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(D) Let the equation of the variable line be $ax + by + c = 0$, where $a^2 + b^2 = 1$.The perpendicular distance from a point $(x_1, y_1)$ to the line

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$(1,1)$

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(D) Let the equation of the variable line be $ax + by + c = 0$, where $a^2 + b^2 = 1$.The perpendicular distance from a point $(x_1, y_1)$ to the line $ax + by + c = 0$ is given by $d = ax_1 + by_1 + c$.The algebraic sum of the distances from the points $(2,0)$, $(0,2)$, and $(1,1)$ is zero:$(2a + 0b + c) + (0a + 2b + c) + (1a + 1b + c) = 0$$3a + 3b + 3c = 0$$a + b + c = 0$Substituting $c = -(a + b)$ into the line equation:$ax + by - (a + b) = 0$$a(x - 1) + b(y - 1) = 0$This equation holds for all $a$ and $b$ if $x - 1 = 0$ and $y - 1 = 0$.Thus, the line always passes through the fixed point $(1,1)$.

If the algebraic sum of the distances from the points $(2,0)$, $(0,2)$, and $(1,1)$ to a variable straight line is zero, then the line passes through the fixed point:

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