Find the area of the triangle whose vertices are :
(2,3), (-1,0), (2,-4)

Answer

To find the area of the triangle with vertices at points A(2, 3), B(-1, 0), and C(2, -4), we can use the formula for the area of a triangle given by its vertices in a coordinate plane. The formula is:

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

Where:
- \( (x_1, y_1) = (2, 3) \) (coordinates of point A)
- \( (x_2, y_2) = (-1, 0) \) (coordinates of point B)
- \( (x_3, y_3) = (2, -4) \) (coordinates of point C)

### Step 1: Substitute the coordinates into the formula

Substituting the coordinates into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 2(0 - (-4)) + (-1)(-4 - 3) + 2(3 - 0) \right| \]

### Step 2: Simplify the expression

Calculating each term step by step:

1. For the first term:
\[ 2(0 - (-4)) = 2(0 + 4) = 2 \times 4 = 8 \]

2. For the second term:
\[ -1(-4 - 3) = -1(-7) = 7 \]

3. For the third term:
\[ 2(3 - 0) = 2 \times 3 = 6 \]

Now, combine these results:
\[ \text{Area} = \frac{1}{2} \left| 8 + 7 + 6 \right| = \frac{1}{2} \left| 21 \right| \]

### Step 3: Calculate the final area

Thus, the area is:
\[ \text{Area} = \frac{1}{2} \times 21 = \frac{21}{2} \]

### Final Result

The area of the triangle is:
\[ \frac{21}{2} \text{ square units} \]

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