Reflecting a point over a line

Question

I am working with my son on a math question:

Triangle $DUC$ with coordinates $D(-3,-1), U(-1,8)$, and $C(8,6)$. After reflecting $U$ over $DC$ prove that the resulting shape $DUCU'$ is a square.

In the answer they jump to the fact that $U' = (6-3)$ , I'm not seeing how that is a given? Please explain

Answer 1

Noting that

$$ U C=\sqrt{9^2+2^2}=\sqrt{85}=\sqrt{2^2+9^2}=UD $$ Therefore the mid-point $M$ of $DC$ is the perpendicular foot from $U$ to $CD$ i.e. $UM \perp CD$.

$$M= \left(\frac{-3+8}{2}, \frac{-1+6}{2}\right)=\left(\frac{5}{2}, \frac{5}{2}\right)$$ Let $U’(a,b)$ be the reflectional image of $U$ on $CD$, then $M$ is also the mid-point of $UU’$, hence $$ \left\{\begin{array} { l } { \frac { - 1 + a } { 2 } = \frac { 5 } { 2 } } \\ { \frac { 8 + b } { 2 } = \frac { 5 } { 2 } } \end{array} \Rightarrow \left\{\begin{array}{l} a=6 \\ b=-3 \end{array}\right.\right. $$ Hence $U’=(6,-3)$.

$$ \textrm{ Diagram for reference }$$

Answer 2

Hint: the midpoint of $CD$ is also the midpoint of $UU'$ (why?).