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

Options

A. `23x-7y+6=0`
B. `7x+4y+3=0`
C. `83x-35y+92=0`
D. none

Answer

To find the equation of the line that is bisected at the point (1, 5) between the lines \(5x - y = 4\) and \(3x + 4y - 4 = 0\), we will follow these steps:

### Step 1: Find the intersection points of the line with the given lines
Let the line we are looking for be denoted as \(L\). We need to find the points where this line intersects the two given lines.

### Step 2: Set up the coordinates of the intersection points
Let the intersection point with the line \(5x - y = 4\) be \(A(\alpha, 5\alpha - 4)\) and the intersection point with the line \(3x + 4y - 4 = 0\) be \(B(\beta, 4 - 3\beta/4)\).

### Step 3: Use the midpoint formula
Since the point (1, 5) bisects the segment \(AB\), we can use the midpoint formula:
\[ \left(\frac{\alpha + \beta}{2}, \frac{(5\alpha - 4) + (4 - \frac{3\beta}{4})}{2}\right) = (1, 5) \]

From this, we can set up two equations:
1. \(\frac{\alpha + \beta}{2} = 1\)
2. \(\frac{(5\alpha - 4) + (4 - \frac{3\beta}{4})}{2} = 5\)

### Step 4: Solve the first equation
From the first equation:
\[ \alpha + \beta = 2 \quad \text{(Equation 1)} \]

### Step 5: Solve the second equation
Multiply the second equation by 2:
\[ (5\alpha - 4) + (4 - \frac{3\beta}{4}) = 10 \]
This simplifies to:
\[ 5\alpha - \frac{3\beta}{4} = 10 \]
Multiply through by 4 to eliminate the fraction:
\[ 20\alpha - 3\beta = 40 \quad \text{(Equation 2)} \]

### Step 6: Substitute Equation 1 into Equation 2
From Equation 1, we can express \(\beta\) in terms of \(\alpha\):
\[ \beta = 2 - \alpha \]
Substituting this into Equation 2:
\[ 20\alpha - 3(2 - \alpha) = 40 \]
Expanding this gives:
\[ 20\alpha - 6 + 3\alpha = 40 \]
Combining like terms:
\[ 23\alpha - 6 = 40 \]
Adding 6 to both sides:
\[ 23\alpha = 46 \]
Dividing by 23:
\[ \alpha = 2 \]

### Step 7: Find \(\beta\)
Substituting \(\alpha = 2\) back into Equation 1:
\[ \beta = 2 - 2 = 0 \]

### Step 8: Find the coordinates of points A and B
Now we can find the coordinates of points \(A\) and \(B\):
- For \(A\):
\[ A(2, 5(2) - 4) = (2, 6) \]
- For \(B\):
\[ B(0, 4 - 3(0)/4) = (0, 4) \]

### Step 9: Find the slope of line segment AB
The slope \(m\) of line segment \(AB\) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 6}{0 - 2} = \frac{-2}{-2} = 1 \]

### Step 10: Use point-slope form to find the equation of line L
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = 1\) and point \(D(1, 5)\):
\[ y - 5 = 1(x - 1) \]
Simplifying this gives:
\[ y - 5 = x - 1 \implies y = x + 4 \]

### Step 11: Convert to standard form
Rearranging gives:
\[ x - y + 4 = 0 \quad \text{or} \quad x - y = -4 \]

### Final Equation
The final equation of the line is:
\[ x - y + 4 = 0 \]