A variable line through the point `(6/5, 6/5)` cuts the coordinates axes in the point `A and B`. If the point `P` divides `AB` internally in the ratio `2:1`, show that the equation to the locus of `P` is :

Options

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

Explanation

To find the locus of the point \( P \) that divides the line segment \( AB \) in the ratio \( 2:1 \), we can follow these steps:

### Step 1: Define the variable line
Let the variable line that passes through the point \( \left(\frac{6}{5}, \frac{6}{5}\right) \) be represented in intercept form as:
\[ \frac{x}{A} + \frac{y}{B} = 1 \]
where \( A \) and \( B \) are the x-intercept and y-intercept of the line, respectively.

### Step 2: Identify the coordinates of points A and B
From the intercept form, the coordinates of point \( A \) (where the line intersects the x-axis) are \( (A, 0) \) and the coordinates of point \( B \) (where the line intersects the y-axis) are \( (0, B) \).

### Step 3: Use the section formula to find the coordinates of point P
Let the coordinates of point \( P \) be \( (H, K) \). Since \( P \) divides \( AB \) in the ratio \( 2:1 \), we can use the section formula:
\[ H = \frac{2 \cdot 0 + 1 \cdot A}{2 + 1} = \frac{A}{3} \]
\[ K = \frac{2 \cdot B + 1 \cdot 0}{2 + 1} = \frac{2B}{3} \]

### Step 4: Express A and B in terms of H and K
From the equations derived:
\[ A = 3H \]
\[ B = \frac{3K}{2} \]

### Step 5: Substitute A and B into the line equation
Since the line passes through the point \( \left(\frac{6}{5}, \frac{6}{5}\right) \), we can substitute \( A \) and \( B \) into the line equation:
\[ \frac{6/5}{A} + \frac{6/5}{B} = 1 \]
Substituting \( A \) and \( B \):
\[ \frac{6/5}{3H} + \frac{6/5}{\frac{3K}{2}} = 1 \]
This simplifies to:
\[ \frac{6}{15H} + \frac{12}{15K} = 1 \]
Multiplying through by 15 gives:
\[ 6K + 12H = 15HK \]

### Step 6: Rearranging the equation
Rearranging the equation gives:
\[ 15HK - 6K - 12H = 0 \]
Factoring out common terms:
\[ 3(5HK - 2K - 4H) = 0 \]
Thus, we have:
\[ 5HK - 2K - 4H = 0 \]

### Step 7: Final equation for the locus
To express this in terms of \( x \) and \( y \) (where \( H = x \) and \( K = y \)):
\[ 5xy - 2y - 4x = 0 \]
This is the equation of the locus of point \( P \).