A line L is such that its segment between the straight lines 5x - y - 4 = 0 and 3 x + 4y - 4 = 0 is bisected at the point (1,5) . Obtain the equation.

Answer

To find the equation of the line \( L \) that is bisected at the point \( (1, 5) \) between the two given lines, we will follow these steps:

### Step 1: Identify the equations of the given lines
The equations of the two lines are:
1. \( 5x - y - 4 = 0 \) (Line 1)
2. \( 3x + 4y - 4 = 0 \) (Line 2)

### Step 2: Find the points of intersection of the line \( L \) with the given lines
Let the points where line \( L \) intersects Line 1 and Line 2 be \( A(\alpha_1, \beta_1) \) and \( B(\alpha_2, \beta_2) \) respectively.

### Step 3: Use the midpoint formula
Since the midpoint \( P(1, 5) \) bisects the segment \( AB \), we can use the midpoint formula:
\[ \left( \frac{\alpha_1 + \alpha_2}{2}, \frac{\beta_1 + \beta_2}{2} \right) = (1, 5) \]
From this, we can derive two equations:
1. \( \alpha_1 + \alpha_2 = 2 \) (Equation 1)
2. \( \beta_1 + \beta_2 = 10 \) (Equation 2)

### Step 4: Express \( \alpha_2 \) and \( \beta_2 \) in terms of \( \alpha_1 \) and \( \beta_1 \)
From Equation 1:
\[ \alpha_2 = 2 - \alpha_1 \]
From Equation 2:
\[ \beta_2 = 10 - \beta_1 \]

### Step 5: Substitute \( \alpha_2 \) and \( \beta_2 \) into the equations of the lines
Since point \( A \) lies on Line 1:
\[ 5\alpha_1 - \beta_1 - 4 = 0 \implies 5\alpha_1 - \beta_1 = 4 \quad \text{(Equation 3)} \]

Since point \( B \) lies on Line 2:
\[ 3\alpha_2 + 4\beta_2 - 4 = 0 \]
Substituting \( \alpha_2 \) and \( \beta_2 \):
\[ 3(2 - \alpha_1) + 4(10 - \beta_1) - 4 = 0 \]
Expanding this gives:
\[ 6 - 3\alpha_1 + 40 - 4\beta_1 - 4 = 0 \implies 42 - 3\alpha_1 - 4\beta_1 = 0 \]
Rearranging gives:
\[ 3\alpha_1 + 4\beta_1 = 42 \quad \text{(Equation 4)} \]

### Step 6: Solve the system of equations
Now we have two equations:
1. \( 5\alpha_1 - \beta_1 = 4 \) (Equation 3)
2. \( 3\alpha_1 + 4\beta_1 = 42 \) (Equation 4)

We can solve these equations simultaneously. From Equation 3, we can express \( \beta_1 \):
\[ \beta_1 = 5\alpha_1 - 4 \]
Substituting this into Equation 4:
\[ 3\alpha_1 + 4(5\alpha_1 - 4) = 42 \]
Expanding gives:
\[ 3\alpha_1 + 20\alpha_1 - 16 = 42 \implies 23\alpha_1 = 58 \implies \alpha_1 = \frac{58}{23} = \frac{26}{23} \]
Now substituting \( \alpha_1 \) back to find \( \beta_1 \):
\[ \beta_1 = 5\left(\frac{26}{23}\right) - 4 = \frac{130}{23} - \frac{92}{23} = \frac{38}{23} \]

### Step 7: Find \( \alpha_2 \) and \( \beta_2 \)
Using \( \alpha_1 \) and \( \beta_1 \):
\[ \alpha_2 = 2 - \frac{26}{23} = \frac{46}{23} - \frac{26}{23} = \frac{20}{23} \]
\[ \beta_2 = 10 - \frac{38}{23} = \frac{230}{23} - \frac{38}{23} = \frac{192}{23} \]

### Step 8: Find the equation of line \( L \)
The line passes through points \( (1, 5) \) and \( \left(\frac{26}{23}, \frac{38}{23}\right) \). Using the point-slope form of the equation of a line:
\[ y - 5 = \frac{\beta_1 - 5}{\alpha_1 - 1}(x - 1) \]
Substituting \( \alpha_1 \) and \( \beta_1 \):
\[ y - 5 = \frac{\frac{38}{23} - 5}{\frac{26}{23} - 1}(x - 1) \]
Calculating the slope:
\[ = \frac{\frac{38}{23} - \frac{115}{23}}{\frac{26}{23} - \frac{23}{23}} = \frac{-\frac{77}{23}}{\frac{3}{23}} = -\frac{77}{3} \]
Thus, the equation becomes:
\[ y - 5 = -\frac{77}{3}(x - 1) \]
Rearranging gives:
\[ 3y - 15 = -77x + 77 \implies 77x + 3y = 92 \]

### Final Equation
The equation of line \( L \) is:
\[ 77x - 3y = 92 \]