The perpendicular bisector of the line segment joining the points `A(1,5)and B(4,6)` cuts the y-axis at which point?

Answer

To find the point where the perpendicular bisector of the line segment joining points A(1,5) and B(4,6) cuts the y-axis, we will follow these steps:

### Step 1: Find the Midpoint of AB
The midpoint \( P \) of the line segment \( AB \) can be calculated using the midpoint formula:
\[ P\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of points \( A(1, 5) \) and \( B(4, 6) \):
\[ P\left( \frac{1 + 4}{2}, \frac{5 + 6}{2} \right) = P\left( \frac{5}{2}, \frac{11}{2} \right) \]

### Step 2: Calculate the Slope of AB
The slope \( m_{AB} \) of the line segment \( AB \) is given by:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 5}{4 - 1} = \frac{1}{3} \]

### Step 3: Find the Slope of the Perpendicular Bisector
The slope of the perpendicular bisector \( m_L \) is the negative reciprocal of the slope of \( AB \):
\[ m_L = -\frac{1}{m_{AB}} = -\frac{1}{\frac{1}{3}} = -3 \]

### Step 4: Write the Equation of the Perpendicular Bisector
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting the midpoint \( P\left( \frac{5}{2}, \frac{11}{2} \right) \) and the slope \( -3 \):
\[ y - \frac{11}{2} = -3\left(x - \frac{5}{2}\right) \]
Expanding this:
\[ y - \frac{11}{2} = -3x + \frac{15}{2} \]
Rearranging gives:
\[ y = -3x + \frac{15}{2} + \frac{11}{2} = -3x + 13 \]

### Step 5: Find the Intersection with the Y-axis
The y-axis is defined by \( x = 0 \). To find the intersection point \( M \):
\[ y = -3(0) + 13 = 13 \]
Thus, the coordinates of point \( M \) where the perpendicular bisector cuts the y-axis are \( (0, 13) \).

### Final Answer
The perpendicular bisector of the line segment joining points A(1,5) and B(4,6) cuts the y-axis at the point \( (0, 13) \).

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