The coordinates of \( \Delta T W O \) are \( T(2,2), W(4,5), O(8,3) \). The triangle is reflected \( R_{y-x+4}(\Delta T W O) \). What are the coordinates of the image \( \Delta T^{\prime} W^{\prime} O^{\prime} \) ?

To find the coordinates of the image \( \Delta T' W' O' \) after reflecting the triangle \( \Delta T W O \) across the line \( y - x + 4 = 0 \), we will follow these steps: 1. **Identify the line of reflection**: The line \( y - x + 4 = 0 \) can be rewritten in slope-intercept form as \( y = x - 4 \). This line has a slope of 1 and a y-intercept of -4. 2. **Find the reflection of each vertex**: To reflect a point across a line, we can use the formula for the reflection of a point \( (x_1, y_1) \) across the line \( Ax + By + C = 0 \): \[ \left( \frac{x_1(B^2 - A^2) - 2AB y_1 - 2AC}{A^2 + B^2}, \frac{y_1(A^2 - B^2) - 2AB x_1 - 2BC}{A^2 + B^2} \right) \] For our line \( y - x + 4 = 0 \), we have \( A = -1 \), \( B = 1 \), and \( C = 4 \). 3. **Calculate the reflection for each vertex**: - For point \( T(2, 2) \): \[ x' = \frac{2(1^2 - (-1)^2) - 2(-1)(1)(2) - 2(-1)(4)}{(-1)^2 + 1^2} \] \[ y' = \frac{2((-1)^2 - 1^2) - 2(-1)(1)(2) - 2(1)(4)}{(-1)^2 + 1^2} \] - For point \( W(4, 5) \): \[ x' = \frac{4(1^2 - (-1)^2) - 2(-1)(1)(5) - 2(-1)(4)}{(-1)^2 + 1^2} \] \[ y' = \frac{5((-1)^2 - 1^2) - 2(-1)(1)(4) - 2(1)(4)}{(-1)^2 + 1^2} \] - For point \( O(8, 3) \): \[ x' = \frac{8(1^2 - (-1)^2) - 2(-1)(1)(3) - 2(-1)(4)}{(-1)^2 + 1^2} \] \[ y' = \frac{3((-1)^2 - 1^2) - 2(-1)(1)(8) - 2(1)(4)}{(-1)^2 + 1^2} \] Now, let's calculate these reflections step by step. ### Step 1: Reflect point \( T(2, 2) \) Using the reflection formula: \[ x_T' = \frac{2(1^2 - (-1)^2) - 2(-1)(1)(2) - 2(-1)(4)}{(-1)^2 + 1^2} \] \[ = \frac{2(1 - 1) + 4 + 8}{1 + 1} = \frac{12}{2} = 6 \] \[ y_T' = \frac{2((-1)^2 - 1^2) - 2(-1)(1)(2) - 2(1)(4)}{(-1)^2 + 1^2} \] \[ = \frac{2(1 - 1) + 4 - 8}{1 + 1} = \frac{-4}{2} = -2 \] So, the reflection of \( T(2, 2) \) is \( T'(6, -2) \). ### Step 2: Reflect point \( W(4, 5) \) Using the reflection formula: \[ x_W' = \frac{4(1^2 - (-1)^2) - 2(-1)(1)(5) - 2(-1)(4)}{(-1)^2 + 1^2} \] \[ = \frac{4(1 - 1) + 10 + 8}{1 + 1} = \frac{18}{2} = 9 \] \[ y_W' = \frac{5((-1)^2 - 1^2) - 2(-1)(1)(4) - 2(1)(4)}{(-1)^2 + 1^2} \] \[ = \frac{5(1 - 1) + 8 - 8}{1 + 1} = \frac{0}{2} = 0 \] So, the reflection of \( W(4, 5) \) is \( W'(9, 0) \). ### Step 3: Reflect point \( O(8, 3) \) Using the reflection formula: \[ x_O' = \frac{8(1^2 - (-1)^2) - 2(-1)(1)(3) - 2(-1)(4)}{(-1)^2 + 1^2} \] \[ = \frac{8(1 - 1) + 6 + 8}{1 + 1} = \frac{14}{2} = 7 \] \[ y_O' = \frac{3((-1)^2 - 1^2) - 2(-1)(1)(8) - 2(1)(4)}{(-1)^2 + 1^2} \] \[ = \frac{3(1 - 1) + 16 - 8}{1 + 1} = \frac{8}{2} = 4 \] So, the reflection of \( O(8, 3) \) is \( O'(7, 4) \). ### Final Result The coordinates of the image \( \Delta T' W' O' \) after reflection are: - \( T'(6, -2) \) - \( W'(9, 0) \) - \( O'(7, 4) \) Thus, the final coordinates are: \[ \Delta T' W' O' = (6, -2), (9, 0), (7, 4) \]