If -z2-iz1-z1-z2- and -z1-3 and -z2-4- then the area of - ABC- if affixes of A- B- and C are z1- z2- and -z2-iz1-1-i- respectively- is

The correct option is C 254 |z_1+iz_1|=|z_1|+|z_2| ⇒ iz_1, 0+i0 and z_2 are collinear ⇒ arg(iz_1)=arg (z_2) ⇒ arg (z_2) arg(z_1) ...

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Byju's Answer

Standard XII

Mathematics

Distance Formula

If |z2+iz1|...

Question

If $| z_{2} + i z_{1} | = | z_{1} | + | z_{2} |$ and $| z_{1} | = 3$ and $| z_{2} | = 4$ , then the area of $Δ A B C$ , if affixes of A, B, and C are $z_{1}, z_{2},$ and $[(z_{2} - i z_{1}) / (1 - i)]$ respectively, is

\frac{5}{2}

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\frac{25}{2}

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\frac{25}{4}

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Solution

The correct option is C $\frac{25}{4}$
$| z_{1} + i z_{1} | = | z_{1} | + | z_{2} |$
$\Rightarrow i z_{1}, 0 + i 0$ and $z_{2}$ are collinear
$\Rightarrow a r g (i z_{1}) = a r g (z_{2})$
$\Rightarrow a r g (z_{2}) - a r g (z_{1}) = \frac{π}{2}$
Let $z_{3} = \frac{z_{2} - i z_{1}}{1 - i}$
or $(1 - i) z_{3} = z_{2} - i z_{1}$
or $z_{2} - z_{3} = i (z_{1} - z_{3})$
$∴ ∠ A C B = \frac{π}{2}$
and $| z_{1} - z_{3} | = | z_{2} - z_{3} |$
$\Rightarrow A C = B C$
$∵ A B^{2} = A C^{2} + B C^{2}$
$\Rightarrow A C = \frac{5}{\sqrt{2}} (∵ A B = 5)$
Therefore, area of $Δ A B C$ is $(1 / 2) A C \times B C = A C^{2} / 2 = 25 / 4 s q . u n i t s$

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If |z2+iz1|=|z1|+|z2| and |z1|=3 and |z2|=4, then the area of ΔABC, if affixes of A, B, and C are z1,z2, and [(z2−iz1)/(1−i)] respectively, is

If $| z_{2} + i z_{1} | = | z_{1} | + | z_{2} |$ and $| z_{1} | = 3$ and $| z_{2} | = 4$ , then the area of $Δ A B C$ , if affixes of A, B, and C are $z_{1}, z_{2},$ and $[(z_{2} - i z_{1}) / (1 - i)]$ respectively, is