[UPLIFTJEE]

Consider the following two statements :
Statement I : For any two non-zero complex numbers $\mathrm{z}_{1}, \mathrm{z}_{2},\left(\left|\mathrm{z}_{1}\right|+\left|\mathrm{z}_{2}\right|\right)\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}+\frac{\mathrm{z}_{2}}{\left|\mathrm{z}_{2}\right|}\right| \leq 2\left(\left|\mathrm{z}_{1}\right|+\left|\mathrm{z}_{2}\right|\right)$, and
Statement II : If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are three distinct complex numbers and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are three positive real numbers such that $\frac{\mathrm{a}}{|\mathrm{y}-\mathrm{z}|}=\frac{\mathrm{b}}{|\mathrm{z}-\mathrm{x}|}$ $=\frac{\mathrm{c}}{|\mathrm{x}-\mathrm{y}|}$, then $\frac{\mathrm{a}^{2}}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^{2}}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^{2}}{\mathrm{x}-\mathrm{y}}=1$. Between the above two statements, [JEE(Main)5 April 2024 SI]
(A) Statement I is correct but Statement II is incorrect
(B) Both Statement I and Statement II are correct
(C) Statement I is incorrect but Statement II is correct
(D) Both Statement I and Statement II are incorrect

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[UPLIFTJEE]

Solution

Since $\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}+\frac{\mathrm{z}_{2}}{\left|\mathrm{z}_{2}\right|}\right| \leq\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}\right|+\left|\frac{\mathrm{z}_{2}}{\left|\mathrm{z}_{2}\right|}\right|=\frac{\left|\mathrm{z}_{1}\right|}{\left|\mathrm{z}_{1}\right|}+\frac{\left|\mathrm{z}_{2}\right|}{\left|\mathrm{z}_{2}\right|} \quad \Rightarrow\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}+\frac{\mathrm{z}_{2} \mid}{\left|\mathrm{z}_{2}\right|}\right| \leq 2$
$\left(\left|\mathrm{z}_{1}\right|+\left|\mathrm{z}_{2}\right|\right)\left(\left|\frac{\mathrm{z}_{1}}{\left|\mathrm{z}_{1}\right|}+\frac{\mathrm{z}_{2}}{\left|\mathrm{z}_{2}\right|}\right|\right) \leq 2\left(\left|\mathrm{z}_{1}\right|+\left|\mathrm{z}_{2}\right|\right) \therefore$ statement I is correct
For Statement II : $\because \frac{\mathrm{a}}{|\mathrm{y}-\mathrm{z}|}=\frac{\mathrm{b}}{|\mathrm{z}-\mathrm{x}|}=\frac{\mathrm{c}}{|\mathrm{x}-\mathrm{y}|} \Rightarrow \frac{\mathrm{a}^{2}}{|\mathrm{y}-\mathrm{z}|^{2}}=\frac{\mathrm{b}^{2}}{|\mathrm{z}-\mathrm{x}|^{2}}=\frac{\mathrm{c}^{2}}{|\mathrm{x}-\mathrm{y}|^{2}}=\lambda$
$\mathrm{a}^{2}=\lambda\left(|\mathrm{y}-\mathrm{z}|^{2}\right)=\lambda(\mathrm{y}-\mathrm{z})(\overline{\mathrm{y}}-\overline{\mathrm{z}}) \Rightarrow \mathrm{b}^{2}=\lambda(\mathrm{z}-\mathrm{x})(\overline{\mathrm{z}}-\overline{\mathrm{x}})$ and $\mathrm{c}^{2}=\lambda(\mathrm{x}-\mathrm{y})(\overline{\mathrm{x}}-\overline{\mathrm{y}})$
Now $\frac{\mathrm{a}^{2}}{\mathrm{y}-\mathrm{z}}+\frac{\mathrm{b}^{2}}{\mathrm{z}-\mathrm{x}}+\frac{\mathrm{c}^{2}}{\mathrm{x}-\mathrm{y}}=\lambda(\overline{\mathrm{y}}-\overline{\mathrm{z}}+\overline{\mathrm{z}}-\overline{\mathrm{x}}+\overline{\mathrm{x}}-\overline{\mathrm{y}})=0$ Statement II is false

[UPLIFTJEE]

Answer

A

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