Let $z_1$and $z_2$ be non-zero complex numbers satisfying the equation $z_1{ }^2-2z_1{z}_2+2{z_2}^2=20$ , the geometrical nature of the triangle whose vertices are the origin and the points representing $z_1$  and $z_2$ is

The given equation is $z_1{ }^2-2z_1{z}_2+2z_2{ }^2=0$ 

$\Rightarrow {\left(\frac{z_1}{z_2}\right)}^2-2\left(\frac{z_1}{z_2}\right)+2=0$

$\Rightarrow \frac{z_1}{z_2}=\frac{2\pm \sqrt{4-8}}{2}=\left(1\pm i\right)$ 

  $\Rightarrow z_1=\left(1\pm i\right)z_2$

$\Rightarrow \left|z_1\right|=\sqrt{2}\left|z_2\right|\mathrm{.................}\left(1\right)$

$\text{again},z_1=z_2\pm i{z}_2$

$\Rightarrow z_1-z_2=\pm i{z}_2$             

$\Rightarrow (z_1-z_2)is\perp to{z}_2$ …………..(2)             $$\begin{gathered} [ \sin ce arg \left(z_1-z_2\right) = arg \left(\pm i{z}_2\right) = arg \left(\pm i\right) +arg z_{2 }=\pm \frac{\mathrm{\pi }}{2} +arg z_2 \\ \Rightarrow arg \left(z_1-z_2\right)-arg{z}_2 =\pm \frac{\mathrm{\pi }}{2}] \end{gathered}$$

$\text{and}\left|z_1-z_2\right|=\left|z_2\right|$

 $\Rightarrow \left|z_1-z_2\right|=\left|0-z_2\right|$

$\Rightarrow \text{O},z_1\text{and}{z}_2$  from an isosceles right angled triangle.