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Let $A(1,1)$ be a point. $B$ is the image of $A$ with respect to the line $x+2 y+2=0$. If $C$ is the foot of the perpendicular from $B$ on the line $3 x+4 y-10=0$, then $A C$ is equal to
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$\begin{aligned} & \text { For } B, \frac{x-1}{1}=\frac{y-1}{2}=\frac{-2(5)}{5} \\ & \Rightarrow x-1=-2 \quad \Rightarrow y-1=-4 \\ & \Rightarrow \quad x=-1 \quad \Rightarrow \quad y=-3 \\ & \therefore \quad B \equiv(-1,-3)\end{aligned}$
$\begin{aligned} & \text { For } C, \frac{x+1}{3}=\frac{y+3}{4}=\frac{-(-25)}{25} \\ & \Rightarrow \quad \frac{x+1}{3}=1, \frac{y+3}{4}=1 \\ & \Rightarrow \quad x=2, y=1 \\ & \therefore \quad C \quad \equiv(2,1) \\ & \text { Hence, } A C=\sqrt{(1-2)^2+(1-1)^2}=1\end{aligned}$
$\begin{aligned} & \text { For } C, \frac{x+1}{3}=\frac{y+3}{4}=\frac{-(-25)}{25} \\ & \Rightarrow \quad \frac{x+1}{3}=1, \frac{y+3}{4}=1 \\ & \Rightarrow \quad x=2, y=1 \\ & \therefore \quad C \quad \equiv(2,1) \\ & \text { Hence, } A C=\sqrt{(1-2)^2+(1-1)^2}=1\end{aligned}$
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