The portion of the line `x+3y-1 =0` intersepted between the lines `ax+y+1 = 0` and `x + 3y =0` subtend a right angle at origin , then the value of |a| is

Answer

To solve the problem step by step, we will follow the outlined approach to find the value of |a|.

### Step 1: Identify the lines and their equations
We have the following lines:
1. Line 1: \( x + 3y - 1 = 0 \)
2. Line 2: \( ax + y + 1 = 0 \)
3. Line 3: \( x + 3y = 0 \)

### Step 2: Find the points of intersection
To find the points of intersection of the lines, we will solve the equations of Line 1 and Line 2, and then Line 1 and Line 3.

#### Intersection of Line 1 and Line 2:
We can rewrite Line 2 as \( y = -ax - 1 \).

Substituting \( y \) from Line 2 into Line 1:
\[ x + 3(-ax - 1) - 1 = 0 \]
\[ x - 3ax - 3 - 1 = 0 \]
\[ (1 - 3a)x - 4 = 0 \]
\[ x = \frac{4}{1 - 3a} \]

Now substituting \( x \) back into Line 1 to find \( y \):
\[ y = \frac{1 - x}{3} = \frac{1 - \frac{4}{1 - 3a}}{3} \]
\[ y = \frac{(1 - 3a) - 4}{3(1 - 3a)} = \frac{-3 - 3a}{3(1 - 3a)} = \frac{-1 - a}{1 - 3a} \]

Thus, the point of intersection \( A \) is:
\[ A\left(\frac{4}{1 - 3a}, \frac{-1 - a}{1 - 3a}\right) \]

#### Intersection of Line 1 and Line 3:
From Line 3, we can express \( y \) as:
\[ y = -\frac{1}{3}x \]

Substituting this into Line 1:
\[ x + 3\left(-\frac{1}{3}x\right) - 1 = 0 \]
\[ x - x - 1 = 0 \implies -1 = 0 \text{ (no solution)} \]

This means the lines are parallel and do not intersect. We will find the point of intersection between Line 2 and Line 3 instead.

#### Intersection of Line 2 and Line 3:
Substituting \( y = -\frac{1}{3}x \) into Line 2:
\[ ax + \left(-\frac{1}{3}x\right) + 1 = 0 \]
\[ \left(a - \frac{1}{3}\right)x + 1 = 0 \]
\[ x = -\frac{3}{3a - 1} \]

Now substituting \( x \) back to find \( y \):
\[ y = -\frac{1}{3}\left(-\frac{3}{3a - 1}\right) = \frac{1}{3a - 1} \]

Thus, the point of intersection \( B \) is:
\[ B\left(-\frac{3}{3a - 1}, \frac{1}{3a - 1}\right) \]

### Step 3: Determine the slopes of lines OA and OB
Let \( O \) be the origin \( (0, 0) \).

The slope \( m_1 \) of line \( OA \):
\[ m_1 = \frac{\frac{-1 - a}{1 - 3a}}{\frac{4}{1 - 3a}} = \frac{-1 - a}{4} \]

The slope \( m_2 \) of line \( OB \):
\[ m_2 = \frac{\frac{1}{3a - 1}}{-\frac{3}{3a - 1}} = -\frac{1}{3} \]

### Step 4: Use the right angle condition
Since the lines subtend a right angle at the origin:
\[ m_1 \cdot m_2 = -1 \]
Substituting the values:
\[ \left(\frac{-1 - a}{4}\right) \cdot \left(-\frac{1}{3}\right) = -1 \]
\[ \frac{1 + a}{12} = -1 \]
\[ 1 + a = -12 \]
\[ a = -13 \]

### Step 5: Find the value of |a|
Thus, the value of \( |a| \) is:
\[ |a| = |-13| = 13 \]

### Final Answer
The value of \( |a| \) is \( 13 \).