Choose the correct answer from the following :
The verices of a triangle (2,-4), (-6,3) and (3,5). The area of triangle is :

Options

A. `79/2"sq.units"`
B. `81/2"sq.units"`
C. `75/5"sq.units"`
D. `85/2"sq.units"`

Answer

To find the area of the triangle with vertices at the points (2, -4), (-6, 3), and (3, 5), we can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

### Step-by-Step Solution:

1. **Identify the coordinates**:
- Let \( (x_1, y_1) = (2, -4) \)
- Let \( (x_2, y_2) = (-6, 3) \)
- Let \( (x_3, y_3) = (3, 5) \)

2. **Substitute the coordinates into the area formula**:
\[ \text{Area} = \frac{1}{2} \left| 2(3 - 5) + (-6)(5 + 4) + 3(-4 - 3) \right| \]

3. **Calculate each term**:
- First term: \( 2(3 - 5) = 2(-2) = -4 \)
- Second term: \( -6(5 + 4) = -6(9) = -54 \)
- Third term: \( 3(-4 - 3) = 3(-7) = -21 \)

4. **Combine the terms**:
\[ \text{Area} = \frac{1}{2} \left| -4 - 54 - 21 \right| = \frac{1}{2} \left| -79 \right| \]

5. **Calculate the absolute value and finalize the area**:
\[ \text{Area} = \frac{1}{2} \times 79 = \frac{79}{2} \]

Thus, the area of the triangle is:

\[ \text{Area} = \frac{79}{2} \text{ square units} \]