For each pair of linear equations given below, drawn graphs and then state, whether the lines drawn are parallel or perpendicular to each other.
`2x-3y=6`
`x/2+y/3=1`

Answer

To determine whether the lines represented by the equations \(2x - 3y = 6\) and \(\frac{x}{2} + \frac{y}{3} = 1\) are parallel or perpendicular, we will follow these steps:

### Step 1: Convert the equations to slope-intercept form (y = mx + b)

**Equation 1: \(2x - 3y = 6\)**

1. Rearrange the equation to isolate \(y\):
\[ -3y = -2x + 6 \]
\[ y = \frac{2}{3}x - 2 \]
Here, the slope \(m_1 = \frac{2}{3}\).

**Equation 2: \(\frac{x}{2} + \frac{y}{3} = 1\)**

2. Multiply through by 6 to eliminate the fractions:
\[ 3x + 2y = 6 \]
Rearranging gives:
\[ 2y = -3x + 6 \]
\[ y = -\frac{3}{2}x + 3 \]
Here, the slope \(m_2 = -\frac{3}{2}\).

### Step 2: Determine the relationship between the slopes

3. Check if the lines are parallel or perpendicular:
- Two lines are parallel if their slopes are equal: \(m_1 = m_2\).
- Two lines are perpendicular if the product of their slopes is \(-1\): \(m_1 \cdot m_2 = -1\).

Calculating the product of the slopes:
\[ m_1 \cdot m_2 = \left(\frac{2}{3}\right) \cdot \left(-\frac{3}{2}\right) = -1 \]

### Conclusion

Since the product of the slopes is \(-1\), the lines are perpendicular to each other.

### Summary of Steps:
1. Convert each equation to slope-intercept form.
2. Identify the slopes of both lines.
3. Check the relationship between the slopes to determine if they are parallel or perpendicular.