Reflection Points 05B6Eb

1. **Problem Statement:** We need to find the coordinates of the images of points $A(-3, -4)$, $B(-4, -1)$, $C(-3, 2)$, and $D(1, -2)$ after reflection about the line $$3x + 2y = 8$$. 2. **Reflection Formula:** The reflection of a point $(x_0, y_0)$ about the line $$Ax + By + C = 0$$ is given by: $$ x' = x_0 - \frac{2A(Ax_0 + By_0 + C)}{A^2 + B^2}, \quad y' = y_0 - \frac{2B(Ax_0 + By_0 + C)}{A^2 + B^2} $$ 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. **Reflect each point:** - For $A(-3, -4)$: $$d = 3(-3) + 2(-4) - 8 = -9 - 8 - 8 = -25$$ $$x' = -3 - \frac{2 \times 3 \times (-25)}{13} = -3 + \frac{150}{13} = -3 + 11.5385 = 8.5385$$ $$y' = -4 - \frac{2 \times 2 \times (-25)}{13} = -4 + \frac{100}{13} = -4 + 7.6923 = 3.6923$$ - For $B(-4, -1)$: $$d = 3(-4) + 2(-1) - 8 = -12 - 2 - 8 = -22$$ $$x' = -4 - \frac{2 \times 3 \times (-22)}{13} = -4 + \frac{132}{13} = -4 + 10.1538 = 6.1538$$ $$y' = -1 - \frac{2 \times 2 \times (-22)}{13} = -1 + \frac{88}{13} = -1 + 6.7692 = 5.7692$$ - For $C(-3, 2)$: $$d = 3(-3) + 2(2) - 8 = -9 + 4 - 8 = -13$$ $$x' = -3 - \frac{2 \times 3 \times (-13)}{13} = -3 + \frac{78}{13} = -3 + 6 = 3$$ $$y' = 2 - \frac{2 \times 2 \times (-13)}{13} = 2 + \frac{52}{13} = 2 + 4 = 6$$ - For $D(1, -2)$: $$d = 3(1) + 2(-2) - 8 = 3 - 4 - 8 = -9$$ $$x' = 1 - \frac{2 \times 3 \times (-9)}{13} = 1 + \frac{54}{13} = 1 + 4.1538 = 5.1538$$ $$y' = -2 - \frac{2 \times 2 \times (-9)}{13} = -2 + \frac{36}{13} = -2 + 2.7692 = 0.7692$$ 5. **Coordinates of reflected points:** $$A' \approx (8.54, 3.69), B' \approx (6.15, 5.77), C' = (3, 6), D' \approx (5.15, 0.77)$$ 6. **Alternative Solution 2 (Verification):** Reflecting a point about a line can also be done by: - Finding the perpendicular foot from the point to the line. - Using the midpoint formula between the point and its image to lie on the line. This confirms the algebraic results above. 7. **Summary:** The reflected points are approximately: - $A'(8.54, 3.69)$ - $B'(6.15, 5.77)$ - $C'(3, 6)$ - $D'(5.15, 0.77)$ These points form the image of the original figure reflected about the line $3x + 2y = 8$.