Let `l` be the line belonging to the family of straight lines `(a + 2b)x+ (a - 3b)y +a-8b = 0, a, b in R`, which is farthest from the point `(2, 2),` then area enclosed by the line `L` and the coordinate axes is

Options

A. x+4y+7=0
B. 2x+3y+4=0
C. 4x-y-6=0
D. none of these

Explanation

To solve the problem step by step, we will follow the outlined process to find the area enclosed by the line \( L \) and the coordinate axes.

### Step 1: Write the equation of the family of lines
The given family of straight lines is:
\[ (a + 2b)x + (a - 3b)y + (a - 8b) = 0 \]

### Step 2: Rearranging the equation
We can rearrange the equation as follows:
\[ a(x + y + 1) + b(2x - 3y - 8) = 0 \]
This shows that the lines in this family will pass through the intersection of the lines defined by \( a \) and \( b \).

### Step 3: Find the intersection point of the two lines
To find the point of intersection, we can set:
1. \( 2x - 3y - 8 = 0 \) (Line 1)
2. \( 3x + 3y + 3 = 0 \) (Line 2, obtained by multiplying the first equation by 3)

Now, we solve these two equations simultaneously.

From Line 1:
\[ 2x - 3y = 8 \quad \text{(1)} \]

From Line 2:
\[ 3x + 3y = -3 \quad \text{(2)} \]

### Step 4: Solve the equations
Multiply equation (1) by 3:
\[ 6x - 9y = 24 \quad \text{(3)} \]

Now, add equations (2) and (3):
\[ 6x - 9y + 3x + 3y = 24 - 3 \]
\[ 9x - 6y = 21 \]
\[ 3x - 2y = 7 \quad \text{(4)} \]

Now, we can solve for \( x \) and \( y \) using equations (1) and (4).

From equation (4):
\[ x = \frac{7 + 2y}{3} \]

Substituting \( x \) into equation (1):
\[ 2\left(\frac{7 + 2y}{3}\right) - 3y = 8 \]
\[ \frac{14 + 4y}{3} - 3y = 8 \]
Multiply through by 3:
\[ 14 + 4y - 9y = 24 \]
\[ -5y = 10 \implies y = -2 \]

Substituting \( y = -2 \) back into equation (1):
\[ 2x - 3(-2) = 8 \]
\[ 2x + 6 = 8 \implies 2x = 2 \implies x = 1 \]

Thus, the point of intersection is \( (1, -2) \).

### Step 5: Find the slope of line \( AB \)
We need to find the line that is farthest from the point \( B(2, 2) \). The slope of line \( AB \) is given by:
\[ m_{AB} = \frac{2 - (-2)}{2 - 1} = \frac{4}{1} = 4 \]

### Step 6: Find the slope of the line perpendicular to \( AB \)
The slope of the line perpendicular to \( AB \) is:
\[ m = -\frac{1}{4} \]

### Step 7: Write the equation of the line through point \( A(1, -2) \)
Using point-slope form:
\[ y + 2 = -\frac{1}{4}(x - 1) \]
Simplifying:
\[ y + 2 = -\frac{1}{4}x + \frac{1}{4} \]
\[ 4y + 8 = -x + 1 \]
\[ x + 4y + 7 = 0 \]

### Step 8: Find the intercepts
To find the x-intercept, set \( y = 0 \):
\[ x + 4(0) + 7 = 0 \implies x = -7 \]

To find the y-intercept, set \( x = 0 \):
\[ 0 + 4y + 7 = 0 \implies 4y = -7 \implies y = -\frac{7}{4} \]

### Step 9: Calculate the area of the triangle formed by the line and the axes
The area \( A \) of the triangle formed by the x-intercept and y-intercept is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is \( 7 \) and the height is \( \frac{7}{4} \):
\[ A = \frac{1}{2} \times 7 \times \frac{7}{4} = \frac{49}{8} \]

### Final Answer
The area enclosed by the line \( L \) and the coordinate axes is:
\[ \frac{49}{8} \]