Sure! Let's tackle this problem step-by-step to find the area of the isosceles right triangle formed by points \(\( A(-8, 7) \)\), \(\( B(-4, 5) \)\), and \(\( C(-2, 9) \)\). Here’s a detailed solution:
### Step 1: Plot and Label Points
1. Plot point \(\( A(-8, 7) \)\).
2. Plot point \(\( B(-4, 5) \)\).
3. Plot point \(\( C(-2, 9) \)\).
Connect these points to form the triangle ABC.
### Step 2: Find the lengths of the sides
Using the distance formula:
\(\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]\)
Calculate the distances:
1. Distance \(\( AB \)\):
\(\[ AB = \sqrt{((-4) - (-8))^2 + (5 - 7)^2} \]\)
\(\[ AB = \sqrt{(4)^2 + (-2)^2} \]\)
\(\[ AB = \sqrt{16 + 4} \]\)
\(\[ AB = \sqrt{20} \]\)
\(\[ AB = 4.47213595499958 \]\)
2. Distance \(\( BC \)\):
\(\[ BC = \sqrt{((-2) - (-4))^2 + (9 - 5)^2} \]\)
\(\[ BC = \sqrt{(2)^2 + (4)^2} \]\)
\(\[ BC = \sqrt{4 + 16} \]\)
\(\[ BC = \sqrt{20} \]\)
\(\[ BC = 4.47213595499958 \]\)
3. Distance \(\( CA \)\):
\(\[ CA = \sqrt{((-2) - (-8))^2 + (9 - 7)^2} \]\)
\(\[ CA = \sqrt{(6)^2 + (2)^2} \]\)
\(\[ CA = \sqrt{36 + 4} \]\)
\(\[ CA = \sqrt{40} \]\)
\(\[ CA = 6.324555320336759 \]\)
### Step 3: Identify the isosceles right triangle's legs
By examining the side lengths, we see that \(\( AB \)\) and \(\( BC \)\) are equal, thus indicating that these are the legs of the isosceles right triangle.
### Step 4: Calculate the Area of the Triangle
For an isosceles right triangle, the area can be found using:
\(\[ \text{Area} = \frac{1}{2} \times (\text{leg})^2 \]\)
Here, each leg (AB or BC) has a length of \(\( 4.47213595499958 \)\).
\(\[\text{Area} = \frac{1}{2} \times (4.47213595499958)^2 \]\)
\(\[\text{Area} = \frac{1}{2} \times 20 \]\)
\(\[\text{Area} = 10 \]\)
Therefore, the area of the right isosceles triangle formed by the points \(\( A(-8, 7) \)\), \(\( B(-4, 5) \)\), and \(\( C(-2, 9) \)\) is \(\( 10 \)\) square units.
