$\begin {array} {1 1} (A)\;(0,0) & \quad (B)\;(1,1) \\ (C)\;(-1, -1) & \quad (D)\;\text{None of the above} \end {array}$

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- Length of the perpendicular from $(x_1, y_1)$ to the line $ax+by+c=0$ is $ p = \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$

Step 1 :

Let the variable line be $ ax+by=1$

It is given that the algebraic sum of the perpendicular from the points $(2,0), (0,2 ), (1,1)$ to this line is zero.

$ \therefore \large\frac{(2)a+(0)b-1}{\sqrt{a^2+b^2}}$$+ \large\frac{(0)(a)+(2)b-1}{\sqrt{a^2+b^2}}$$+ \large\frac{a+b-1}{\sqrt{a^2+b^2}}$$=0$

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

$ \Rightarrow a+b-1=0$

$\Rightarrow a +b=1$

This is the linear relation between the parameter $a$ and $b$.

So the equation $ax+by=1$ represents a family of straight lines passing through a fixed point.

Comparing $ax+by=1$ and $a+b=1$

We obtain that the coordinates of the fixed point are (1,1)

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