To find the area of the triangle with vertices at $(5, 0)$, $(-2, 6)$, and $(-2, 0)$, we can follow these steps:

Step 1: Check if the Triangle is Isosceles

A triangle is isosceles if two of its sides are of equal length. We calculate the lengths of the sides using the distance formula:

$\text{Distance between } (x_1, y_1) \text{ and } (x_2, y_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

1. Length of side between $(5, 0)$ and $(-2, 6)$: $d_1 = \sqrt{((-2) - 5)^2 + (6 - 0)^2} = \sqrt{(-7)^2 + 6^2} = \sqrt{49 + 36} = \sqrt{85}$

2. Length of side between $(5, 0)$ and $(-2, 0)$: $d_2 = \sqrt{((-2) - 5)^2 + (0 - 0)^2} = \sqrt{(-7)^2} = 7$

3. Length of side between $(-2, 6)$ and $(-2, 0)$: $d_3 = \sqrt{((-2) - (-2))^2 + (6 - 0)^2} = \sqrt{0 + 6^2} = 6$

The sides $d_2$ and $d_3$ are not equal to $d_1$, and they aren't equal to each other either. So, the triangle is not isosceles, and we don't need the midpoint formula.

Step 2: Use the Area Formula for a Triangle

For a general triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area $A$ can be calculated using the following formula:

$A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

Substitute the coordinates of the points into the formula:

$A = \frac{1}{2} \left| 5(6 - 0) + (-2)(0 - 0) + (-2)(0 - 6) \right|$

$A = \frac{1}{2} \left| 5(6) + 0 + (-2)(-6) \right|$

$A = \frac{1}{2} \left| 30 + 12 \right| = \frac{1}{2} \times 42 = 21$

Final Answer:

The area of the triangle is 21 square units.

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Would you like more details on any part of the solution, or do you have any further questions?

Related Questions:

1. How can we determine if a triangle is isosceles using only the coordinates?
2. What is the formula to calculate the perimeter of a triangle given its vertices?
3. How do you find the centroid of a triangle using its vertices?
4. What is the significance of the determinant in calculating the area of a triangle?
5. How would the area formula change if the triangle were in 3D space?

Tip:

When dealing with coordinate geometry problems, drawing a rough sketch of the points on the coordinate plane can often help you visualize the problem better and avoid mistakes.