Question

If $z_1,z_2,z_3$ are $3$ distinct complex numbers such that $\frac{3}{|z_2-z_3|}=\frac{4}{|z_3-z_1|}=\frac{5}{|z_1-z_2|}$, then the value of $\frac{9}{z_2-z_3}+\frac{16}{z_3-z_1}+\frac{25}{z_1-z_2}$ equals

• A. $0$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $0$

According tothe question:

$\begin{array}{c}Givenrelationno:\\ \frac{3}{|z_2-z_3|}=\frac{4}{|z_3-z_1|}=\frac{5}{|z_1-z_2|}=k|kisconstant\\ then,andsquaringbothside:\\ \Rightarrow 3=k\left(\left|z_2-z_3\right|\right),\Rightarrow 9=k^2(z_2-z_3{)}^2[{\left|z\right|}^2=z\overline{z}\\ \Rightarrow 9=k^2(z_2-z_3)(\overline{z_2}-\overline{z_3})-----\left(1\right)\\ \Rightarrow 4=k\left(\left|z_3-z_1\right|\right)\Rightarrow 16=k^2(z_3-z_1{)}^2\\ \Rightarrow 16=k^2(z_3-z_1)(\overline{z_3}-\overline{z_1})-----\left(2\right)\\ \Rightarrow 5=k\left(\left|z_1-z_2\right|\right)\Rightarrow 25=k^2(z_2-z_3{)}^2\\ \Rightarrow 25=k^2(z_1-z_2)(\overline{z_1}-\overline{z_2})------\left(3\right)\\ \underset{–}{fromthequestion,wefindthevalue}:\\ \frac{9}{z_2-z_3}+\frac{16}{z_3-z_1}+\frac{25}{z_1-z_2}\\ Nowuseingequ\left(1\right),\left(2\right)\&\left(3\right)\\ \Rightarrow \frac{k^2(z_2-z_3)(\overline{z_2}-\overline{z_3})}{\left(z_2-z_3\right)}+\frac{k^2(z_3-z_1)(\overline{z_3}-\overline{z_1})}{\left(z_3-z_1\right)}+\frac{k^2(z_1-z_2)(\overline{z_1}-\overline{z_2})}{\left(z_1-z_2\right)}\\ \Rightarrow k^2(\overline{z_2}-\overline{z_3})+k^2(\overline{z_3}-\overline{z_1})+k^2(\overline{z_1}-\overline{z_2})\\ \Rightarrow k^2\overline{z_2}-\overline{z_3}+\overline{z_3}-\overline{z_1}+\overline{z_1}-\overline{z_2}\\ \Rightarrow k^2\times 0\\ \therefore 0\\ SothatthecorrectoptionisA.\end{array}$