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(D) Let the equation of the variable line be $ax + by + c = 0$,where $a^2 + b^2 
eq 0$. The perpendicular distance from a point $(x_1, y_1)$ to the l

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$(1,1)$

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(D) Let the equation of the variable line be $ax + by + c = 0$,where $a^2 + b^2 
eq 0$. The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is given by $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$. Since the algebraic sum of the perpendicular distances is zero,we consider the signed distances: $d_1 = \frac{2a + c}{\sqrt{a^2 + b^2}}$,$d_2 = \frac{2b + c}{\sqrt{a^2 + b^2}}$,and $d_3 = \frac{a + b + c}{\sqrt{a^2 + b^2}}$. Given $d_1 + d_2 + d_3 = 0$,we have: $\frac{2a + c + 2b + c + a + b + c}{\sqrt{a^2 + b^2}} = 0$ $3a + 3b + 3c = 0$ $a + b + c = 0$ Substituting $c = -(a + b)$ into the line equation $ax + by + c = 0$: $ax + by - (a + b) = 0$ $a(x - 1) + b(y - 1) = 0$ For this to hold for all $a$ and $b$,we must have $x - 1 = 0$ and $y - 1 = 0$. Thus,the fixed point is $(1, 1)$.

If the algebraic sum of the perpendicular distances from the points $(2,0)$,$(0,2)$,and $(1,1)$ to a variable line is zero,then the variable line always passes through a fixed point. The coordinates of that point are

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