Let z1=2+3i and z2=3+4i . The set S={z∈ℂ:|z−z1|2−|z−z2|2=|z1−z2|2} represents a

|x−2|2+(y−3)2−|x−3|2−(y−4)2=2⇒(x−2)2+(y−3)2−(x−3)2−(y−4)2=1+1⇒−4x+4+9−6y−9+6x−16+8y=2⇒2x+2y=14⇒x+y=7

Let z1=2+3i and z2=3+4i . The set S={z∈ℂ:|z−z1|2−|z−z2|2=|z1−z2|2} represents a

|x−2|2+(y−3)2−|x−3|2−(y−4)2=2⇒(x−2)2+(y−3)2−(x−3)2−(y−4)2=1+1⇒−4x+4+9−6y−9+6x−16+8y=2⇒2x+2y=14⇒x+y=7

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Let $z_{1} = 2 + 3 i$ and $z_{2} = 3 + 4 i$ . The set $S = {z \in ℂ : {| z - z_{1} |}^{2} - {| z - z_{2} |}^{2} = {| z_{1} - z_{2} |}^{2}}$ represents a

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hyperbola with the length of the transverse axis 7

Straight line with the sum of its intercepts on the coordinate axes equals -18

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

hyperbola with eccentricity 2

answer is D.

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Detailed Solution

${| x - 2 |}^{2} + {(y - 3)}^{2} - {| x - 3 |}^{2} - {(y - 4)}^{2} = 2$
$\begin{array}{l} \Rightarrow (x - 2)^{2} + (y - 3)^{2} - (x - 3)^{2} - (y - 4)^{2} = 1 + 1 \\ \Rightarrow - 4 x + 4 + 9 - 6 y - 9 + 6 x - 16 + 8 y = 2 \\ \Rightarrow 2 x + 2 y = 14 \end{array}$
$\Rightarrow x + y = 7$