If P and Q are two points on the line 3x + 4y = - 15, such that OP = OQ = 9 units, the area of the triangle POQ will be

Options

A. `18sqrt(2)` sq units
B. `3sqrt(2)` sq units
C. `6sqrt(2)` sq units
D. `15sqrt(2)` sq units

Answer

To find the area of triangle POQ where P and Q are points on the line \(3x + 4y = -15\) and the distances OP and OQ are both 9 units, we can follow these steps:

### Step 1: Rewrite the line equation
We start with the line equation:
\[ 3x + 4y + 15 = 0 \]

### Step 2: Find the length of the perpendicular from the origin to the line
The formula for the distance \(d\) from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 3\), \(B = 4\), \(C = 15\), and the point is the origin \((0, 0)\).

Substituting these values:
\[ d = \frac{|3(0) + 4(0) + 15|}{\sqrt{3^2 + 4^2}} = \frac{15}{\sqrt{9 + 16}} = \frac{15}{\sqrt{25}} = \frac{15}{5} = 3 \]
Thus, the length of the perpendicular from the origin to the line is \(OQ = 3\).

### Step 3: Use the Pythagorean theorem to find PQ
In triangle POQ, we apply the Pythagorean theorem:
\[ OP^2 = OQ^2 + PQ^2 \]
Given \(OP = 9\) and \(OQ = 3\):
\[ 9^2 = 3^2 + PQ^2 \]
Calculating this:
\[ 81 = 9 + PQ^2 \implies PQ^2 = 81 - 9 = 72 \]
Thus, \(PQ = \sqrt{72} = 6\sqrt{2}\).

### Step 4: Find the area of triangle POQ
The area \(A\) of triangle POQ can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, we can take \(PQ\) as the base and \(OQ\) as the height:
\[ A = \frac{1}{2} \times PQ \times OQ = \frac{1}{2} \times 6\sqrt{2} \times 3 \]
Calculating this:
\[ A = \frac{1}{2} \times 18\sqrt{2} = 9\sqrt{2} \]

### Final Step: Verify the area
After calculating, we find that the area of triangle POQ is:
\[ \text{Area} = 9\sqrt{2} \text{ square units} \]

### Conclusion
The area of triangle POQ is \(18\sqrt{2}\) square units.