Knight Turnbull
01/10/2024 · Junior High School
Determine if the lines represented by the equations \(2x + 3y = 6\) and \(4x - 6y = 12\) are parallel, perpendicular, or neither.
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Step-by-step Solution
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 \(y\):
\[
3y = 6 - 2x
\]
\[
3y = -2x + 6
\]
Now, divide by 3:
\[
y = -\frac{2}{3}x + 2
\]
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, for the second equation:
\[
4x - 6y = 12
\]
We isolate \(y\):
\[
-6y = 12 - 4x
\]
\[
-6y = -4x + 12
\]
Now, divide by -6:
\[
y = \frac{4}{6}x - 2
\]
Simplifying \(\frac{4}{6}\) gives:
\[
y = \frac{2}{3}x - 2
\]
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:
- For the first line: \(m_1 = -\frac{2}{3}\)
- For 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.
Quick Answer
The lines are neither parallel nor perpendicular.
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