The line x-4y =6` is the perpendicular bisector of the line segment AB. If B= (1, 3), find the co-ordinates of point A.

Answer

To find the coordinates of point A given that the line \( x - 4y = 6 \) is the perpendicular bisector of the line segment AB, where point B is \( (1, 3) \), we can follow these steps:

### Step 1: Understand the properties of the perpendicular bisector
The line \( x - 4y = 6 \) is the perpendicular bisector of the segment AB. This means that:
1. The midpoint M of segment AB lies on the line \( x - 4y = 6 \).
2. The slope of line AB is the negative reciprocal of the slope of the line \( x - 4y = 6 \).

### Step 2: Find the slope of the given line
The equation of the line can be rearranged into slope-intercept form \( y = mx + c \):
\[ x - 4y = 6 \implies 4y = x - 6 \implies y = \frac{1}{4}x - \frac{3}{2} \]
Thus, the slope \( m_1 \) of the line is \( \frac{1}{4} \).

### Step 3: Find the slope of line AB
Since the line \( x - 4y = 6 \) is the perpendicular bisector, the slope \( m_2 \) of line AB is the negative reciprocal of \( m_1 \):
\[ m_2 = -\frac{1}{m_1} = -4 \]

### Step 4: Write the equation of line AB
Using point B \( (1, 3) \) and the slope \( m_2 = -4 \), we can use the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (1, 3) \):
\[ y - 3 = -4(x - 1) \]
Expanding this:
\[ y - 3 = -4x + 4 \implies y = -4x + 7 \]

### Step 5: Find the midpoint M of segment AB
Let the coordinates of point A be \( (a, b) \). The midpoint M of segment AB is given by:
\[ M = \left( \frac{a + 1}{2}, \frac{b + 3}{2} \right) \]
Since M lies on the line \( x - 4y = 6 \), we substitute the coordinates of M into the line equation:
\[ \frac{a + 1}{2} - 4\left(\frac{b + 3}{2}\right) = 6 \]
Multiplying through by 2 to eliminate the fractions:
\[ a + 1 - 4(b + 3) = 12 \]
Simplifying:
\[ a + 1 - 4b - 12 = 12 \implies a - 4b - 11 = 0 \implies a - 4b = 11 \quad \text{(Equation 1)} \]

### Step 6: Use the slope of line AB
From the slope of line AB, we have:
\[ \frac{b - 3}{a - 1} = -4 \]
Cross-multiplying gives:
\[ b - 3 = -4(a - 1) \implies b - 3 = -4a + 4 \implies b + 4a = 7 \quad \text{(Equation 2)} \]

### Step 7: Solve the system of equations
Now we have a system of two equations:
1. \( a - 4b = 11 \)
2. \( 4a + b = 7 \)

We can solve these equations simultaneously. From Equation 1, we can express \( a \) in terms of \( b \):
\[ a = 4b + 11 \]
Substituting this into Equation 2:
\[ 4(4b + 11) + b = 7 \implies 16b + 44 + b = 7 \implies 17b + 44 = 7 \implies 17b = 7 - 44 \implies 17b = -37 \implies b = -\frac{37}{17} \]

Substituting \( b \) back to find \( a \):
\[ a = 4\left(-\frac{37}{17}\right) + 11 = -\frac{148}{17} + \frac{187}{17} = \frac{39}{17} \]

### Final Coordinates
Thus, the coordinates of point A are:
\[ A = \left( \frac{39}{17}, -\frac{37}{17} \right) \]