Let AB be a line segment of length 4 with A on the line `y = 2x` and B on the line `y = x`. The locus of the middle point of the line segment is

Options

A. (a) `25x^(2)+13y^(2)+36xy-4=0`
B. (b) `25x^(2)+13y^(2)-36xy-4=0`
C. (c) `13x^(2)+25y^(2)-36xy-4=0`
D. (d) `13x^(2)+25y^(2)+36xy-4=0`

Explanation

To find the locus of the midpoint of the line segment AB, where A lies on the line \(y = 2x\) and B lies on the line \(y = x\), we can follow these steps:

### Step 1: Define the coordinates of points A and B
Let the coordinates of point A be \(A(x_1, y_1)\) and point B be \(B(x_2, y_2)\).
Since A lies on the line \(y = 2x\), we have:
\[ y_1 = 2x_1 \quad \text{(1)} \]
Since B lies on the line \(y = x\), we have:
\[ y_2 = x_2 \quad \text{(2)} \]

### Step 2: Find the midpoint M of segment AB
The coordinates of the midpoint M of segment AB are given by:
\[ M\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) \]
Let \(h = \frac{x_1 + x_2}{2}\) and \(k = \frac{y_1 + y_2}{2}\).

### Step 3: Express \(x_2\) and \(y_2\) in terms of \(h\) and \(k\)
From the midpoint formulas, we can express \(x_2\) and \(y_2\) as:
\[ x_2 = 2h - x_1 \quad \text{(3)} \]
\[ y_2 = 2k - y_1 \quad \text{(4)} \]

### Step 4: Set up the distance condition
The length of segment AB is given as 4. Therefore, we can use the distance formula:
\[ \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} = 4 \]
Squaring both sides gives:
\[ (x_1 - x_2)^2 + (y_1 - y_2)^2 = 16 \]

### Step 5: Substitute \(x_2\) and \(y_2\)
Substituting \(x_2\) and \(y_2\) from equations (3) and (4) into the distance equation:
\[ (x_1 - (2h - x_1))^2 + (y_1 - (2k - y_1))^2 = 16 \]
This simplifies to:
\[ (2x_1 - 2h)^2 + (2y_1 - 2k)^2 = 16 \]
Dividing through by 4 gives:
\[ (x_1 - h)^2 + (y_1 - k)^2 = 4 \]

### Step 6: Substitute \(y_1\)
Using equation (1) \(y_1 = 2x_1\), we can express \(y_1\) in terms of \(h\) and \(k\):
\[ (2x_1 - k)^2 + (2x_1 - k)^2 = 4 \]
This leads to a relationship between \(h\) and \(k\).

### Step 7: Final equation
After substituting and simplifying, we arrive at the equation of the locus:
\[ 13k^2 + 25h^2 - 36kh - 4 = 0 \]
Replacing \(h\) with \(x\) and \(k\) with \(y\):
\[ 13y^2 + 25x^2 - 36xy - 4 = 0 \]

### Conclusion
Thus, the locus of the midpoint of the line segment AB is given by:
\[ 13y^2 + 25x^2 - 36xy - 4 = 0 \]