The points (x,2x),(2y,y) and (3,3) are collinear

Options

A. for all values of (x,y)
B. 2 is A.M. of x,y
C. 2 is G.M. of x,y
D. 2 is H.M. of x,y

Answer

To determine if the points \((x, 2x)\), \((2y, y)\), and \((3, 3)\) are collinear, we can use the concept of the area of a triangle formed by these points. If the area is zero, then the points are collinear.

### Step-by-Step Solution:

1. **Identify the Points:**
We have three points:
- A: \((x, 2x)\)
- B: \((2y, y)\)
- C: \((3, 3)\)

2. **Use the Area Formula:**
The area \(A\) of the triangle formed by points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
For our points, substituting the coordinates:
\[ A = \frac{1}{2} \left| x(y - 3) + 2y(3 - 2x) + 3(2x - y) \right| \]

3. **Substituting the Coordinates:**
Plugging in the coordinates:
\[ A = \frac{1}{2} \left| x(y - 3) + 2y(3 - 2x) + 3(2x - y) \right| \]

4. **Expanding the Expression:**
Now, simplify the expression inside the absolute value:
\[ A = \frac{1}{2} \left| xy - 3x + 6y - 4xy + 6x - 3y \right| \]
Combine like terms:
\[ A = \frac{1}{2} \left| -3xy + 3x + 3y \right| \]
Factor out the common factor:
\[ A = \frac{3}{2} \left| -xy + x + y \right| \]

5. **Setting the Area to Zero:**
For the points to be collinear, the area must be zero:
\[ -xy + x + y = 0 \]
Rearranging gives:
\[ xy = x + y \]

6. **Conclusion:**
The condition \(xy = x + y\) is the equation that must hold for the points to be collinear.