, 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\$.$