Find the area of the triangle with vertices P(4,5),Q(4,-2) and R(-6,2).

Options

A. 21 sq. units
B. 35 sq. units
C. 30 sq. units
D. 40 sq. units

Answer

To find the area of the triangle with vertices P(4,5), Q(4,-2), and R(-6,2), we can use the determinant formula for the area of a triangle given its vertices. The formula is:

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

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

### Step 1: Identify the coordinates
Let:
- \( P(4, 5) \) → \( (x_1, y_1) = (4, 5) \)
- \( Q(4, -2) \) → \( (x_2, y_2) = (4, -2) \)
- \( R(-6, 2) \) → \( (x_3, y_3) = (-6, 2) \)

### Step 2: Set up the determinant
We can set up the determinant as follows:

\[ \begin{vmatrix} 4 & 5 & 1 \\ 4 & -2 & 1 \\ -6 & 2 & 1 \\ \end{vmatrix} \]

### Step 3: Calculate the determinant
Now, we will calculate the determinant using the formula for a 3x3 matrix:

\[ \text{Determinant} = x_1(y_2 \cdot 1 - y_3 \cdot 1) - y_1(x_2 \cdot 1 - x_3 \cdot 1) + 1(x_2 \cdot y_3 - x_3 \cdot y_2) \]

Substituting the values:

\[ = 4((-2) \cdot 1 - 2 \cdot 1) - 5(4 \cdot 1 - (-6) \cdot 1) + 1(4 \cdot 2 - (-6) \cdot (-2)) \]

Calculating each term:

1. First term: \( 4((-2) - 2) = 4(-4) = -16 \)
2. Second term: \( -5(4 + 6) = -5(10) = -50 \)
3. Third term: \( 1(8 - 12) = 1(-4) = -4 \)

Now, summing these results:

\[ \text{Determinant} = -16 - 50 - 4 = -70 \]

### Step 4: Calculate the area
Now we can find the area:

\[ \text{Area} = \frac{1}{2} \left| -70 \right| = \frac{1}{2} \cdot 70 = 35 \]

Thus, the area of the triangle is \( 35 \) square units.

### Summary of Steps:
1. Identify the coordinates of the vertices.
2. Set up the determinant using the coordinates.
3. Calculate the determinant using the formula for a 3x3 matrix.
4. Find the area using the absolute value of the determinant.