Consider the following two statements : Statement I : For any two non-zero complex numbers $z_1, z_2$,…
Statement II : If $x, y, z$ are three distinct complex numbers and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are three positive real numbers such that $\frac{\mathrm{a}}{|y-z|}=\frac{\mathrm{b}}{|z-x|}=\frac{\mathrm{c}}{|x-y|}$, then $\frac{\mathrm{a}^2}{y-z}+\frac{\mathrm{b}^2}{z-x}+\frac{\mathrm{c}^2}{x-y}=1 .$
Between the above two statements,
- Statement I is correct but Statement II is incorrect.
- both Statement I and Statement II are correct.
- both Statement I and Statement II are incorrect.
- Statement I is incorrect but Statement II is correct.
Solution
Since $\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq\left|\frac{z_1}{\left|z_1\right|}\right|+\left|\frac{z_2}{\left|z_2\right|}\right|$ $\begin{aligned} & \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq \frac{\left|z_1\right|}{\left|z_1\right|}+\frac{\left|z_2\right|}{\left|z_2\right|} \\ & \left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right| \leq 2 \end{aligned}$ $\left(\left|z_1\right|+\left|z_2\right|\right)\left(\left|\frac{z_1}{\left|z_1\right|}+\frac{z_2}{\left|z_2\right|}\right|\right) \leq 2\left(\left|z_1\right|+\left|z_2\right|\right)$ $\therefore$ statement $\mathrm{I}$ is correct For Statement II : $\begin{aligned} & \frac{a}{|y-z|}=\frac{b}{|z-x|}=\frac{c}{|x-y|} \\ & \frac{a^2}{|y-z|^2}=\frac{b^2}{|z-x|^2}=\frac{c^2}{|x-y|^2}=\lambda \\ & a^2=\lambda\left(|y-z|^2\right)=\lambda(y-z)(\bar{y}-\bar{z}) \\ & b^2=\lambda(z-x)(\bar{z}-\bar{x}) \text { and } c^2=\lambda(x-y)(\bar{x}-\bar{y}) \\ & \frac{a^2}{y-z}+\frac{b^2}{z-x}+\frac{c^2}{x-y}=\lambda(\bar{y}-\bar{z}+\bar{z}-\bar{x}+\bar{x}-\bar{y})=0 \end{aligned}$ Statement II is false
Asked in: JEE Main 2024 (05 Apr Shift 1)