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To solve the problem, let's analyze the given information and find the possible locations of point \( F \) such that the triangle \( \Delta DEF \) has an area of 28 square units.

### Step 1: Calculate the Length of Segment \( DE \)
The coordinates of points \( D \) and \( E \) are \( D(-6, 3) \) and \( E(2, 3) \), respectively. Since both points have the same y-coordinate, the segment \( DE \) is horizontal.

The length of \( DE \) can be found using the distance formula for horizontal segments:
\[
\text{Length of } DE = |x_2 - x_1| = |2 - (-6)| = 2 + 6 = 8 \text{ units}
\]

### Step 2: Use the Area Formula for a Triangle
The area of a triangle can be calculated using the formula:
\[
\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}
\]
Given that the area is 28 square units and the base \( DE \) is 8 units:
\[
28 = \frac{1}{2} \times 8 \times \text{Height}
\]
\[
28 = 4 \times \text{Height}
\]
\[
\text{Height} = \frac{28}{4} = 7 \text{ units}
\]
The height is the perpendicular distance from point \( F \) to the line \( DE \).

### Step 3: Determine the Coordinates of \( F \)
The height must be 7 units above or below the line \( DE \), which has a y-coordinate of 3. Therefore, the possible y-coordinates of \( F \) are:
\[
y = 3 + 7 = 10 \quad \text{or} \quad y = 3 - 7 = -4
\]

### Step 4: Check the Possible Points
We now check the provided options:

- **(7, 4):** The y-coordinate is 4, which doesn't match the possible heights (10 or -4). **Not a valid point.**
- **(1, 10):** The y-coordinate is 10, which is 7 units above the line \( DE \). **Valid point.**
- **(-6, 10):** The y-coordinate is 10, which is 7 units above the line \( DE \). **Valid point.**
- **(5, -4):** The y-coordinate is -4, which is 7 units below the line \( DE \). **Valid point.**

### Conclusion:
The valid points for \( F \) where the area of \( \Delta DEF \) equals 28 square units are:
- (1, 10)
- (-6, 10)
- (5, -4)

The selected answers are correct.

Would you like more details or have any questions?

Here are five related questions for further practice:
1. How would the area change if \( F \) were at a different height?
2. What would happen if the base \( DE \) were longer or shorter?
3. How would you find the area of \( \Delta DEF \) if \( F \) had different coordinates?
4. What if the triangle had a different area, say 56 square units?
5. Can you calculate the perimeter of \( \Delta DEF \) with the selected points \( F \)?

**Tip:** Remember that the height in the area formula for a triangle always refers to the perpendicular distance from a vertex to the opposite side.

The query submitted by the user: To solve the problem, let's analyze the given information and find the possible locations of point F such that the triangle ΔDEF has an area of 28 square units. ... The selected answers are correct. Would you like more details or have any questions? Here are five related questions for further practice: 1. How would the area change if F were at a different height? 2. What would happen if the base DE were longer or shorter? 3. How would you find the area of ΔDEF if F had different coordinates? 4. What if the triangle had a different area, say 56 square units? 5. Can you calculate the perimeter of ΔDEF with the selected points F? **Tip:** Remember that the height in the area formula for a triangle always refers to the perpendicular distance from a vertex to the opposite side.

Learn how to find the coordinates of point F such that triangle DEF has an area of 28 square units. This problem involves using the distance formula and the area of a triangle formula in coordinate geometry. Suitable for Grades 7-9.

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