Let P is a point on the line `y+2x=2` and Q and R are two points on the line `3y+6x=3`. If the triangle PQR is an equilateral triangle, then its area (in sq. units) is equal to

Options

A. `(1)/(15)`
B. `(1)/(5sqrt3)`
C. `(1)/(3sqrt5)`
D. `(1)/(2sqrt(15))`

Answer

To solve the problem, we need to find the area of an equilateral triangle PQR, where point P lies on the line \(y + 2x = 2\) and points Q and R lie on the line \(3y + 6x = 3\).

### Step-by-Step Solution:

1. **Identify the lines**:
- The first line is given by \(y + 2x = 2\).
- The second line can be simplified from \(3y + 6x = 3\) to \(y + 2x = 1\).

Thus, we have:
- Line 1: \(y + 2x = 2\)
- Line 2: \(y + 2x = 1\)

These two lines are parallel since they have the same slope.

2. **Find the perpendicular distance between the two lines**:
The formula for the perpendicular distance \(d\) between two parallel lines \(Ax + By + C_1 = 0\) and \(Ax + By + C_2 = 0\) is given by:
\[ d = \frac{|C_2 - C_1|}{\sqrt{A^2 + B^2}} \]
For our lines:
- Line 1: \(y + 2x - 2 = 0\) (thus, \(C_1 = -2\))
- Line 2: \(y + 2x - 1 = 0\) (thus, \(C_2 = -1\))

Plugging in the values:
\[ d = \frac{|-1 - (-2)|}{\sqrt{2^2 + 1^2}} = \frac{1}{\sqrt{4 + 1}} = \frac{1}{\sqrt{5}} \]

3. **Understanding the triangle PQR**:
Since triangle PQR is equilateral, all sides are equal, and the height can be calculated using the formula for the height \(h\) of an equilateral triangle in terms of its side length \(s\):
\[ h = \frac{\sqrt{3}}{2}s \]
The height of the triangle is also equal to the perpendicular distance between the two parallel lines.

4. **Set the height equal to the perpendicular distance**:
Therefore, we have:
\[ \frac{\sqrt{3}}{2}s = \frac{1}{\sqrt{5}} \]
Solving for \(s\):
\[ s = \frac{2}{\sqrt{3}} \cdot \frac{1}{\sqrt{5}} = \frac{2}{\sqrt{15}} \]

5. **Calculate the area of the equilateral triangle**:
The area \(A\) of an equilateral triangle is given by:
\[ A = \frac{\sqrt{3}}{4}s^2 \]
Substituting \(s = \frac{2}{\sqrt{15}}\):
\[ A = \frac{\sqrt{3}}{4} \left(\frac{2}{\sqrt{15}}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{4}{15} = \frac{\sqrt{3}}{15} \]

### Final Answer:
The area of triangle PQR is \(\frac{\sqrt{3}}{15}\) square units.