If algebraic sum of distances of a variable line from points $(2,0),(0,2)$ and $(-2,-2)$ is zero, then the line passes through the fixed point

The correct option is B

$(1,1)$

Explanation for the correct option:

Find the required point:

Let the variable line be $ax+by+c=0$

Given, the sum of the perpendicular from the points $(2,0),(0,2)$ and $(1,1)$ to $ax+by+c=0$ is zero

$\therefore \left|\frac{2\times a+b\times 0+c}{\sqrt{a^2+b^2}}\right|+\left|\frac{a\times 0+b\times 2+c}{\sqrt{a^2+b^2}}\right|+\left|\frac{a\times 1+b\times 1+c}{\sqrt{a^2+b^2}}\right|=0$

$\Rightarrow$ $\pm \left(\frac{2a+c}{\sqrt{a^2+b^2}}\right)\pm \left(\frac{2b+c}{\sqrt{a^2+b^2}}\right)\pm \left(\frac{a+b+c}{\sqrt{a^2+b^2}}\right)=0$

$\Rightarrow$ $2a+c+2b+c+a+b+c=0$

$\Rightarrow$ $3a+3b+3c=0$

$\Rightarrow$ $a+b+c=0$

This is a linear relation between $a,b$ and $c$.

By Comparing $ax+by+c=0$ and $a+b+c=0$, We get

The coordinates of fixed point are $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{1}\mathbf{)}$.

Hence, Option ‘B’ is Correct.