Reflection Points 1Bc74B

1. **State the problem:** 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 across the line $$3x + 2y = 8$$. 2. **Formula and rules for reflection about a line:** The line can be rewritten as $$3x + 2y - 8 = 0$$. For a point $P(x_0,y_0)$, its reflection $P'(x',y')$ 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 $C=-8$. 3. **Calculate the denominator:** $$A^2 + B^2 = 3^2 + 2^2 = 9 + 4 = 13$$ 4. **Reflect each point:** **For $A(-3,-4)$:** Calculate numerator term: $$3(-3) + 2(-4) - 8 = -9 - 8 - 8 = -25$$ Then, $$x' = -3 - \frac{2 \times 3 \times (-25)}{13} = -3 - \frac{-150}{13} = -3 + \frac{150}{13} = \frac{-39 + 150}{13} = \frac{111}{13} \approx 8.54$$ $$y' = -4 - \frac{2 \times 2 \times (-25)}{13} = -4 - \frac{-100}{13} = -4 + \frac{100}{13} = \frac{-52 + 100}{13} = \frac{48}{13} \approx 3.69$$ So, $A' \approx \left(\frac{111}{13}, \frac{48}{13}\right)$. **For $B(-4,-1)$:** Calculate numerator term: $$3(-4) + 2(-1) - 8 = -12 - 2 - 8 = -22$$ Then, $$x' = -4 - \frac{2 \times 3 \times (-22)}{13} = -4 - \frac{-132}{13} = -4 + \frac{132}{13} = \frac{-52 + 132}{13} = \frac{80}{13} \approx 6.15$$ $$y' = -1 - \frac{2 \times 2 \times (-22)}{13} = -1 - \frac{-88}{13} = -1 + \frac{88}{13} = \frac{-13 + 88}{13} = \frac{75}{13} \approx 5.77$$ So, $B' \approx \left(\frac{80}{13}, \frac{75}{13}\right)$. **For $C(-3,2)$:** Calculate numerator term: $$3(-3) + 2(2) - 8 = -9 + 4 - 8 = -13$$ Then, $$x' = -3 - \frac{2 \times 3 \times (-13)}{13} = -3 - \frac{-78}{13} = -3 + 6 = 3$$ $$y' = 2 - \frac{2 \times 2 \times (-13)}{13} = 2 - \frac{-52}{13} = 2 + 4 = 6$$ So, $C' = (3,6)$. **For $D(1,-2)$:** Calculate numerator term: $$3(1) + 2(-2) - 8 = 3 - 4 - 8 = -9$$ Then, $$x' = 1 - \frac{2 \times 3 \times (-9)}{13} = 1 - \frac{-54}{13} = 1 + \frac{54}{13} = \frac{13 + 54}{13} = \frac{67}{13} \approx 5.15$$ $$y' = -2 - \frac{2 \times 2 \times (-9)}{13} = -2 - \frac{-36}{13} = -2 + \frac{36}{13} = \frac{-26 + 36}{13} = \frac{10}{13} \approx 0.77$$ So, $D' \approx \left(\frac{67}{13}, \frac{10}{13}\right)$. 5. **Summary of reflected points:** $$A' \approx \left(8.54, 3.69\right), B' \approx \left(6.15, 5.77\right), C' = (3,6), D' \approx \left(5.15, 0.77\right)$$ 6. **Alternative Solution 2 (Verification):** Alternative Solution 2 usually involves using vector projection or coordinate transformation to verify the reflection. Since the algebraic method is exact, plotting these points and the line will visually confirm the reflection. --- **Final answer:** $$A' = \left(\frac{111}{13}, \frac{48}{13}\right), B' = \left(\frac{80}{13}, \frac{75}{13}\right), C' = (3,6), D' = \left(\frac{67}{13}, \frac{10}{13}\right)$$