For z C, if the minimum value of (|z-3√2|+|z-p√2,i|) is 5√2, then a value of p is

Question

For $z\in\mathbb C$, if the minimum value of $\big(|z-3\sqrt2|+|z-p\sqrt2\,i|\big)$ is $5\sqrt2$, then a value of $p$ is

Accepted Answer

It is sum of distance of z from $$\left( {3\sqrt 2 ,0} \right)$$ and $$\left( {0,p\sqrt 2 } \right)$$

For minimising, z should lie on AB and $$AB = 5\sqrt 2 $$

${(AB)^2} = 18 + 2{p^2}$

$$p = \, \pm \,4$$

Solutions · 1