Question

If $\frac{a}{|z_1-z_2|}=\frac{b}{|z_2-z_3|}=\frac{c}{|z_3-z_1|}\left(\text{a,}\text{b,}\text{c}\in \text{R}\right),$ $\text{then}\frac{a^2}{z_1-z_2}+\frac{b^2}{z_2-z_3}+\frac{c^2}{z_3-z_1}$ is

• A. Independent of a, b and c only
• B. is Independent of $z_1,z_2,z_3$ only
• C. is Independent of all a, b, c, $z_1,z_2,z_3$
• D. 1

Answer 1

The correct option is A Independent of a, b and c only

Given,

$\frac{a}{|z_1-z_2|}=\frac{b}{|z_2-z_3|}=\frac{c}{|z_3-z_1|}=\frac{1}{k}$ (let). Where $k$ is a non-zero real constant.

or, $|z_1-z_2|=ak$.......(1), $|z_1-z_2|=bk$.......(2) and $|z_3-z_1|=ck$......(3).

Now,

$\frac{a^2}{z_1-z_2}+\frac{b^2}{z_2-z_3}+\frac{c^2}{z_3-z_1}$

$=\frac{a^2\left(\overline{z_1-z_2}\right)}{|z_1-z_2{|}^2}+\frac{b^2\overline{(z_2-z_3)}}{|z_2-z_3{|}^2}+\frac{c^2\overline{(z_3-z_1)}}{|z_3-z_1{|}^2}$

$=\frac{1}{k^2}(\overline{z_1}-\overline{z_2}+\overline{z_2}-\overline{z_3}+\overline{z_3}-\overline{z_1})$ [ Using (1), (2) and (3)]

$=0$.

So the value is independent of $a,b,c$.