, 2026

Since $B(2,1)$ and $C(6,1)$ lie along the same horizontal
line, then $BC=6-2=4$, which is the
positive difference between their $x$-coordinates.

Suppose $G$ is the point that lies on
$BC$ vertically below $A(3,4)$.

Then $G$ has the same $x$-coordinate as $A(3,4)$ and the same $y$-coordinate as $B(2,1)$ and $C(6,1)$, and thus has coordinates $(3,1)$. Since $A(3,4)$ and $G(3,1)$ lie along the same vertical line,
then $AG=4-1=3$, the positive
difference between their $y$-coordinates.

The area of $\triangle ABC$ is $\$\dfrac12\times (BC)\times (AG)=\dfrac12\times 4\times$ 3=6$.
The reflection of the point $(x,y)$ in the $y$-axis is the point $(-x,y)$.

Thus, the reflection of $B(2,1)$ in
the $y$-axis is $D(-2,1)$.

Using $G(3,1)$ from part (a), we get
$DC=6-(-2)=8$, $AG=3$, and so the area of $\triangle ADC$ is $\$\dfrac12\times (DC)\times (AG)=\dfrac12\times 8\times$ 3=12$.
Point $A(3,4)$ lies $4-(-2)=6$ units vertically above the line
$y=-2$.

So, the reflection of $A$ in the line
$y=-2$ lies $6$ units vertically below the line $y=-2$, and thus has coordinates $E(3,-2-6)$ or $E(3,-8)$.

We may once again use $G(3,1)$ from
part (a) since $G$ lies on $BC$ and lies on the vertical line through
$E(3,-8)$. Doing so, we get $EG=1-(-8)=9$ and $BC=4$.

Thus, the area of $\triangle EBC$ is
$\$\dfrac12\times (BC)\times (EG)=\dfrac12\times 4\times$ 9=18$.

Using $G(3,1)$ from part (a),
$\triangle FBC$ has base $BC=4$, height $FG$, and area $12$.

Since the area of $\triangle FBC$ is
$\dfrac12\times(BC)\times(FG)=12$,
then $2\times (FG)=12$, and so the
triangle has height $FG=6$.

With $FG=6$ and $F$ vertically above $G(3,1)$, we determine that $F$ has coordinates $(3,1+6)=(3,7)$.
With $FG=6$ and $F$ vertically below $G(3,1)$, we determine that $F$ has coordinates $(3,1-6)=(3,-5)$.

Since $F$ is the image of the
point $A(3,4)$ after it is reflected
in the horizontal line $y=k$, then
the distance from the line to $F$ is
equal to the distance from the line to $A$.

This tells us that $k$ is equal to
the average of the $y$-coordinates of
$F$ and $A$.

The average of the $y$-coordinates of
$F(3,7)$ and $A(3,4)$ is $\dfrac{7+4}{2}=\dfrac{11}{2}$.

The average of the $y$-coordinates of
$F(3,-5)$ and $A(3,4)$ is $\dfrac{-5+4}{2}=-\dfrac{1}{2}$.

The two values of $k$ for which the
area of $\triangle FBC$ is $12$ are
$k=\dfrac{11}{2}$ and $k=-\dfrac12$.