Question

If $z_1,z_2$ & $z_3$ are the affixes of three points $A,B$ & $C$ respectively and satisfy the condition $|z_1-z_2|=\left|z_1\right|+\left|z_2\right|$ and $|(2-i)z_1+i{z}_3|=\left|z_1\right|+|(1-i)z_1+i{z}_3|$ then prove that $△ABC$ in a right angled.

Answer 1

$|z_1-z_2|=\left|z_1\right|+\left|z_2\right|$

$\Rightarrow$ $z_1,z_2$ and origin will be collinear and $z_1,z_2$ will be opposite side of origin

Similarly $|(2-i)z_1+i{z}_3|=\left|z_1\right|+|(1-i)z_1+i{z}_3|$

$\Rightarrow$ $z_1$ and $(1-i)z_1+i{z}_3=z_4$ say, are collinear with origin and lies on same side of origin.

Let $z_4=\lambda z_1,\lambda$ real

then $(1-i)z_1+i{z}_3=\lambda z_1$

$\Rightarrow$ $i(z_3-z_1)=(\lambda -1)z_1$

$\Rightarrow$ $\frac{(z_3-z_1)}{-z_1}=(\lambda -1)i$

$\Rightarrow$ $\frac{z_3-z_1}{0-z_1}=m{e}^{i\pi /2},m=\lambda -1$

$\Rightarrow$ $z_3-z_1$ is perpendicular to the vector $0-z_1.$

i.e. also $z_2$ is on line joining origin and $z_1$

so we can say the triangle formed by $z_1,z_2$ and $z_3$ is right angled.