Complex Numbers 4 Question 15
15. Complex numbers $z _1, z _2, z _3$ are the vertices $A, B, C$ respectively of an isosceles right angled triangle with right angle at $C$. Show that
$\left(z _1-z _2\right)^{2}=2\left(z _1-z _3\right)\left(z _3-z _2\right)$.
(1986, $2 \frac{1}{2}$ M)
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Solution:
- Since, triangle is a right angled isosceles triangle.
$\therefore \quad$ Rotating $z _2$ about $z _3$ in anti-clockwise direction through an angle of $\pi / 2$, we get
where, $\left|z _2-z _3\right|=\left|z _1-z _3\right|$
$$ \Rightarrow \quad\left(z _2-z _3\right)=i\left(z _1-z _3\right) $$
On squaring both sides, we get
$$ \begin{aligned} & \left(z _2-z _3\right)^{2}=-\left(z _1-z _3\right)^{2} \\ & \Rightarrow \quad z _2^{2}+z _3^{2}-2 z _2 z _3=-z _1^{2}-z _3^{2}+2 z _1 z _3 \\ & \Rightarrow \quad z _1^{2}+z _2^{2}-2 z _1 z _2=2 z _1 z _3+2 z _2 z _3-2 z _3^{2}-2 z _1 z _2 \end{aligned} $$
$$ \begin{array}{ll} \Rightarrow & \left(z _1-z _2\right)^{2}=2{\left(z _1 z _3-z _3^{2}\right)+\left(z _2 z _3-z _1 z _2\right) } \\ \Rightarrow & \left(z _1-z _2\right)^{2}=2\left(z _1-z _3\right)\left(z _3-z _2\right) \end{array} $$
