The equation of the locus of the foot of perpendicular drawn from (5, 6) on the family of lines `(x-2)+lambda(y-3)=0` (where `lambda in R`) is

Options

A. `(x-1)(x-3)+(y-2)(y-6)=0`
B. `(x-5)(x-6)+(y-2)(y-6)=0`
C. `(x-2)(x-5)+(y-3)(y-6)=0`
D. `(x+2)(x+5)+(y+3)(y+6)=0`

Answer

To find the equation of the locus of the foot of the perpendicular drawn from the point (5, 6) to the family of lines given by the equation \( (x - 2) + \lambda (y - 3) = 0 \), we will follow these steps:

### Step 1: Understand the family of lines
The equation of the family of lines can be rewritten as:
\[ x - 2 + \lambda (y - 3) = 0 \]
This represents a family of lines passing through the point (2, 3) for different values of \(\lambda\).

### Step 2: Identify the foot of the perpendicular
Let the foot of the perpendicular from the point (5, 6) to the line be denoted as \( P(h, k) \). The line \( AP \) (where \( A \) is (5, 6)) is perpendicular to the line represented by the equation above.

### Step 3: Use the condition for perpendicularity
The slopes of the lines can be used to set up the condition for perpendicularity. The slope of line \( AP \) is given by:
\[ \text{slope of } AP = \frac{k - 6}{h - 5} \]
The slope of the line \( (x - 2) + \lambda (y - 3) = 0 \) can be rearranged to:
\[ y = \frac{1}{\lambda}(2 - x) + 3 \]
Thus, the slope of this line is:
\[ \text{slope of line} = -\frac{1}{\lambda} \]
For the lines to be perpendicular:
\[ \frac{k - 6}{h - 5} \cdot \left(-\frac{1}{\lambda}\right) = -1 \]
This simplifies to:
\[ (k - 6) = \lambda (h - 5) \]

### Step 4: Substitute \(\lambda\) in terms of \(h\) and \(k\)
From the equation \( (x - 2) + \lambda (y - 3) = 0 \), we can express \(\lambda\) as:
\[ \lambda = \frac{(2 - h)}{(k - 3)} \]
Substituting this back into the perpendicularity condition gives:
\[ k - 6 = \frac{(2 - h)(h - 5)}{(k - 3)} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying yields:
\[ (k - 3)(k - 6) = (2 - h)(h - 5) \]
Expanding both sides:
\[ k^2 - 9k + 18 = 2h - 10 - h^2 + 5h \]
Rearranging gives:
\[ h^2 - 7h + k^2 - 9k + 28 = 0 \]

### Step 6: Identify the locus
This equation represents a conic section. To express it in standard form, we can complete the square for both \(h\) and \(k\).

### Final Step: Write the locus equation
The equation can be rearranged to represent the locus of the foot of the perpendicular:
\[ (h - \frac{7}{2})^2 + (k - \frac{9}{2})^2 = r^2 \]
This indicates that the locus is a circle.