4.6 Find the equation of a line (Page 3/7)

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Find an equation of a line given two points

Find an equation of a line that contains the points $(5, 4)$ and $(3, 6)$ . Write the equation in slope–intercept form.

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. In the first row of the table, the first cell on the left reads: “Step 1. Find the slope using the given points.” The text in the second cell reads: “To use the point-slope form, we first find the slope.” The third cell contains the slope of a line formula: m equals y superscript 2 minus y superscript 1 divided by x superscript 2 minus x superscript 1. Below this is m equals 6 minus 4 divided by 3 minus 5. Below this is m equals 2 divided by negative 2. Below this is m equals negative 1.

In the second row, the first cell reads: “Step 2. Choose one point.” The second cell reads: “Choose either point.” The third cell contains the ordered pair (5, 4) with a superscript x subscript 1 over 5 and a superscript y subscript 1 over 4.

In the third row, the first cell reads: “Step 3. Substitute the values into the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses.” The top line of the second cell is left blank. The third cell contains the point-slope form, y minus y subscript 1 equals m times x minus x subscript 1 in parentheses. Below this is the point-slope form with 5 substituted for x subscript 1, 4 substituted for y subscript 1, and negative 1 substituted for m: y minus 4 equals negative 1 times x minus 5 in parentheses. Below this is y minus 4 equals negative x plus 5.

In the fourth row, the first cell reads: “Step 4. Write the equation in slope-intercept form.” The second cell is blank. The third cell contains y equals negative x plus 9.

Use the point $(3, 6)$ and see that you get the same equation.

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Find an equation of a line containing the points $(3, 1)$ and $(5, 6)$ .

$y = \frac{5}{2} x - \frac{13}{2}$

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Find an equation of a line containing the points $(1, 4)$ and $(6, 2)$ .

$y = - \frac{2}{5} x + \frac{22}{5}$

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Find an equation of a line given two points.

Find the slope using the given points.
Choose one point.
Substitute the values into the point-slope form, $y - y_{1} = m (x - x_{1})$ .
Write the equation in slope–intercept form.

Find an equation of a line that contains the points $(−3, −1)$ and $(2, −2)$ . Write the equation in slope–intercept form.

Solution

Since we have two points, we will find an equation of the line using the point–slope form. The first step will be to find the slope.

Find the slope of the line through (−3, −1) and (2, −2).



Choose either point.
Substitute the values into $y - y_{1} = m (x - x_{1}) .$


Write in slope–intercept form.

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Find an equation of a line containing the points $(−2, −4)$ and $(1, −3)$ .

$y = \frac{1}{3} x - \frac{10}{3}$

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Find an equation of a line containing the points $(−4, −3)$ and $(1, −5)$ .

$y = - \frac{2}{5} x - \frac{23}{5}$

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Find an equation of a line that contains the points $(−2, 4)$ and $(−2, −3)$ . Write the equation in slope–intercept form.

Again, the first step will be to find the slope.

$\begin{matrix} \begin{matrix} Find the slope of the line through (−2, 4) and (−2, −3) . \end{matrix} & \begin{matrix} m & = & \frac{y_{2} - y_{1}}{x_{2} - x_{1}} \\ m & = & \frac{−3 - 4}{−2 - (−2)} \\ m & = & \frac{−7}{0} \end{matrix} \\ The slope is undefined. \end{matrix}$

This tells us it is a vertical line. Both of our points have an x -coordinate of $−2$ . So our equation of the line is $x = −2$ . Since there is no $y$ , we cannot write it in slope–intercept form.

You may want to sketch a graph using the two given points. Does the graph agree with our conclusion that this is a vertical line?

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Find an equation of a line containing the points $(5, 1)$ and $(5, −4)$ .

$x = 5$

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Find an equaion of a line containing the points $(−4, 4)$ and $(−4, 3)$ .

$x = −4$

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We have seen that we can use either the slope–intercept form or the point–slope form to find an equation of a line. Which form we use will depend on the information we are given. This is summarized in [link] .

To Write an Equation of a Line
If given:	Use:	Form:
Slope and y -intercept	slope–intercept	$y = m x + b$
Slope and a point	point–slope	$y - y_{1} = m (x - x_{1})$
Two points	point–slope	$y - y_{1} = m (x - x_{1})$

Find an equation of a line parallel to a given line

Suppose we need to find an equation of a line that passes through a specific point and is parallel to a given line. We can use the fact that parallel lines have the same slope. So we will have a point and the slope—just what we need to use the point–slope equation.

First let’s look at this graphically.

The graph shows the graph of $y = 2 x - 3$ . We want to graph a line parallel to this line and passing through the point $(−2, 1)$ .

The graph shows the x y-coordinate plane. The x and y-axes each run from negative 7 to 7. The line whose equation is y equals 2x minus 3 intercepts the y-axis at (0, negative 3) and intercepts the x-axis at (3 halves, 0). Elsewhere on the graph, the point (negative 2, 1) is plotted.

We know that parallel lines have the same slope. So the second line will have the same slope as $y = 2 x - 3$ . That slope is $m_{∥} = 2$ . We’ll use the notation $m_{∥}$ to represent the slope of a line parallel to a line with slope $m$ . (Notice that the subscript $∥$ looks like two parallel lines.)

The second line will pass through $(−2, 1)$ and have $m = 2$ . To graph the line, we start at $(−2, 1)$ and count out the rise and run. With $m = 2$ (or $m = \frac{2}{1}$ ), we count out the rise 2 and the run 1. We draw the line.

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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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