^2 = (x - 3)^2 + (y - 4)^2$.Given $|z_1 - z_2|^2 = |(2 - 3) + i(3 - 4)|

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straight line with sum of its intercepts on the coordinate axes equals $14$

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(A) Let $z = x + iy$. Then $|z - z_1|^2 = (x - 2)^2 + (y - 3)^2$ and $|z - z_2|^2 = (x - 3)^2 + (y - 4)^2$.Given $|z_1 - z_2|^2 = |(2 - 3) + i(3 - 4)|^2 = |-1 - i|^2 = (-1)^2 + (-1)^2 = 2$.The equation is $(x - 2)^2 + (y - 3)^2 - ((x - 3)^2 + (y - 4)^2) = 2$.Expanding the terms: $(x^2 - 4x + 4 + y^2 - 6y + 9) - (x^2 - 6x + 9 + y^2 - 8y + 16) = 2$.$(x^2 + y^2 - 4x - 6y + 13) - (x^2 + y^2 - 6x - 8y + 25) = 2$.$2x + 2y - 12 = 2$.$2x + 2y = 14 \Rightarrow x + y = 7$.This is a straight line with $x$-intercept $7$ and $y$-intercept $7$.The sum of the intercepts is $7 + 7 = 14$.

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

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