Lines `x+y=1 and 3y=x+3` intersect the ellipse `x^(2)+9y^(2)=9` at the points P,Q,R. the area of the triangles PQR is

Options

A. `(36)/(5)`
B. `(18)/(5)`
C. `(9)/(5)`
D. `(1)/(5)`

Answer

To find the area of triangle PQR formed by the intersection of the lines \( x + y = 1 \) and \( 3y = x + 3 \) with the ellipse \( x^2 + 9y^2 = 9 \), we will follow these steps:

### Step 1: Find the intersection points of the lines with the ellipse.

1. **Equation of the ellipse**:
\[ x^2 + 9y^2 = 9 \]

2. **First line**:
\[ x + y = 1 \quad \Rightarrow \quad y = 1 - x \]

Substitute \( y \) in the ellipse equation:
\[ x^2 + 9(1 - x)^2 = 9 \]
Expanding:
\[ x^2 + 9(1 - 2x + x^2) = 9 \]
\[ x^2 + 9 - 18x + 9x^2 = 9 \]
\[ 10x^2 - 18x = 0 \]
Factor out \( x \):
\[ x(10x - 18) = 0 \]
Thus, \( x = 0 \) or \( x = \frac{18}{10} = \frac{9}{5} \).

For \( x = 0 \):
\[ y = 1 \quad \Rightarrow \quad P(0, 1) \]

For \( x = \frac{9}{5} \):
\[ y = 1 - \frac{9}{5} = -\frac{4}{5} \quad \Rightarrow \quad Q\left(\frac{9}{5}, -\frac{4}{5}\right) \]

3. **Second line**:
\[ 3y = x + 3 \quad \Rightarrow \quad y = \frac{x}{3} + 1 \]

Substitute \( y \) in the ellipse equation:
\[ x^2 + 9\left(\frac{x}{3} + 1\right)^2 = 9 \]
Expanding:
\[ x^2 + 9\left(\frac{x^2}{9} + \frac{2x}{3} + 1\right) = 9 \]
\[ x^2 + x^2 + 6x + 9 = 9 \]
\[ 2x^2 + 6x = 0 \]
Factor out \( 2x \):
\[ 2x(x + 3) = 0 \]
Thus, \( x = 0 \) or \( x = -3 \).

For \( x = 0 \):
\[ y = 1 \quad \Rightarrow \quad P(0, 1) \quad \text{(already found)} \]

For \( x = -3 \):
\[ y = \frac{-3}{3} + 1 = 0 \quad \Rightarrow \quad R(-3, 0) \]

### Step 2: Identify the points of intersection.

The points of intersection are:
- \( P(0, 1) \)
- \( Q\left(\frac{9}{5}, -\frac{4}{5}\right) \)
- \( R(-3, 0) \)

### Step 3: Calculate the area of triangle PQR.

Using the formula for the area of a triangle given vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

Substituting the points \( P(0, 1) \), \( Q\left(\frac{9}{5}, -\frac{4}{5}\right) \), \( R(-3, 0) \):
\[ \text{Area} = \frac{1}{2} \left| 0\left(-\frac{4}{5} - 0\right) + \frac{9}{5}(0 - 1) + (-3)(1 - (-\frac{4}{5})) \right| \]
\[ = \frac{1}{2} \left| 0 + \frac{9}{5}(-1) + (-3)(1 + \frac{4}{5}) \right| \]
\[ = \frac{1}{2} \left| -\frac{9}{5} - 3 \cdot \frac{9}{5} \right| \]
\[ = \frac{1}{2} \left| -\frac{9}{5} - \frac{27}{5} \right| \]
\[ = \frac{1}{2} \left| -\frac{36}{5} \right| = \frac{18}{5} \]

### Final Answer:
The area of triangle PQR is \( \frac{18}{5} \) square units.

---