A line passes through the points `A(2,-3) and B(6,3)` . Find the slopes of the lines which are ,
(i) parallel to AB (ii) perpendicular to AB

Answer

To find the slopes of the lines that are parallel and perpendicular to the line segment AB, which passes through the points A(2, -3) and B(6, 3), we will follow these steps:

### Step 1: Find the slope of line AB
The formula for the slope (m) between two points (x1, y1) and (x2, y2) is given by:

\[ m = \frac{y2 - y1}{x2 - x1} \]

Here, the coordinates of point A are (x1, y1) = (2, -3) and the coordinates of point B are (x2, y2) = (6, 3).

Substituting the values into the formula:

\[ m = \frac{3 - (-3)}{6 - 2} = \frac{3 + 3}{6 - 2} = \frac{6}{4} = \frac{3}{2} \]

### Step 2: Find the slope of the line parallel to AB
Lines that are parallel have the same slope. Therefore, the slope of the line parallel to AB is also:

\[ m_{parallel} = \frac{3}{2} \]

### Step 3: Find the slope of the line perpendicular to AB
For two lines to be perpendicular, the product of their slopes must equal -1. If the slope of line AB is \( m = \frac{3}{2} \), then the slope of the line perpendicular to AB (let's call it \( m_{perpendicular} \)) can be found using the relationship:

\[ m_{parallel} \cdot m_{perpendicular} = -1 \]

Substituting the known slope:

\[ \frac{3}{2} \cdot m_{perpendicular} = -1 \]

To find \( m_{perpendicular} \), we rearrange the equation:

\[ m_{perpendicular} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3} \]

### Final Results
- The slope of the line parallel to AB is \( \frac{3}{2} \).
- The slope of the line perpendicular to AB is \( -\frac{2}{3} \).