{"id":1062,"date":"2015-11-12T18:35:32","date_gmt":"2015-11-12T18:35:32","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1062"},"modified":"2017-03-31T21:53:42","modified_gmt":"2017-03-31T21:53:42","slug":"write-the-point-slope-form-of-an-equation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/write-the-point-slope-form-of-an-equation\/","title":{"raw":"Write the point-slope form of an equation","rendered":"Write the point-slope form of an equation"},"content":{"raw":"

Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the point-slope form<\/strong>.<\/p>\r\n\r\n

[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\n

The point-slope form is derived from the slope formula.<\/p>\r\n\r\n

[latex]\\begin{cases}{m}=\\frac{y-{y}_{1}}{x-{x}_{1}}\\hfill & \\text{assuming }{ x }\\ne {x}_{1}\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=\\frac{y-{y}_{1}}{x-{x}_{1}}\\left(x-{x}_{1}\\right)\\hfill & \\text{Multiply both sides by }\\left(x-{x}_{1}\\right)\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=y-{y}_{1}\\hfill & \\text{Simplify}\\hfill \\\\ y-{y}_{1}={ m }\\left(x-{x}_{1}\\right)\\hfill &\\text{Rearrange}\\hfill \\end{cases}[\/latex]<\/div>\r\n

Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form, [latex]y - 4=-\\frac{1}{2}\\left(x - 6\\right)[\/latex] . We can convert it to the slope-intercept form as shown.<\/p>\r\n\r\n

[latex]\\begin{cases}y - 4=-\\frac{1}{2}\\left(x - 6\\right)\\hfill & \\hfill \\\\ y - 4=-\\frac{1}{2}x+3\\hfill & \\text{Distribute the }-\\frac{1}{2}.\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+7\\hfill & \\text{Add 4 to each side}.\\hfill \\end{cases}[\/latex]<\/div>\r\n

Therefore, the same line can be described in slope-intercept form as [latex]y=-\\frac{1}{2}x+7[\/latex].<\/p>\r\n\r\n

\r\n

A General Note: Point-Slope Form of a Linear Equation<\/h3>\r\n

The point-slope form<\/strong> of a linear equation takes the form<\/p>\r\n\r\n

[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\n

where [latex]m[\/latex]\u00a0is the slope, [latex]{x}_{1 }\\text{and} {y}_{1}[\/latex]\u00a0are the [latex]x\\text{ and }y[\/latex]\u00a0coordinates of a specific point through which the line passes.<\/p>\r\n\r\n<\/div>\r\n

\r\n

\r\n

Writing the Equation of a Line Using a Point and the Slope<\/span><\/h2>\r\n

The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told that a line has a slope of 2 and passes through the point [latex]\\left(4,1\\right)[\/latex].\u00a0We know that [latex]m=2[\/latex]\u00a0and that [latex]{x}_{1}=4[\/latex]\u00a0and [latex]{y}_{1}=1[\/latex]. We can substitute these values into the general point-slope equation.<\/p>\r\n\r\n

[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 4\\right)\\end{cases}[\/latex]<\/div>\r\n

If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques.<\/p>\r\n\r\n

[latex]\\begin{cases}y - 1=2\\left(x - 4\\right)\\hfill & \\hfill \\\\ y - 1=2x - 8\\hfill & \\text{Distribute the }2.\\hfill \\\\ \\text{ }y=2x - 7\\hfill & \\text{Add 1 to each side}.\\hfill \\end{cases}[\/latex]<\/div>\r\n

Both equations, [latex]y - 1=2\\left(x - 4\\right)[\/latex]\u00a0and [latex]y=2x - 7[\/latex], describe the same line. See Figure 6.\r\n<\/span><\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 6<\/b>[\/caption]\r\n\r\n

\r\n
\r\n
\r\n

Example 5: Writing Linear Equations Using a Point and the Slope<\/h3>\r\n

Write the point-slope form of an equation of a line with a slope of 3 that passes through the point [latex]\\left(6,-1\\right)[\/latex]. Then rewrite it in the slope-intercept form.<\/p>\r\n\r\n<\/div>\r\n

\r\n

Solution<\/h3>\r\n

Let\u2019s figure out what we know from the given information. The slope is 3, so m\u00a0<\/em>= 3. We also know one point, so we know [latex]{x}_{1}=6[\/latex] and [latex]{y}_{1}=-1[\/latex]. Now we can substitute these values into the general point-slope equation.<\/p>\r\n\r\n

[latex]\\begin{cases}\\text{ }y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill & \\hfill \\\\ y-\\left(-1\\right)=3\\left(x - 6\\right)\\hfill & \\text{Substitute known values}.\\hfill \\\\ \\text{ }y+1=3\\left(x - 6\\right)\\hfill & \\text{Distribute }-1\\text{ to find point-slope form}.\\hfill \\end{cases}[\/latex]<\/div>\r\n

Then we use algebra to find the slope-intercept form.<\/p>\r\n\r\n

[latex]\\begin{cases}y+1=3\\left(x - 6\\right)\\hfill & \\hfill \\\\ y+1=3x - 18\\hfill & \\text{Distribute 3}.\\hfill \\\\ \\text{ }y=3x - 19\\hfill & \\text{Simplify to slope-intercept form}.\\hfill \\end{cases}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
\r\n

Try It 3<\/h3>\r\n

Write the point-slope form of an equation of a line with a slope of \u20132 that passes through the point [latex]\\left(-2,\\text{ }2\\right)[\/latex]. Then rewrite it in the slope-intercept form.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n<\/section>

\r\n

\r\n

Writing the Equation of a Line Using Two Points<\/span><\/h2>\r\n

The point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points [latex]\\left(0,\\text{ }1\\right)[\/latex]\u00a0and [latex]\\left(3,\\text{ }2\\right)[\/latex]. We can use the coordinates of the two points to find the slope.<\/p>\r\n\r\n

[latex]\\begin{cases}{m}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\\ \\text{ }=\\frac{2 - 1}{3 - 0}\\hfill \\\\ \\text{ }=\\frac{1}{3}\\hfill \\end{cases}[\/latex]<\/div>\r\n

Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point.<\/p>\r\n\r\n

[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=\\frac{1}{3}\\left(x - 0\\right)\\end{cases}[\/latex]<\/div>\r\n

As before, we can use algebra to rewrite the equation in the slope-intercept form.<\/p>\r\n\r\n

[latex]\\begin{cases}y - 1=\\frac{1}{3}\\left(x - 0\\right)\\hfill & \\hfill \\\\ y - 1=\\frac{1}{3}x\\hfill & \\text{Distribute the }\\frac{1}{3}.\\hfill \\\\ \\text{ }y=\\frac{1}{3}x+1\\hfill & \\text{Add 1 to each side}.\\hfill \\end{cases}[\/latex]<\/div>\r\n

Both equations describe the line shown in Figure 7.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 7<\/b>[\/caption]\r\n\r\n

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\r\n

Example 6: Writing Linear Equations Using Two Points<\/h3>\r\n

Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form.<\/p>\r\n\r\n<\/div>\r\n

\r\n

Solution<\/h3>\r\n

Let\u2019s begin by finding the slope.<\/p>\r\n\r\n

[latex]\\begin{cases}{m}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{ }=\\frac{7 - 1}{8 - 5}\\hfill \\\\ \\text{ }=\\frac{6}{3}\\hfill \\\\ \\text{ }=2\\hfill \\end{cases}[\/latex]<\/div>\r\n

So [latex]m=2[\/latex]. Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use [latex]\\left(5,1\\right)[\/latex].<\/p>\r\n\r\n

[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 5\\right)\\end{cases}[\/latex]<\/div>\r\n

The point-slope equation of the line is [latex]{y}_{2}-1=2\\left({x}_{2}-5\\right)[\/latex]. To rewrite the equation in slope-intercept form, we use algebra.<\/p>\r\n\r\n

[latex]\\begin{cases}y - 1=2\\left(x - 5\\right)\\hfill \\\\ y - 1=2x - 10\\hfill \\\\ \\text{ }y=2x - 9\\hfill \\end{cases}[\/latex]<\/div>\r\n

The slope-intercept equation of the line is [latex]y=2x - 9[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

\r\n

Try It 4<\/h3>\r\n

Write the point-slope form of an equation of a line that passes through the points [latex]\\left(-1,3\\right)\\\\ [\/latex] and [latex]\\left(0,0\\right)[\/latex]. Then rewrite it in the slope-intercept form.<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>\r\n<\/section><\/section>","rendered":"

\n

Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the point-slope form<\/strong>.<\/p>\n

[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n

The point-slope form is derived from the slope formula.<\/p>\n

[latex]\\begin{cases}{m}=\\frac{y-{y}_{1}}{x-{x}_{1}}\\hfill & \\text{assuming }{ x }\\ne {x}_{1}\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=\\frac{y-{y}_{1}}{x-{x}_{1}}\\left(x-{x}_{1}\\right)\\hfill & \\text{Multiply both sides by }\\left(x-{x}_{1}\\right)\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=y-{y}_{1}\\hfill & \\text{Simplify}\\hfill \\\\ y-{y}_{1}={ m }\\left(x-{x}_{1}\\right)\\hfill &\\text{Rearrange}\\hfill \\end{cases}[\/latex]<\/div>\n

Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form, [latex]y - 4=-\\frac{1}{2}\\left(x - 6\\right)[\/latex] . We can convert it to the slope-intercept form as shown.<\/p>\n

[latex]\\begin{cases}y - 4=-\\frac{1}{2}\\left(x - 6\\right)\\hfill & \\hfill \\\\ y - 4=-\\frac{1}{2}x+3\\hfill & \\text{Distribute the }-\\frac{1}{2}.\\hfill \\\\ \\text{ }y=-\\frac{1}{2}x+7\\hfill & \\text{Add 4 to each side}.\\hfill \\end{cases}[\/latex]<\/div>\n

Therefore, the same line can be described in slope-intercept form as [latex]y=-\\frac{1}{2}x+7[\/latex].<\/p>\n

\n

A General Note: Point-Slope Form of a Linear Equation<\/h3>\n

The point-slope form<\/strong> of a linear equation takes the form<\/p>\n

[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n

where [latex]m[\/latex]\u00a0is the slope, [latex]{x}_{1 }\\text{and} {y}_{1}[\/latex]\u00a0are the [latex]x\\text{ and }y[\/latex]\u00a0coordinates of a specific point through which the line passes.<\/p>\n<\/div>\n

\n

\n

Writing the Equation of a Line Using a Point and the Slope<\/span><\/h2>\n

The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, we are told that a line has a slope of 2 and passes through the point [latex]\\left(4,1\\right)[\/latex].\u00a0We know that [latex]m=2[\/latex]\u00a0and that [latex]{x}_{1}=4[\/latex]\u00a0and [latex]{y}_{1}=1[\/latex]. We can substitute these values into the general point-slope equation.<\/p>\n

[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 4\\right)\\end{cases}[\/latex]<\/div>\n

If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques.<\/p>\n

[latex]\\begin{cases}y - 1=2\\left(x - 4\\right)\\hfill & \\hfill \\\\ y - 1=2x - 8\\hfill & \\text{Distribute the }2.\\hfill \\\\ \\text{ }y=2x - 7\\hfill & \\text{Add 1 to each side}.\\hfill \\end{cases}[\/latex]<\/div>\n

Both equations, [latex]y - 1=2\\left(x - 4\\right)[\/latex]\u00a0and [latex]y=2x - 7[\/latex], describe the same line. See Figure 6.
\n<\/span><\/p>\n

\"\"<\/p>\n

Figure 6<\/b><\/p>\n<\/div>\n

\n
\n
\n

Example 5: Writing Linear Equations Using a Point and the Slope<\/h3>\n

Write the point-slope form of an equation of a line with a slope of 3 that passes through the point [latex]\\left(6,-1\\right)[\/latex]. Then rewrite it in the slope-intercept form.<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

Let\u2019s figure out what we know from the given information. The slope is 3, so m\u00a0<\/em>= 3. We also know one point, so we know [latex]{x}_{1}=6[\/latex] and [latex]{y}_{1}=-1[\/latex]. Now we can substitute these values into the general point-slope equation.<\/p>\n

[latex]\\begin{cases}\\text{ }y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill & \\hfill \\\\ y-\\left(-1\\right)=3\\left(x - 6\\right)\\hfill & \\text{Substitute known values}.\\hfill \\\\ \\text{ }y+1=3\\left(x - 6\\right)\\hfill & \\text{Distribute }-1\\text{ to find point-slope form}.\\hfill \\end{cases}[\/latex]<\/div>\n

Then we use algebra to find the slope-intercept form.<\/p>\n

[latex]\\begin{cases}y+1=3\\left(x - 6\\right)\\hfill & \\hfill \\\\ y+1=3x - 18\\hfill & \\text{Distribute 3}.\\hfill \\\\ \\text{ }y=3x - 19\\hfill & \\text{Simplify to slope-intercept form}.\\hfill \\end{cases}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
\n

Try It 3<\/h3>\n

Write the point-slope form of an equation of a line with a slope of \u20132 that passes through the point [latex]\\left(-2,\\text{ }2\\right)[\/latex]. Then rewrite it in the slope-intercept form.<\/p>\n

Solution<\/a><\/p>\n<\/div>\n<\/section>\n

\n

\n

Writing the Equation of a Line Using Two Points<\/span><\/h2>\n

The point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points [latex]\\left(0,\\text{ }1\\right)[\/latex]\u00a0and [latex]\\left(3,\\text{ }2\\right)[\/latex]. We can use the coordinates of the two points to find the slope.<\/p>\n

[latex]\\begin{cases}{m}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\\ \\text{ }=\\frac{2 - 1}{3 - 0}\\hfill \\\\ \\text{ }=\\frac{1}{3}\\hfill \\end{cases}[\/latex]<\/div>\n

Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point.<\/p>\n

[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=\\frac{1}{3}\\left(x - 0\\right)\\end{cases}[\/latex]<\/div>\n

As before, we can use algebra to rewrite the equation in the slope-intercept form.<\/p>\n

[latex]\\begin{cases}y - 1=\\frac{1}{3}\\left(x - 0\\right)\\hfill & \\hfill \\\\ y - 1=\\frac{1}{3}x\\hfill & \\text{Distribute the }\\frac{1}{3}.\\hfill \\\\ \\text{ }y=\\frac{1}{3}x+1\\hfill & \\text{Add 1 to each side}.\\hfill \\end{cases}[\/latex]<\/div>\n

Both equations describe the line shown in Figure 7.<\/p>\n

\"\"<\/p>\n

Figure 7<\/b><\/p>\n<\/div>\n

\n
\n
\n

Example 6: Writing Linear Equations Using Two Points<\/h3>\n

Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form.<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

Let\u2019s begin by finding the slope.<\/p>\n

[latex]\\begin{cases}{m}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{ }=\\frac{7 - 1}{8 - 5}\\hfill \\\\ \\text{ }=\\frac{6}{3}\\hfill \\\\ \\text{ }=2\\hfill \\end{cases}[\/latex]<\/div>\n

So [latex]m=2[\/latex]. Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use [latex]\\left(5,1\\right)[\/latex].<\/p>\n

[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 5\\right)\\end{cases}[\/latex]<\/div>\n

The point-slope equation of the line is [latex]{y}_{2}-1=2\\left({x}_{2}-5\\right)[\/latex]. To rewrite the equation in slope-intercept form, we use algebra.<\/p>\n

[latex]\\begin{cases}y - 1=2\\left(x - 5\\right)\\hfill \\\\ y - 1=2x - 10\\hfill \\\\ \\text{ }y=2x - 9\\hfill \\end{cases}[\/latex]<\/div>\n

The slope-intercept equation of the line is [latex]y=2x - 9[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

\n

Try It 4<\/h3>\n

Write the point-slope form of an equation of a line that passes through the points [latex]\\left(-1,3\\right)\\\\[\/latex] and [latex]\\left(0,0\\right)[\/latex]. Then rewrite it in the slope-intercept form.<\/p>\n

Solution<\/a><\/p>\n<\/div>\n<\/section>\n<\/section>\n\n\t\t\t

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Candela Citations<\/h3>\n\t\t\t\t\t
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