If (a, 3a) is a variable point lying above the straight line 2x+y+4 =0 and below the line x+4y-8=0, then find the values of a.

Explanation

To solve the problem, we need to determine the values of \( a \) for which the point \( (a, 3a) \) lies above the line \( L_1: 2x + y + 4 = 0 \) and below the line \( L_2: x + 4y - 8 = 0 \).

### Step 1: Determine the condition for the point to lie above line \( L_1 \)

The line \( L_1 \) is given by the equation:
\[ 2x + y + 4 = 0 \]
To find the condition for the point \( (a, 3a) \) to lie above this line, we substitute \( x = a \) and \( y = 3a \) into the equation:
\[ 2a + 3a + 4 > 0 \]
This simplifies to:
\[ 5a + 4 > 0 \]
Now, we solve for \( a \):
\[ 5a > -4 \implies a > -\frac{4}{5} \]

### Step 2: Determine the condition for the point to lie below line \( L_2 \)

The line \( L_2 \) is given by the equation:
\[ x + 4y - 8 = 0 \]
To find the condition for the point \( (a, 3a) \) to lie below this line, we substitute \( x = a \) and \( y = 3a \) into the equation:
\[ a + 4(3a) - 8 < 0 \]
This simplifies to:
\[ a + 12a - 8 < 0 \]
Combining like terms gives:
\[ 13a - 8 < 0 \]
Now, we solve for \( a \):
\[ 13a < 8 \implies a < \frac{8}{13} \]

### Step 3: Combine the inequalities

From the two conditions we derived:
1. \( a > -\frac{4}{5} \)
2. \( a < \frac{8}{13} \)

We can combine these inequalities to find the range of \( a \):
\[ -\frac{4}{5} < a < \frac{8}{13} \]

### Final Answer

Thus, the values of \( a \) for which the point \( (a, 3a) \) lies above the line \( L_1 \) and below the line \( L_2 \) are:
\[ -\frac{4}{5} < a < \frac{8}{13} \]