Find the area of the triangle whose vertices are `A(4,4),B(3,-16) and C(3,-2)`

Options

A. `9sq. units`
B. `7sq. units`
C. `5sq. units`
D. `4sq. units`

Explanation

To find the area of the triangle with vertices A(4, 4), B(3, -16), and C(3, -2), we can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

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

### Step 1: Identify the coordinates of the vertices
- \(A(4, 4)\) gives us \(x_1 = 4\) and \(y_1 = 4\)
- \(B(3, -16)\) gives us \(x_2 = 3\) and \(y_2 = -16\)
- \(C(3, -2)\) gives us \(x_3 = 3\) and \(y_3 = -2\)

### Step 2: Substitute the coordinates into the area formula
Substituting the values into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 4(-16 - (-2)) + 3(-2 - 4) + 3(4 - (-16)) \right| \]

### Step 3: Simplify the expression inside the absolute value
Calculate each term step by step:

1. \(y_2 - y_3 = -16 + 2 = -14\)
2. \(y_3 - y_1 = -2 - 4 = -6\)
3. \(y_1 - y_2 = 4 + 16 = 20\)

Now substitute these results back into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 4(-14) + 3(-6) + 3(20) \right| \]

### Step 4: Calculate each term
Now calculate each term:

1. \(4 \times -14 = -56\)
2. \(3 \times -6 = -18\)
3. \(3 \times 20 = 60\)

Now substitute these values back into the area formula:

\[ \text{Area} = \frac{1}{2} \left| -56 - 18 + 60 \right| \]

### Step 5: Combine the terms
Combine the terms inside the absolute value:

\[ -56 - 18 + 60 = -14 \]

### Step 6: Calculate the area
Now, substitute this back into the area formula:

\[ \text{Area} = \frac{1}{2} \left| -14 \right| = \frac{1}{2} \times 14 = 7 \]

Thus, the area of the triangle is \(7\) square units.

### Final Answer
The area of the triangle is \(7\) square units.