Question

Line s passes through points (7, 1) and (10, 6). Line t passes through points (9, 6) and (4, 9). Are line s and line t parallel or perpendicular?

Answer 1

To determine whether lines \( s \) and \( t \) are parallel, perpendicular, or neither, we need to calculate the slopes of both lines.

### Finding the slope of line \( s \)

Line \( s \) passes through the points \((7, 1)\) and \((10, 6)\). The formula for the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

Substituting in the points for line \( s \):

\[
m_s = \frac{6 - 1}{10 - 7} = \frac{5}{3}
\]

### Finding the slope of line \( t \)

Line \( t \) passes through the points \((9, 6)\) and \((4, 9)\). Using the same slope formula:

\[
m_t = \frac{9 - 6}{4 - 9} = \frac{3}{-5} = -\frac{3}{5}
\]

### Determining the relationship between the slopes

Now that we have the slopes:
- Slope of line \( s \): \( m_s = \frac{5}{3} \)
- Slope of line \( t \): \( m_t = -\frac{3}{5} \)

Two lines are:
- **Parallel** if their slopes are equal (\( m_s = m_t \)).
- **Perpendicular** if the product of their slopes is \(-1\) (\( m_s \cdot m_t = -1 \)).

Calculating the product of the slopes:

\[
m_s \cdot m_t = \left(\frac{5}{3}\right) \cdot \left(-\frac{3}{5}\right) = -1
\]

Since the product of the slopes is \(-1\), the lines \( s \) and \( t \) are **perpendicular**.