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. 4/3 sq. units
B. 9/2 sq. units
C. 49/8 sq. units
D. none of these

Explanation

To solve the problem step by step, we will follow the reasoning presented in the video transcript:

### Step 1: Understand the Family of Lines
The given family of lines is:
\[ (a + 2b)x + (a - 3b)y + (a - 8b) = 0 \]
where \(a, b \in \mathbb{R}\).

### Step 2: Rearranging the Equation
We can rearrange the equation into a more manageable form:
\[ a(x + y + 1) + b(2x - 3y - 8) = 0 \]
This indicates that any line in this family passes through the intersection of the lines defined by \(a\) and \(b\).

### Step 3: Find the Intersection Point
To find the intersection point of the lines \(2x - 3y - 8 = 0\) and \(3x + 3y + 3 = 0\), we can solve these equations simultaneously.

1. From \(2x - 3y - 8 = 0\):
\[ 2x - 3y = 8 \quad \text{(1)} \]

2. From \(3x + 3y + 3 = 0\):
\[ 3x + 3y = -3 \quad \text{(2)} \]

Multiply equation (2) by 1:
\[ 3x + 3y = -3 \]

Now, we can solve equations (1) and (2):
- Multiply equation (1) by 3:
\[ 6x - 9y = 24 \quad \text{(3)} \]

- Multiply equation (2) by 2:
\[ 6x + 6y = -6 \quad \text{(4)} \]

Now, subtract equation (4) from equation (3):
\[ (6x - 9y) - (6x + 6y) = 24 + 6 \]
\[ -15y = 30 \implies y = -2 \]

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

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

### Step 4: Finding the Slope of Line AB
The point \(B\) is given as \(B(2, 2)\). The slope \(m_{AB}\) of line \(AB\) is:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - (-2)}{2 - 1} = \frac{4}{1} = 4 \]

### Step 5: Slope of the Required Line
The slope of the line we need (which is perpendicular to \(AB\)) is:
\[ m = -\frac{1}{m_{AB}} = -\frac{1}{4} \]

### Step 6: Equation of the Required Line
Using point-slope form of the line equation with point \(A(1, -2)\):
\[ y - (-2) = -\frac{1}{4}(x - 1) \]
\[ y + 2 = -\frac{1}{4}x + \frac{1}{4} \]
\[ y = -\frac{1}{4}x - \frac{7}{4} \]

Rearranging gives:
\[ x + 4y + 7 = 0 \]

### Step 7: Finding Intercepts
To find the x-intercept, set \(y = 0\):
\[ x + 4(0) + 7 = 0 \implies x = -7 \]
So the x-intercept is \((-7, 0)\).

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

### Step 8: Area of the Triangle
The area \(A\) of the triangle formed by the line and the axes is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7 \times \frac{7}{4} = \frac{49}{8} \]

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