4.5 Use the slope–intercept form of an equation of a line (Page 7/12)

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Use slopes and y -intercepts to determine if the lines $x = 1$ and $x = −5$ are parallel.

parallel

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Use slopes and y -intercepts to determine if the lines $x = 8$ and $x = −6$ are parallel.

parallel

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Use slopes and y -intercepts to determine if the lines $y = 2 x - 3$ and $−6 x + 3 y = −9$ are parallel. You may want to graph these lines, too, to see what they look like.

Solution

$\begin{matrix} \begin{matrix} The first equation is already in slope–intercept form. \\ Solve the second equation for y . \end{matrix} & \begin{matrix} y & = & 2 x - 3 \\ y & = & 2 x - 3 \\ - 6 x + 3 y & = & −9 \\ 3 y & = & 6 x - 9 \\ \frac{3 y}{3} & = & \frac{6 x - 9}{3} \end{matrix} & \begin{matrix} and \end{matrix} & \begin{matrix} −6 x + 3 y = −9 \end{matrix} \\ \begin{matrix} The second equation is now in \\ slope–intercept form. \\ Identify the slope and y -intercept of both lines. \end{matrix} & \begin{matrix} y & = & 2 x - 3 \\ y & = & 2 x - 3 \\ y & = & m x + b \\ m & = & 2 \end{matrix} & \begin{matrix} y & = & 2 x - 3 \\ y & = & m x + b \\ m & = & 2 \end{matrix} \\ y -intercept is (0 ,−3) & y -intercept is (0, −3) \end{matrix}$

The lines have the same slope, but they also have the same y -intercepts. Their equations represent the same line. They are not parallel; they are the same line.

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Use slopes and y -intercepts to determine if the lines $y = - \frac{1}{2} x - 1$ and $x + 2 y = 2$ are parallel.

not parallel; same line

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Use slopes and y -intercepts to determine if the lines $y = \frac{3}{4} x - 3$ and $3 x - 4 y = 12$ are parallel.

not parallel; same line

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Use slopes to identify perpendicular lines

Let’s look at the lines whose equations are $y = \frac{1}{4} x - 1$ and $y = −4 x + 2$ , shown in [link] .

The figure shows two lines graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 8 to 8. The y-axis of the plane runs from negative 8 to 8. One line is labeled with the equation y equals negative 4x plus 2 and goes through the points (0,2) and (1, negative 2). The other line is labeled with the equation y equals one fourth x minus 1 and goes through the points (0, negative 1) and (4,0).

These lines lie in the same plane and intersect in right angles. We call these lines perpendicular .

What do you notice about the slopes of these two lines? As we read from left to right, the line $y = \frac{1}{4} x - 1$ rises, so its slope is positive. The line $y = −4 x + 2$ drops from left to right, so it has a negative slope. Does it make sense to you that the slopes of two perpendicular lines will have opposite signs?

If we look at the slope of the first line, $m_{1} = \frac{1}{4}$ , and the slope of the second line, $m_{2} = −4$ , we can see that they are negative reciprocals of each other. If we multiply them, their product is $−1 .$

\begin{matrix} m_{1} \cdot m_{2} \\ \frac{1}{4} (−4) \\ - 1 \end{matrix}

This is always true for perpendicular lines and leads us to this definition.

Perpendicular lines

Perpendicular lines are lines in the same plane that form a right angle.

If $m_{1} and m_{2}$ are the slopes of two perpendicular lines, then:

m_{1} \cdot m_{2} = −1 and m_{1} = \frac{−1}{m_{2}}

Vertical lines and horizontal lines are always perpendicular to each other.

We were able to look at the slope–intercept form of linear equations and determine whether or not the lines were parallel. We can do the same thing for perpendicular lines.

We find the slope–intercept form of the equation, and then see if the slopes are negative reciprocals. If the product of the slopes is $−1$ , the lines are perpendicular. Perpendicular lines may have the same y -intercepts.

Use slopes to determine if the lines, $y = −5 x - 4$ and $x - 5 y = 5$ are perpendicular.

Solution

$\begin{matrix} The first equation is in slope–intercept form. & y & = & −5 x - 4 \\ Solve the second equation for y . & x - 5 y & = & 5 \\ - 5 y & = & − x + 5 \\ \frac{−5 y}{−5} & = & \frac{− x + 5}{−5} \\ y & = & \frac{1}{5} x - 1 \\ Identify the slope of each line. & y & = & −5 x - 4 & y & = & \frac{1}{5} x - 1 \\ y & = & m x + b & y & = & m x + b \\ m_{1} & = & −5 & m_{2} & = & \frac{1}{5} \end{matrix}$

The slopes are negative reciprocals of each other, so the lines are perpendicular. We check by multiplying the slopes,

\begin{matrix} m_{1} \cdot m_{2} \\ - 5 (\frac{1}{5}) \\ - 1 ✓ \end{matrix}

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Use slopes to determine if the lines $y = −3 x + 2$ and $x - 3 y = 4$ are perpendicular.

perpendicular

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Use slopes to determine if the lines $y = 2 x - 5$ and $x + 2 y = −6$ are perpendicular.

perpendicular

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Use slopes to determine if the lines, $7 x + 2 y = 3$ and $2 x + 7 y = 5$ are perpendicular.

Solution

$\begin{matrix} Solve the equations for y . & 7 x + 2 y & = & 3 & 2 x + 7 y & = & 5 \\ 2 y & = & −7 x + 3 & 7 y & = & −2 x + 5 \\ \frac{2 y}{2} & = & \frac{−7 x + 3}{2} & \frac{7 y}{7} & = & \frac{−2 x + 5}{7} \\ y & = & - \frac{7}{2} x + \frac{3}{2} & y & = & - \frac{2}{7} x + \frac{5}{7} \\ Identify the slope of each line. & y & = & m x + b & y & = & m x + b \\ m_{1} & = & - \frac{7}{2} & m_{2} & = & - \frac{2}{7} \end{matrix}$

The slopes are reciprocals of each other, but they have the same sign. Since they are not negative reciprocals, the lines are not perpendicular.

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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