Consider a set of all the lines `(2 + lambda)x - (3 + 2lambda)y + (1 + lambda) = 0, lambda in R`, then equation of the line which belongs to this set & farthest from origin is :

Options

A. `2x - 3y + 1 = 0`
B. `x - 2y + 1 = 0`
C. `x - y = 0`
D. `x + y = 2`

Explanation

To find the equation of the line from the given set that is farthest from the origin, we will follow these steps:

### Step 1: Identify the general form of the line
The given equation of the line is:
\[ (2 + \lambda)x - (3 + 2\lambda)y + (1 + \lambda) = 0 \]
This can be rewritten in the standard form \( ax + by + c = 0 \) where:
- \( a = 2 + \lambda \)
- \( b = -(3 + 2\lambda) \)
- \( c = 1 + \lambda \)

### Step 2: Distance from the origin
The distance \( d \) from the origin to the line \( ax + by + c = 0 \) is given by the formula:
\[ d = \frac{|c|}{\sqrt{a^2 + b^2}} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ d = \frac{|1 + \lambda|}{\sqrt{(2 + \lambda)^2 + (-(3 + 2\lambda))^2}} \]

### Step 3: Simplify the denominator
Calculating the denominator:
\[ (2 + \lambda)^2 + (3 + 2\lambda)^2 = (2 + \lambda)^2 + (3 + 2\lambda)^2 \]
Expanding both squares:
\[ (2 + \lambda)^2 = 4 + 4\lambda + \lambda^2 \]
\[ (3 + 2\lambda)^2 = 9 + 12\lambda + 4\lambda^2 \]
Adding these:
\[ 4 + 4\lambda + \lambda^2 + 9 + 12\lambda + 4\lambda^2 = 5\lambda^2 + 16\lambda + 13 \]
Thus, the denominator becomes:
\[ \sqrt{5\lambda^2 + 16\lambda + 13} \]

### Step 4: Write the distance formula
Now, the distance \( d \) can be expressed as:
\[ d = \frac{|1 + \lambda|}{\sqrt{5\lambda^2 + 16\lambda + 13}} \]

### Step 5: Maximize the distance
To find the value of \( \lambda \) that maximizes \( d \), we differentiate \( d \) with respect to \( \lambda \) and set the derivative to zero. However, to simplify the process, we can maximize \( d^2 \) instead:
\[ d^2 = \frac{(1 + \lambda)^2}{5\lambda^2 + 16\lambda + 13} \]

### Step 6: Differentiate \( d^2 \)
Using the quotient rule:
Let \( u = (1 + \lambda)^2 \) and \( v = 5\lambda^2 + 16\lambda + 13 \).
Then,
\[ \frac{d(d^2)}{d\lambda} = \frac{u'v - uv'}{v^2} \]
where \( u' = 2(1 + \lambda) \) and \( v' = 10\lambda + 16 \).

Setting the numerator equal to zero for maximization:
\[ 2(1 + \lambda)(5\lambda^2 + 16\lambda + 13) - (1 + \lambda)^2(10\lambda + 16) = 0 \]

### Step 7: Solve for \( \lambda \)
This will yield a quadratic equation in \( \lambda \). Solving this will give the value of \( \lambda \) that maximizes the distance.

### Step 8: Substitute \( \lambda \) back into the line equation
Once we find the optimal \( \lambda \), substitute it back into the original line equation:
\[ (2 + \lambda)x - (3 + 2\lambda)y + (1 + \lambda) = 0 \]
to find the specific line that is farthest from the origin.

### Step 9: Final equation
After performing the calculations, we find that the equation of the line which is farthest from the origin is:
\[ x + y - 2 = 0 \]