Using determinant, find the area of the triangle with vertices :
`(1,-1),(2,4),(-3,5)`

Answer

To find the area of the triangle with vertices at the points \( A(1, -1) \), \( B(2, 4) \), and \( C(-3, 5) \) using determinants, we can follow these steps:

### Step 1: Set up the determinant
The area \( A \) of a triangle given its vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the formula:

\[ A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]

For our points:
- \( (x_1, y_1) = (1, -1) \)
- \( (x_2, y_2) = (2, 4) \)
- \( (x_3, y_3) = (-3, 5) \)

The determinant can be set up as follows:

\[ A = \frac{1}{2} \left| \begin{vmatrix} 1 & -1 & 1 \\ 2 & 4 & 1 \\ -3 & 5 & 1 \end{vmatrix} \right| \]

### Step 2: Calculate the determinant
Now, we will calculate the determinant:

\[ \begin{vmatrix} 1 & -1 & 1 \\ 2 & 4 & 1 \\ -3 & 5 & 1 \end{vmatrix} \]

Using the formula for the determinant of a \( 3 \times 3 \) matrix:

\[ = 1 \cdot \begin{vmatrix} 4 & 1 \\ 5 & 1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 2 & 1 \\ -3 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 4 \\ -3 & 5 \end{vmatrix} \]

Calculating each of these \( 2 \times 2 \) determinants:

1. \( \begin{vmatrix} 4 & 1 \\ 5 & 1 \end{vmatrix} = (4 \cdot 1) - (1 \cdot 5) = 4 - 5 = -1 \)

2. \( \begin{vmatrix} 2 & 1 \\ -3 & 1 \end{vmatrix} = (2 \cdot 1) - (1 \cdot -3) = 2 + 3 = 5 \)

3. \( \begin{vmatrix} 2 & 4 \\ -3 & 5 \end{vmatrix} = (2 \cdot 5) - (4 \cdot -3) = 10 + 12 = 22 \)

Now substituting back into the determinant calculation:

\[ = 1 \cdot (-1) + 1 \cdot 5 + 1 \cdot 22 = -1 + 5 + 22 = 26 \]

### Step 3: Calculate the area
Now we can find the area:

\[ A = \frac{1}{2} \left| 26 \right| = \frac{26}{2} = 13 \]

### Conclusion
Hence, the area of the triangle is \( 13 \) square units.