Vertices of a triangle are (0, 5), (0, 0), (4, 0) then find area of triangle.

Answer

To find the area of the triangle with vertices at (0, 5), (0, 0), and (4, 0), we can use the determinant method. Here’s a step-by-step solution:

### Step 1: Identify the vertices
The vertices of the triangle are given as:
- \( A(0, 5) \)
- \( B(0, 0) \)
- \( C(4, 0) \)

### Step 2: Set up the determinant
The formula for the area \( A \) of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) using determinants is:

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

Substituting the coordinates of the vertices into the determinant:

\[ A = \frac{1}{2} \left| \begin{vmatrix} 0 & 5 & 1 \\ 0 & 0 & 1 \\ 4 & 0 & 1 \end{vmatrix} \right| \]

### Step 3: Calculate the determinant
Now, we will calculate the determinant:

\[ \begin{vmatrix} 0 & 5 & 1 \\ 0 & 0 & 1 \\ 4 & 0 & 1 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:

\[ = 0 \cdot \begin{vmatrix} 0 & 1 \\ 0 & 1 \end{vmatrix} - 5 \cdot \begin{vmatrix} 0 & 1 \\ 4 & 1 \end{vmatrix} + 1 \cdot \begin{vmatrix} 0 & 0 \\ 4 & 0 \end{vmatrix} \]

Calculating the smaller 2x2 determinants:

1. \( \begin{vmatrix} 0 & 1 \\ 0 & 1 \end{vmatrix} = 0 \)

2. \( \begin{vmatrix} 0 & 1 \\ 4 & 1 \end{vmatrix} = 0 \cdot 1 - 1 \cdot 4 = -4 \)

3. \( \begin{vmatrix} 0 & 0 \\ 4 & 0 \end{vmatrix} = 0 \)

Putting these back into the determinant calculation:

\[ = 0 - 5 \cdot (-4) + 0 = 20 \]

### Step 4: Calculate the area
Now, substituting back into the area formula:

\[ A = \frac{1}{2} \left| 20 \right| = \frac{20}{2} = 10 \]

### Final Answer
The area of the triangle is \( 10 \) square units.

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