Triangle $ABC$ with vertices $A(-2, 0)$, $B(1, 4)$ and $C(-3, 2)$ is reflected over the $y$-axis to form triangle $A'B'C'$. What is the length of a segment drawn from $C$ to $C'$?
Reflecting a point over the $y$-axis negates the $x$-coordinate.  So if $C$ is $(-3,2)$, $C'$ will be $(3,2)$.  The segment is a horizontal line of length $3+3=\boxed{6}$.
6
Level 1
Geometry
Problem: Triangle $ABC$ with vertices $A(-2, 0)$, $B(1, 4)$ and $C(-3, 2)$ is reflected over the $y$-axis to form triangle $A'B'C'$. What is the length of a segment drawn from $C$ to $C'$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the length of the segment from point C to its reflection C' over the y-axis, we first determine the coordinates of C' after the reflection. The reflection of a point (x, y) over the y-axis results in the point (-x, y). 

Given the coordinates of point C as (-3, 2), we apply the reflection transformation:
C' = (-(-3), 2) = (3, 2).

Next, we calculate the distance between points C and C' using the distance formula, which is given by:
Distance = √[(x2 - x1)² + (y2 - y1)²].

Substituting the coordinates of C (-3, 2) and C' (3, 2) into the formula:
Distance = √[(3 - (-3))² + (2 - 2)²]
= √[(3 + 3)² + (0)²]
= √[6² + 0]
= √[36]
= 6.

Thus, the length of the segment drawn from C to C' is 6.<|END_OF_LATENT|> The final answer is $6$.