Recall from Algebra 1 that the slope of a nonvertical line is the ratio of the vertical change (rise) to the horizontal change (run). If the line passes through any two points \((x_1, y_1)\) and \((x_2, y_2)\) on a coordinate plane, then the slope is represented by the variable \(m\), is given by
Slopes of Lines in the Coordinate Plane
Line with a positive slope: falls to the right, line o Line with a negative slope: falls to the left; line n Horizontal Line: has a zero slope; line m Vertical Line: has an undefined slope or no slope; line l |
Slopes of Parallel and Perpendicular Lines
Slopes of Parallel Lines
In a coordinate plane, two nonvertical lines are parallel if and only if they have the same slope. Any two vertical lines are parallel. Slopes of Perpendicular Lines
In a coordinate plane, two nonvertical lines are perpendicular if and only if their slopes are opposite reciprocals of each other (or the product of their slopes is -1). Horizontal lines are perpendicular to vertical lines. |
Example 1:
Find the slope of each line. Are the lines parallel?
Find the slope of each line. Are the lines parallel?
Solution
Using the slope formula for each pair of coordinates, calculate the slope for each line. The slope for \(m_2\) and \(m_3\) are the same so we can conclude \(m_2\parallel m_3\).
slope \(m_1:\Large\frac{3 - (-2)}{-6 - (-3)}=\frac{5}{-3}=-\frac{5}{3}\)
slope \(m_2:\Large\frac{9 - 1}{-5 - (-1)}= \frac{8}{-4}\normalsize = -2\)
slope \(m_3:\Large\frac{7 - 1}{0 - 3}=\frac{6}{-3}\normalsize = -2\)
Example 3:
Determine whether \(m_1\) and \(m_2\) are perpendicular.
Using the slope formula for each pair of coordinates, calculate the slope for each line. The slope for \(m_2\) and \(m_3\) are the same so we can conclude \(m_2\parallel m_3\).
slope \(m_1:\Large\frac{3 - (-2)}{-6 - (-3)}=\frac{5}{-3}=-\frac{5}{3}\)
slope \(m_2:\Large\frac{9 - 1}{-5 - (-1)}= \frac{8}{-4}\normalsize = -2\)
slope \(m_3:\Large\frac{7 - 1}{0 - 3}=\frac{6}{-3}\normalsize = -2\)
Example 3:
Determine whether \(m_1\) and \(m_2\) are perpendicular.
Using the slope formula for each pair of coordinates, calculate the slope for each line. To determine if the lines are perpendicular, we see if the slopes are opposite reciprocals.
slope \(m_1 = \Large\frac{-1 - 4}{5 - 0} = \frac{-5}{5} \normalsize = -1\)
slope \(m_2 = \Large\frac{3 - 0}{-1 - 3} = \frac{3}{-4} = - \frac{3}{4}\)
Since the slopes are not opposite reciprocals, these are not perpendicular, \(m_1\not\perp m_2\).
Linear equations can be written in many different forms. In Algebra 1, you’ve learned to write the equation of lines in slope-intercept form given the slope and y-intercept. You have also learned point-slope form that uses a given point and slope. To review writing equations of lines, use this link to the Algebra 1 Digi. (add link)
slope \(m_1 = \Large\frac{-1 - 4}{5 - 0} = \frac{-5}{5} \normalsize = -1\)
slope \(m_2 = \Large\frac{3 - 0}{-1 - 3} = \frac{3}{-4} = - \frac{3}{4}\)
Since the slopes are not opposite reciprocals, these are not perpendicular, \(m_1\not\perp m_2\).
Linear equations can be written in many different forms. In Algebra 1, you’ve learned to write the equation of lines in slope-intercept form given the slope and y-intercept. You have also learned point-slope form that uses a given point and slope. To review writing equations of lines, use this link to the Algebra 1 Digi. (add link)
Slope Intercept Form |
Point Slope Form |
\(y = mx + b\) |
\(y - y_1 = m(x - x_1)\) |
\(m =\) slope, \(b = y\)-intercept \(m =\) slope, \((x_1, y_1)\) = point
Example 4:
Write an equation of the line that passes through the point \((4, -5)\) and is parallel to \(y = -2x + 6\) in point-slope and slope-intercept forms.
Solution:
Use the slope from the parallel line \(m = -2\) because parallel lines have the same slope. Substitute the point and the slope into the point-slope equation of the line.
\(\begin{align*}\\
y - y_1 &= m(x - x_1)\\
y + 5 &= -2(x - 4)
\end{align*}\)
To write this equation in slope-intercept form, apply mathematical operations to the point-slope equation.
\(\begin{align*}\\
y + 5 &= -2(x - 4)\\
y + 5 &= -2x + 8\\
y + 5 - 5 &= -2x + 8 - 5\\
y &= -2x + 3
\end{align*}\)
Example 5:
Write an equation of a line that passes through the point \((-6, 1)\) and is perpendicular to the line \(y = \Large\frac{3}{4}\normalsize x + 6\).
Solution:
Perpendicular lines have opposite reciprocal slopes so the perpendicular slope will be \(m=-\Large\frac{4}{3}\). Substitute the point and the slope into the point-slope equation of the line.
\(\begin{align*}\\
y - y_1 &= m(x - x_1)\\
y - 1 &= -\frac{4}{3}\normalsize(x + 6)
\end{align*}\)
To write this equation in slope-intercept form, apply mathematical operations to the point-slope equation.
\(\begin{align}\\
y - 1 &= -\frac{4}{3}\normalsize(x + 6)\\
y - 1 + 1 &= -\frac{4}{3}\normalsize x - 8 + 1\\
y &= -\frac{4}{3}\normalsize x - 7
\end{align}\)
Example 6:
Write an equation of a line parallel to the graphed line.
Write an equation of the line that passes through the point \((4, -5)\) and is parallel to \(y = -2x + 6\) in point-slope and slope-intercept forms.
Solution:
Use the slope from the parallel line \(m = -2\) because parallel lines have the same slope. Substitute the point and the slope into the point-slope equation of the line.
\(\begin{align*}\\
y - y_1 &= m(x - x_1)\\
y + 5 &= -2(x - 4)
\end{align*}\)
To write this equation in slope-intercept form, apply mathematical operations to the point-slope equation.
\(\begin{align*}\\
y + 5 &= -2(x - 4)\\
y + 5 &= -2x + 8\\
y + 5 - 5 &= -2x + 8 - 5\\
y &= -2x + 3
\end{align*}\)
Example 5:
Write an equation of a line that passes through the point \((-6, 1)\) and is perpendicular to the line \(y = \Large\frac{3}{4}\normalsize x + 6\).
Solution:
Perpendicular lines have opposite reciprocal slopes so the perpendicular slope will be \(m=-\Large\frac{4}{3}\). Substitute the point and the slope into the point-slope equation of the line.
\(\begin{align*}\\
y - y_1 &= m(x - x_1)\\
y - 1 &= -\frac{4}{3}\normalsize(x + 6)
\end{align*}\)
To write this equation in slope-intercept form, apply mathematical operations to the point-slope equation.
\(\begin{align}\\
y - 1 &= -\frac{4}{3}\normalsize(x + 6)\\
y - 1 + 1 &= -\frac{4}{3}\normalsize x - 8 + 1\\
y &= -\frac{4}{3}\normalsize x - 7
\end{align}\)
Example 6:
Write an equation of a line parallel to the graphed line.
Solution:
First, find the slope of the line from the graph. The slope is \(3\) so the slope of the parallel line will have the same slope. Then the equation of a parallel line becomes \(y = 3x + b\).
The \(y-\)intercept can be any real number except 1, since that would make the same line as the one graphed, and not a parallel line. For example: \(y = 3x - 17\).
First, find the slope of the line from the graph. The slope is \(3\) so the slope of the parallel line will have the same slope. Then the equation of a parallel line becomes \(y = 3x + b\).
The \(y-\)intercept can be any real number except 1, since that would make the same line as the one graphed, and not a parallel line. For example: \(y = 3x - 17\).
Quick Check
1) Given the lines \(\overleftrightarrow{LI}, \overleftrightarrow{NE}, \overleftrightarrow{AR}\) through the points below, which lines are parallel?
\(L (-1, 4)\) and \(I (5, 2)\)
\(N (-2, -5)\) and \(E (-5, -7)\)
\(A (3, 7)\) and \(R (0, 8)\)
2) Determine whether the lines are perpendicular.
\(-2x + y = 4\)
\(y = \Large\frac{1}{2}\normalsize x - 6\)
Quick Check Solutions
1) Given the lines \(\overleftrightarrow{LI}, \overleftrightarrow{NE}, \overleftrightarrow{AR}\) through the points below, which lines are parallel?
\(L (-1, 4)\) and \(I (5, 2)\)
\(N (-2, -5)\) and \(E (-5, -7)\)
\(A (3, 7)\) and \(R (0, 8)\)
2) Determine whether the lines are perpendicular.
\(-2x + y = 4\)
\(y = \Large\frac{1}{2}\normalsize x - 6\)
Quick Check Solutions