Question

Let $A(0,1),B(1,1),C(1,-1)$ and $D(-1,0)$ be four points. If $P$ is any other point, then the minimum value of $PA+PB+PC+PD$ is equal to

• A. $\sqrt{5}$
• B. $2\sqrt{5}$
• C. $3\sqrt{5}$
• D. $4\sqrt{5}$

Answer 1

The correct option is B $2\sqrt{5}$

We need $PA+PB+PC+PD$ to be minimum
Let's form a triangle using any two coordinates out of $4$ say $A$ and $C$ and third vertex as $P$
In $\Delta PAC$
$PA+PC\ge AC...\left(1\right)$
then taking coordinates $B$ and $D$
In $\Delta PBD$
$PB+PD\ge BD...\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$ we get
$PA+PB+PC+PD\ge AC+BD...\left(3\right)$

min value of $PA+PB+PC+PD=AC+BD$
$AC=\sqrt{(1-0{)}^2+(-1-1{)}^2}=\sqrt{5}$
$BD=\sqrt{(-1-1{)}^2+(0-1{)}^2}=\sqrt{5}$
$AC+BD=2\sqrt{5}$
min value of $PA+PB+PC+PD=2\sqrt{5}$