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 $$ \Delta T^{\prime} W^{\prime} O^{\prime} $$ ?

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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:
$$
\Delta T' W' O' = (6, -2), (9, 0), (7, 4)
$$
The coordinates of the reflected triangle $$ \Delta T' W' O' $$ are $$ T'(6, -2) $$, $$ W'(9, 0) $$, and $$ O'(7, 4) $$.

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 \( \Delta T^{\prime} W^{\prime} O^{\prime} \) ?

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