Find the area of the triangle whose vertices are (3,8), (-4,2) and (5, -1).

Answer

To find the area of the triangle with vertices at (3, 8), (-4, 2), and (5, -1), we can use the formula for the area of a triangle given its vertices:

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Where:
- \( (x_1, y_1) = (3, 8) \)
- \( (x_2, y_2) = (-4, 2) \)
- \( (x_3, y_3) = (5, -1) \)

### Step 1: Substitute the values into the formula

Substituting the coordinates into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 3(2 - (-1)) + (-4)(-1 - 8) + 5(8 - 2) \right| \]

### Step 2: Simplify the expressions inside the absolute value

Calculating each term:

1. \( 3(2 + 1) = 3 \times 3 = 9 \)
2. \( -4(-1 - 8) = -4 \times -9 = 36 \)
3. \( 5(8 - 2) = 5 \times 6 = 30 \)

Now, substituting these values back into the equation:

\[ \text{Area} = \frac{1}{2} \left| 9 + 36 + 30 \right| \]

### Step 3: Calculate the total inside the absolute value

Adding the values together:

\[ 9 + 36 + 30 = 75 \]

### Step 4: Final calculation of the area

Now, substituting back into the area formula:

\[ \text{Area} = \frac{1}{2} \times 75 = \frac{75}{2} = 37.5 \]

Thus, the area of the triangle is:

\[ \text{Area} = 37.5 \text{ square units} \]