Using determinants, find the area of triangle with vertices (0,0),(2,3),(0,5).

Options

A. 10 sq.units
B. 6 sq.units
C. 5' sq.units
D. 12 sq.units

Answer

To find the area of the triangle with vertices at (0,0), (2,3), and (0,5) 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 triangle, the vertices are:
- \((x_1, y_1) = (0, 0)\)
- \((x_2, y_2) = (2, 3)\)
- \((x_3, y_3) = (0, 5)\)

So we can set up the determinant as follows:

\[ A = \frac{1}{2} \left| \begin{vmatrix} 0 & 0 & 1 \\ 2 & 3 & 1 \\ 0 & 5 & 1 \end{vmatrix} \right| \]

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

\[ \begin{vmatrix} 0 & 0 & 1 \\ 2 & 3 & 1 \\ 0 & 5 & 1 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we expand it:

\[ = 0 \cdot \begin{vmatrix} 3 & 1 \\ 5 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 2 & 1 \\ 0 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} \]

Calculating the remaining 2x2 determinant:

\[ \begin{vmatrix} 2 & 3 \\ 0 & 5 \end{vmatrix} = (2 \cdot 5) - (3 \cdot 0) = 10 \]

Thus, the determinant simplifies to:

\[ = 0 - 0 + 10 = 10 \]

### Step 3: Calculate the area
Now we substitute back into the area formula:

\[ A = \frac{1}{2} \left| 10 \right| = \frac{10}{2} = 5 \]

### Conclusion
The area of the triangle with vertices (0,0), (2,3), and (0,5) is \( 5 \) square units.

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