If C is the reflection of A(2, 4) in the x-ax | vedclass

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(C) Given $A \equiv (2, 4)$.Since $C$ is the reflection of $A$ in the $x$-axis, the coordinates of $C$ are $(2, -4)$.Since $B$ is the reflection of $C

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$4 \sqrt{5}$

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(C) Given $A \equiv (2, 4)$.Since $C$ is the reflection of $A$ in the $x$-axis, the coordinates of $C$ are $(2, -4)$.Since $B$ is the reflection of $C$ in the $y$-axis, the coordinates of $B$ are $(-2, -4)$.Now, we calculate the distance $|AB|$ using the distance formula:$|AB| = \sqrt{(2 - (-2))^2 + (4 - (-4))^2}$$|AB| = \sqrt{(2 + 2)^2 + (4 + 4)^2}$$|AB| = \sqrt{4^2 + 8^2}$$|AB| = \sqrt{16 + 64}$$|AB| = \sqrt{80}$$|AB| = \sqrt{16 	imes 5} = 4 \sqrt{5}$

If $C$ is the reflection of $A(2, 4)$ in the $x$-axis and $B$ is the reflection of $C$ in the $y$-axis, then $|AB|$ is

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