A variable line through the point `((6)/(5),(6)/(5))` cuts the coordinate axes at the points A and B respectively. If the point P divides AB internally in the ratio `2:1`, then the equation of the locus of P is

Options

A. `5xy=(2x+y)`
B. `5xy=2(2x+y)`
C. `5xy=(x+2y)`
D. `5xy=2(x+2y)`

Answer

To solve the problem, we need to find the locus of point P that divides the line segment AB in the ratio 2:1, where A and B are the points where a variable line through the point \((\frac{6}{5}, \frac{6}{5})\) intersects the x-axis and y-axis, respectively.

### Step-by-Step Solution:

1. **Identify Points A and B**:
- Let the line cut the x-axis at point A, which has coordinates \(A(a, 0)\).
- Let the line cut the y-axis at point B, which has coordinates \(B(0, b)\).

2. **Equation of the Line**:
- The equation of the line in intercept form is given by:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
- Since the line passes through the point \((\frac{6}{5}, \frac{6}{5})\), we can substitute these coordinates into the equation:
\[ \frac{\frac{6}{5}}{a} + \frac{\frac{6}{5}}{b} = 1 \]
- Multiplying through by \(5ab\) gives:
\[ 6b + 6a = 5ab \]
- Rearranging this, we get:
\[ \frac{1}{a} + \frac{1}{b} = \frac{5}{6} \]

3. **Coordinates of Point P**:
- Point P divides AB in the ratio 2:1. Using the section formula, the coordinates of P \((h, k)\) are given by:
\[ h = \frac{1 \cdot a + 2 \cdot 0}{1 + 2} = \frac{a}{3} \]
\[ k = \frac{1 \cdot 0 + 2 \cdot b}{1 + 2} = \frac{2b}{3} \]

4. **Expressing a and b in terms of h and k**:
- From \(h = \frac{a}{3}\), we find:
\[ a = 3h \]
- From \(k = \frac{2b}{3}\), we find:
\[ b = \frac{3k}{2} \]

5. **Substituting a and b into the equation**:
- Substitute \(a\) and \(b\) into the equation \(\frac{1}{a} + \frac{1}{b} = \frac{5}{6}\):
\[ \frac{1}{3h} + \frac{2}{3k} = \frac{5}{6} \]
- Multiply through by \(6hk\) to eliminate the denominators:
\[ 2k + 4h = 5hk \]
- Rearranging gives:
\[ 5hk - 2k - 4h = 0 \]

6. **Final Locus Equation**:
- Rearranging the above equation:
\[ 5hk - 2k = 4h \]
- Factoring out \(k\):
\[ k(5h - 2) = 4h \]
- Thus, the locus of point P is:
\[ k = \frac{4h}{5h - 2} \]

### Final Equation:
- The final equation of the locus of point P can be expressed as:
\[ 2y + 4x = 5xy \]
or equivalently,
\[ 5xy - 2y - 4x = 0 \]