What is the area of a triangle whose vertices are A(-4, -2), B(-3, -5) and C(3, -2)?

Options

A. 12 sq units
B. 10 sq units
C. 7.5 sq units
D. 10.5 sq units

Answer

To find the area of a triangle given its vertices, we can use the formula:

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

Where \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) are the coordinates of the vertices of the triangle.

Given the vertices:
- \(A(-4, -2)\) → \(x_1 = -4\), \(y_1 = -2\)
- \(B(-3, -5)\) → \(x_2 = -3\), \(y_2 = -5\)
- \(C(3, -2)\) → \(x_3 = 3\), \(y_3 = -2\)

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

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

### Step 2: Simplify each term inside the absolute value

1. Calculate \(y_2 - y_3\):
\[ -5 - (-2) = -5 + 2 = -3 \]
So, the first term becomes:
\[ -4(-3) = 12 \]

2. Calculate \(y_3 - y_1\):
\[ -2 - (-2) = -2 + 2 = 0 \]
So, the second term becomes:
\[ -3(0) = 0 \]

3. Calculate \(y_1 - y_2\):
\[ -2 - (-5) = -2 + 5 = 3 \]
So, the third term becomes:
\[ 3(3) = 9 \]

### Step 3: Combine the results

Now we can combine all the terms:
\[ \text{Area} = \frac{1}{2} \left| 12 + 0 + 9 \right| = \frac{1}{2} \left| 21 \right| = \frac{21}{2} \]

### Step 4: Calculate the final area

\[ \text{Area} = \frac{21}{2} = 10.5 \text{ square units} \]

Thus, the area of the triangle is \(10.5\) square units.