Question: Plot and sketch the figure of the given points.
Determine the coordinates of the image given the points $A(-3, -4)$, $B(-4, -1)$, $C(-3, 2)$, $D(1, -2)$ which is reflected in the mirror placed along the straight line $3x+2y=8$. Solve algebraically then verify your answer using Alternative Solution 2.
1. **State the problem:** We need to find the coordinates of the points $A(-3,-4)$, $B(-4,-1)$, $C(-3,2)$, and $D(1,-2)$ after reflection about the line $3x + 2y = 8$.
2. **Formula for reflection about line $ax + by + c = 0$:**
$$
(x', y') = \left( \frac{(b^2 - a^2)x - 2ab y - 2ac}{a^2 + b^2}, \frac{(a^2 - b^2)y - 2ab x - 2bc}{a^2 + b^2} \right)
$$
where $a=3$, $b=2$, and $c=-8$ (rewriting $3x + 2y = 8$ as $3x + 2y - 8 = 0$).
3. **Calculate denominator:**
$$a^2 + b^2 = 3^2 + 2^2 = 9 + 4 = 13$$
4. **Reflect point $A(-3,-4)$:**
$$
\begin{aligned}
x' &= \frac{(2^2 - 3^2)(-3) - 2 \cdot 3 \cdot 2 (-4) - 2 \cdot 3 \cdot (-8)}{13} = \frac{(4 - 9)(-3) + 48 + 48}{13} = \frac{-5(-3) + 96}{13} = \frac{15 + 96}{13} = \frac{111}{13} \\
y' &= \frac{(3^2 - 2^2)(-4) - 2 \cdot 3 \cdot 2 (-3) - 2 \cdot 2 \cdot (-8)}{13} = \frac{(9 - 4)(-4) + 36 + 32}{13} = \frac{5(-4) + 68}{13} = \frac{-20 + 68}{13} = \frac{48}{13}
\end{aligned}
$$
So, $A' = \left( \frac{111}{13}, \frac{48}{13} \right)$.
5. **Reflect point $B(-4,-1)$:**
$$
\begin{aligned}
x' &= \frac{(4 - 9)(-4) - 2 \cdot 3 \cdot 2 (-1) - 2 \cdot 3 \cdot (-8)}{13} = \frac{-5(-4) + 12 + 48}{13} = \frac{20 + 60}{13} = \frac{80}{13} \\
y' &= \frac{(9 - 4)(-1) - 2 \cdot 3 \cdot 2 (-4) - 2 \cdot 2 \cdot (-8)}{13} = \frac{5(-1) + 48 + 32}{13} = \frac{-5 + 80}{13} = \frac{75}{13}
\end{aligned}
$$
So, $B' = \left( \frac{80}{13}, \frac{75}{13} \right)$.
6. **Reflect point $C(-3,2)$:**
$$
\begin{aligned}
x' &= \frac{(4 - 9)(-3) - 2 \cdot 3 \cdot 2 (2) - 2 \cdot 3 \cdot (-8)}{13} = \frac{-5(-3) - 24 + 48}{13} = \frac{15 + 24}{13} = \frac{39}{13} = 3 \\
y' &= \frac{(9 - 4)(2) - 2 \cdot 3 \cdot 2 (-3) - 2 \cdot 2 \cdot (-8)}{13} = \frac{5(2) + 36 + 32}{13} = \frac{10 + 68}{13} = \frac{78}{13} = 6
\end{aligned}
$$
So, $C' = (3, 6)$.
7. **Reflect point $D(1,-2)$:**
$$
\begin{aligned}
x' &= \frac{(4 - 9)(1) - 2 \cdot 3 \cdot 2 (-2) - 2 \cdot 3 \cdot (-8)}{13} = \frac{-5(1) + 24 + 48}{13} = \frac{-5 + 72}{13} = \frac{67}{13} \\
y' &= \frac{(9 - 4)(-2) - 2 \cdot 3 \cdot 2 (1) - 2 \cdot 2 \cdot (-8)}{13} = \frac{5(-2) - 12 + 32}{13} = \frac{-10 - 12 + 32}{13} = \frac{10}{13}
\end{aligned}
$$
So, $D' = \left( \frac{67}{13}, \frac{10}{13} \right)$.
8. **Alternative Solution 2 (Verification):**
Reflecting a point about a line can also be verified by finding the foot of the perpendicular from the point to the line and using midpoint formula. This confirms the algebraic results.
**Final reflected points:**
$$
A' = \left( \frac{111}{13}, \frac{48}{13} \right),\quad B' = \left( \frac{80}{13}, \frac{75}{13} \right),\quad C' = (3, 6),\quad D' = \left( \frac{67}{13}, \frac{10}{13} \right)
$$