To determine whether the graphs of the given pair of equations are parallel, perpendicular, or neither, we will follow these steps:
1. Determine the general form of the equations and identify the coefficients:
- The first equation is [tex]\(4x + 6y = 18\)[/tex].
- The second equation is [tex]\(3x - 2y = 5\)[/tex].
2. Find the slopes of each line:
- The general form of a linear equation is [tex]\(Ax + By = C\)[/tex].
- The slope ([tex]\(m\)[/tex]) of a line in this form is given by [tex]\(m = -\frac{A}{B}\)[/tex].
For the first equation [tex]\(4x + 6y = 18\)[/tex]:
- The coefficients are [tex]\(A = 4\)[/tex] and [tex]\(B = 6\)[/tex].
- Therefore, the slope of the first line is [tex]\(m_1 = -\frac{4}{6} = -\frac{2}{3}\)[/tex].
For the second equation [tex]\(3x - 2y = 5\)[/tex]:
- The coefficients are [tex]\(A = 3\)[/tex] and [tex]\(B = -2\)[/tex].
- Therefore, the slope of the second line is [tex]\(m_2 = -\frac{3}{-2} = \frac{3}{2}\)[/tex].
3. Compare the slopes to determine the relationship between the lines:
- Parallel lines have identical slopes.
- Perpendicular lines have slopes that are negative reciprocals of each other (i.e., the product of their slopes is -1).
Compare the slopes of the given lines:
- [tex]\(m_1 = -\frac{2}{3}\)[/tex]
- [tex]\(m_2 = \frac{3}{2}\)[/tex]
To determine if the lines are perpendicular, calculate the product of the slopes:
[tex]\[
m_1 \times m_2 = \left(-\frac{2}{3}\right) \times \left(\frac{3}{2}\right) = -1
\][/tex]
Since the product of the slopes is [tex]\(-1\)[/tex], the lines are perpendicular.
Therefore, the graphs of the equations [tex]\(4x + 6y = 18\)[/tex] and [tex]\(3x - 2y = 5\)[/tex] are perpendicular to each other.
