Find the area of the triangle whose vertices are given below `(-3,-4), (-2,-7), (-1,-9)`

Options

A. `2`sq.units
B. `1`sq.units
C. `1/2`sq.units
D. `1/4`sq.units

Answer

To find the area of the triangle with vertices at the points \((-3, -4)\), \((-2, -7)\), and \((-1, -9)\), we can use the formula for the area of a triangle given by the coordinates of its vertices using determinants. The formula is:

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

### Step 1: Set up the determinant
Given the vertices:
- \( (x_1, y_1) = (-3, -4) \)
- \( (x_2, y_2) = (-2, -7) \)
- \( (x_3, y_3) = (-1, -9) \)

We can set up the determinant as follows:

\[ \text{Area} = \frac{1}{2} \left| \begin{vmatrix} -3 & -4 & 1 \\ -2 & -7 & 1 \\ -1 & -9 & 1 \end{vmatrix} \right| \]

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

\[ \begin{vmatrix} -3 & -4 & 1 \\ -2 & -7 & 1 \\ -1 & -9 & 1 \end{vmatrix} = -3 \begin{vmatrix} -7 & 1 \\ -9 & 1 \end{vmatrix} - (-4) \begin{vmatrix} -2 & 1 \\ -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} -2 & -7 \\ -1 & -9 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \(\begin{vmatrix} -7 & 1 \\ -9 & 1 \end{vmatrix} = (-7)(1) - (1)(-9) = -7 + 9 = 2\)

2. \(\begin{vmatrix} -2 & 1 \\ -1 & 1 \end{vmatrix} = (-2)(1) - (1)(-1) = -2 + 1 = -1\)

3. \(\begin{vmatrix} -2 & -7 \\ -1 & -9 \end{vmatrix} = (-2)(-9) - (-7)(-1) = 18 - 7 = 11\)

Now substituting these back into the determinant calculation:

\[ = -3(2) + 4(-1) + 1(11) = -6 - 4 + 11 = 1 \]

### Step 3: Calculate the area
Now, substituting the value of the determinant back into the area formula:

\[ \text{Area} = \frac{1}{2} \left| 1 \right| = \frac{1}{2} \]

Thus, the area of the triangle is:

\[ \text{Area} = \frac{1}{2} \text{ square units} \]