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(D) Step $1$: Find the image $B$ of point $A(1, 1)$ with respect to the line $x + 2y + 2 = 0$. Using the formula $\frac{x - x_1}{a} = \frac{y - y_1}{b

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(D) Step $1$: Find the image $B$ of point $A(1, 1)$ with respect to the line $x + 2y + 2 = 0$. Using the formula $\frac{x - x_1}{a} = \frac{y - y_1}{b} = -2 \frac{ax_1 + by_1 + c}{a^2 + b^2}$: $\frac{x - 1}{1} = \frac{y - 1}{2} = -2 \frac{1(1) + 2(1) + 2}{1^2 + 2^2} = -2 \frac{5}{5} = -2$. $x - 1 = -2 \Rightarrow x = -1$. $y - 1 = -4 \Rightarrow y = -3$. So,$B = (-1, -3)$. Step $2$: Find the foot of the perpendicular $C$ from $B(-1, -3)$ to the line $3x + 4y - 10 = 0$. Using the formula $\frac{x - x_1}{a} = \frac{y - y_1}{b} = - \frac{ax_1 + by_1 + c}{a^2 + b^2}$: $\frac{x - (-1)}{3} = \frac{y - (-3)}{4} = - \frac{3(-1) + 4(-3) - 10}{3^2 + 4^2} = - \frac{-3 - 12 - 10}{25} = - \frac{-25}{25} = 1$. $x + 1 = 3 \Rightarrow x = 2$. $y + 3 = 4 \Rightarrow y = 1$. So,$C = (2, 1)$. Step $3$: Calculate the distance $AC$ between $A(1, 1)$ and $C(2, 1)$. $AC = \sqrt{(2 - 1)^2 + (1 - 1)^2} = \sqrt{1^2 + 0^2} = 1$.

Let $A(1, 1)$ be a point. $B$ is the image of $A$ with respect to the line $x + 2y + 2 = 0$. If $C$ is the foot of the perpendicular from $B$ on the line $3x + 4y - 10 = 0$,then $AC$ is equal to

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