Question

If in Argand plane points $z_1,z_2,z_3$ are the vertices of an isosceles triangle right angled at $z_2$, then

• A. ${z_1}^2+2{z_2}^2+{z_3}^2=2z_2(z_1+z_3)$
• B. ${z_1}^2+{z_2}^2+{z_3}^2=2z_2(z_1+z_2)$
• C. ${z_1}^2+{z_2}^2+2{z_3}^2=2z_2(z_1+z_2)$
• D. $2{z_1}^2+2{z_2}^2+{z_3}^2=2z_2(z_1+z_2)$

Answer 1

The correct option is A ${z_1}^2+2{z_2}^2+{z_3}^2=2z_2(z_1+z_3)$

Let $BA=BC\Rightarrow {|z_1-z_2|}^2={|z_3-z_2|}^2$

$\Rightarrow (z_1-z_2)(\overline{z_1}-\overline{z_2})=(z_3-z_2)(\overline{z_3}-\overline{z_2})$ ....(1)

Again, $\therefore \angle ABC={90}^0\Rightarrow arg\left(\frac{BA}{BC}\right)={90}^0$

$\Rightarrow arg\left(\frac{z_1-z_2}{z_3-z_2}\right)={90}^0\Rightarrow$ real part of $\frac{z_1-z_2}{z_3-z_2}=0$

$\Rightarrow \frac{1}{2}[\frac{z_1-z_2}{z_3-z_2}+\frac{\overline{z_1}-\overline{z_2}}{\overline{z_3}-\overline{z_2}}]=0$

$\Rightarrow \frac{z_1-z_2}{z_3-z_2}=-\frac{\overline{z_1}-\overline{z_2}}{\overline{z_3}-\overline{z_2}}\Rightarrow \frac{z_1-z_2}{\overline{z_1}-\overline{z_2}}=\frac{z_2-z_3}{\overline{z_3}-\overline{z_2}}$ ...(2)

Multiplying (1) by (2), we get

${(z_1-z_2)}^2=-{(z_2-z_3)}^2$

$\Rightarrow {z_1}^2+{z_2}^2-2z_1{z}_2=-({z_2}^2+{z_3}^2-2z_2{z}_3)$

$\Rightarrow {z_1}^2+{2z_2}^2+{z_3}^2=2z_2(z_1+z_3)$