Find the area of the triangle whose vertices are `P(-(2)/(5),6),Q(2,8)` and `R(-4,3)`

Options

A. 0 sq. Unit
B. `5sqrt(2)` sq. Units
C. 9 sq. Units
D. 15 sq. Units

Answer

To find the area of the triangle with vertices P(-2/5, 6), Q(2, 8), and R(-4, 3), 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:
- \( (x_1, y_1) = P(-\frac{2}{5}, 6) \)
- \( (x_2, y_2) = Q(2, 8) \)
- \( (x_3, y_3) = R(-4, 3) \)

### Step 1: Identify the coordinates
Assign the coordinates:
- \( x_1 = -\frac{2}{5}, y_1 = 6 \)
- \( x_2 = 2, y_2 = 8 \)
- \( x_3 = -4, y_3 = 3 \)

### Step 2: Substitute into the area formula
Substituting the values into the area formula:

\[ \text{Area} = \frac{1}{2} \left| -\frac{2}{5}(8 - 3) + 2(3 - 6) + (-4)(6 - 8) \right| \]

### Step 3: Calculate the differences
Calculate the differences in the parentheses:
- \( 8 - 3 = 5 \)
- \( 3 - 6 = -3 \)
- \( 6 - 8 = -2 \)

### Step 4: Substitute the differences back into the formula
Now substitute these values back into the formula:

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

### Step 5: Simplify each term
Now simplify each term:
- \( -\frac{2}{5} \times 5 = -2 \)
- \( 2 \times -3 = -6 \)
- \( -4 \times -2 = 8 \)

### Step 6: Combine the terms
Combine the results:

\[ \text{Area} = \frac{1}{2} \left| -2 - 6 + 8 \right| \]

### Step 7: Calculate the final value
Calculate inside the absolute value:

\[ -2 - 6 + 8 = 0 \]

Thus, the area becomes:

\[ \text{Area} = \frac{1}{2} \left| 0 \right| = 0 \]

### Conclusion
The area of the triangle is \( 0 \) square units, which indicates that the points are collinear.