From any point P on the line x = 2y perpendicular is drawn on y = x. Let foot of perpendicular is Q. Find the locus of mid point of PQ.

Options

A. 5x - 7y = 0
B. 2x - 3y = 0
C. 3x - 2y = 0
D. 7x - 5y = 0

Answer

To find the locus of the midpoint of segment PQ, where P is a point on the line \( x = 2y \) and Q is the foot of the perpendicular from P to the line \( y = x \), we can follow these steps:

### Step 1: Define the Point P
Let the coordinates of point P on the line \( x = 2y \) be represented as:
\[ P(2\alpha, \alpha) \]
Here, \( \alpha \) is a parameter that varies along the line.

### Step 2: Find the Foot of the Perpendicular Q
The line \( y = x \) has a slope of 1. The slope of the line perpendicular to this will be -1. Therefore, the equation of the line passing through P and perpendicular to \( y = x \) can be expressed as:
\[ y - \alpha = -1(x - 2\alpha) \]
This simplifies to:
\[ y = -x + 3\alpha \]

### Step 3: Find the Intersection Point Q
To find the coordinates of Q, we need to solve the equations of the line \( y = x \) and the perpendicular line:
\[ x = -x + 3\alpha \]
Solving for \( x \):
\[ 2x = 3\alpha \implies x = \frac{3\alpha}{2} \]
Substituting back to find \( y \):
\[ y = x = \frac{3\alpha}{2} \]
Thus, the coordinates of Q are:
\[ Q\left(\frac{3\alpha}{2}, \frac{3\alpha}{2}\right) \]

### Step 4: Find the Midpoint M of PQ
The midpoint M of the segment PQ can be calculated as:
\[ M\left(\frac{2\alpha + \frac{3\alpha}{2}}{2}, \frac{\alpha + \frac{3\alpha}{2}}{2}\right) \]
Calculating the x-coordinate of M:
\[ M_x = \frac{2\alpha + \frac{3\alpha}{2}}{2} = \frac{\frac{4\alpha}{2} + \frac{3\alpha}{2}}{2} = \frac{\frac{7\alpha}{2}}{2} = \frac{7\alpha}{4} \]
Calculating the y-coordinate of M:
\[ M_y = \frac{\alpha + \frac{3\alpha}{2}}{2} = \frac{\frac{2\alpha}{2} + \frac{3\alpha}{2}}{2} = \frac{\frac{5\alpha}{2}}{2} = \frac{5\alpha}{4} \]
Thus, the coordinates of M are:
\[ M\left(\frac{7\alpha}{4}, \frac{5\alpha}{4}\right) \]

### Step 5: Eliminate the Parameter α
Let \( h = \frac{7\alpha}{4} \) and \( k = \frac{5\alpha}{4} \). We can express \( \alpha \) in terms of \( h \) and \( k \):
\[ \alpha = \frac{4h}{7} \quad \text{and} \quad \alpha = \frac{4k}{5} \]
Setting these equal gives:
\[ \frac{4h}{7} = \frac{4k}{5} \]
Cross-multiplying:
\[ 5h = 7k \]
This can be rearranged to:
\[ 5h - 7k = 0 \]

### Final Answer
The locus of the midpoint M of PQ is given by the equation:
\[ 5x - 7y = 0 \]