Question: 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. **Problem Statement:**
Find the coordinates of the images of points $A(-3, -4)$, $B(-4, -1)$, $C(-3, 2)$, and $D(1, -2)$ reflected about the line $3x + 2y = 8$.
2. **Reflection Formula:**
For a point $(x,y)$ reflected about the line $ax + by + c = 0$, the image $(x', y')$ is given by:
$$
x' = \frac{(b^2 - a^2)x - 2ab y - 2ac}{a^2 + b^2}, \quad y' = \frac{(a^2 - b^2)y - 2ab x - 2bc}{a^2 + b^2}
$$
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. **Calculate each reflected point:**
- For $A(-3, -4)$:
$$
x' = \frac{(2^2 - 3^2)(-3) - 2 \cdot 3 \cdot 2 (-4) - 2 \cdot 3 (-8)}{13} = \frac{(4 - 9)(-3) + 48 + 48}{13} = \frac{-5(-3) + 96}{13} = \frac{15 + 96}{13} = \frac{111}{13}$$
$$
y' = \frac{(3^2 - 2^2)(-4) - 2 \cdot 3 \cdot 2 (-3) - 2 \cdot 2 (-8)}{13} = \frac{(9 - 4)(-4) + 36 + 32}{13} = \frac{5(-4) + 68}{13} = \frac{-20 + 68}{13} = \frac{48}{13}$$
- For $B(-4, -1)$:
$$
x' = \frac{(4 - 9)(-4) - 2 \cdot 3 \cdot 2 (-1) - 2 \cdot 3 (-8)}{13} = \frac{-5(-4) + 12 + 48}{13} = \frac{20 + 60}{13} = \frac{80}{13}$$
$$
y' = \frac{(9 - 4)(-1) - 2 \cdot 3 \cdot 2 (-4) - 2 \cdot 2 (-8)}{13} = \frac{5(-1) + 48 + 32}{13} = \frac{-5 + 80}{13} = \frac{75}{13}$$
- For $C(-3, 2)$:
$$
x' = \frac{(4 - 9)(-3) - 2 \cdot 3 \cdot 2 (2) - 2 \cdot 3 (-8)}{13} = \frac{-5(-3) - 24 + 48}{13} = \frac{15 + 24}{13} = \frac{39}{13} = 3$$
$$
y' = \frac{(9 - 4)(2) - 2 \cdot 3 \cdot 2 (-3) - 2 \cdot 2 (-8)}{13} = \frac{5(2) + 36 + 32}{13} = \frac{10 + 68}{13} = \frac{78}{13} = 6$$
- For $D(1, -2)$:
$$
x' = \frac{(4 - 9)(1) - 2 \cdot 3 \cdot 2 (-2) - 2 \cdot 3 (-8)}{13} = \frac{-5(1) + 24 + 48}{13} = \frac{-5 + 72}{13} = \frac{67}{13}$$
$$
y' = \frac{(9 - 4)(-2) - 2 \cdot 3 \cdot 2 (1) - 2 \cdot 2 (-8)}{13} = \frac{5(-2) - 12 + 32}{13} = \frac{-10 - 12 + 32}{13} = \frac{10}{13}$$
5. **Final reflected points:**
$$
A' = \left(\frac{111}{13}, \frac{48}{13}\right), \quad B' = \left(\frac{80}{13}, \frac{75}{13}\right), \quad C' = (3, 6), \quad D' = \left(\frac{67}{13}, \frac{10}{13}\right)
$$
6. **Verification (Alternative Solution 2):**
Reflecting a point about a line can also be verified by checking that the midpoint of the segment joining the point and its image lies on the line and the segment is perpendicular to the line.
This completes the algebraic solution and verification.