Determine if the lines represented by the equations \(2x + 3y = 6\) and \(4x - 6y = 12\) are parallel, perpendicular, or neither.

To determine the relationship between the lines represented by the equations \(2x + 3y = 6\) and \(4x - 6y = 12\), we can first convert each equation into slope-intercept form, \(y = mx + b\), where \(m\) is the slope. **Step 1: Convert the first equation to slope-intercept form.** Starting with the first equation: \[ 2x + 3y = 6 \] We can isolate \(y\) by first subtracting \(2x\) from both sides: \[ 3y = -2x + 6 \] Next, divide every term by \(3\): \[ y = -\frac{2}{3}x + 2 \] From this, we can see that the slope \(m_1\) of the first line is: \[ m_1 = -\frac{2}{3} \] **Step 2: Convert the second equation to slope-intercept form.** Now, let's convert the second equation: \[ 4x - 6y = 12 \] We can isolate \(y\) by first subtracting \(4x\) from both sides: \[ -6y = -4x + 12 \] Next, divide every term by \(-6\): \[ y = \frac{4}{6}x - 2 \] This simplifies to: \[ y = \frac{2}{3}x - 2 \] From this, we can see that the slope \(m_2\) of the second line is: \[ m_2 = \frac{2}{3} \] **Step 3: Determine the relationship between the slopes.** Now we have the slopes of both lines: - Slope of the first line: \(m_1 = -\frac{2}{3}\) - Slope of the second line: \(m_2 = \frac{2}{3}\) To check if the lines are parallel, we see if the slopes are equal: \[ m_1 \neq m_2 \quad \text{(since } -\frac{2}{3} \neq \frac{2}{3}\text{)} \] To check if the lines are perpendicular, we see if the product of the slopes is \(-1\): \[ m_1 \cdot m_2 = \left(-\frac{2}{3}\right) \cdot \left(\frac{2}{3}\right) = -\frac{4}{9} \] Since \(-\frac{4}{9} \neq -1\), the lines are not perpendicular. **Conclusion:** The lines represented by the equations \(2x + 3y = 6\) and \(4x - 6y = 12\) are neither parallel nor perpendicular.