Question

The minimum value of $\left|z-1\right|+\left|z-2\right|+\left|z-3\right|+\left|z-4\right|+\left|z-5\right|$ is

• A. $3$
• B. $5$
• C. $7$
• D. $6$

Answer 1

The correct option is D $6$

Min value of $|z-1|+|z-2|+|z-3|+|z-4|+|z-5|$

Taking $x+iy=z$

We have to find :-

min value of :-

$\sqrt{(x-1{)}^2+y^2}+\sqrt{(x-2{)}^2+y^2}+\sqrt{(x-3{)}^2+y^2}+\sqrt{(x-4{)}^2+y^2}+\sqrt{(x-5{)}^2+y^2}$

As we have to find min value we can safely take $y=0$

as $y^2\ge 0$, and it just increase the value

$\therefore$ Finally, we have to min value of

$|x-1|+|x-2|+|x-3|+|x-4|+|x-5|$

Now taking cases :-

Case $1$ :

$\therefore$ equation becomes,

$1-x+2-x+3-x+4-x+5-x$

$15-x5$ ; (Now min value of this $=10$)

Case $2$; $1<x\le 2$

$\therefore$ equation becomes ; $x-1+2-x+3-x+4-x+5-x$

$=13-3x;$ (Now min value of this $=7$)

Case $3$ ; $2<x\le 3$

$\therefore$ Equation becomes $x-1+x-2+3-x+4-x+5-x$

$=9-x$; (Now min value of this $=6$)

Case $4$ ; $3\le x\le 4$

$\therefore$ Equation becomes

$x-1+x-2+x-3+4-x+5-x$

$=x+3$ ; (Now min value $=6$)

Case $5$ ; $4\le x\le 5$

$\therefore$ Equation becomes,

$x-1+x-2+x-3+x-4+5-x$

$=3x-5$; (Now min value of this $=7$)

Case $6$ ; $x\ge 5$

$\therefore$ Equation becomes,

$x-1+x-2+x-3+x-4+x-5$

$=5x-15$ (Now min value of $=10$ )

Hence min value is $6$