{"id":4646,"date":"2017-06-07T18:54:05","date_gmt":"2017-06-07T18:54:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-point-slope-form\/"},"modified":"2017-08-16T03:16:03","modified_gmt":"2017-08-16T03:16:03","slug":"read-point-slope-form","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-point-slope-form\/","title":{"raw":"Equation of a Line Given Point and Slope","rendered":"Equation of a Line Given Point and Slope"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Write and solve equations of lines using slope and a point on the line\r\n<ul>\r\n \t<li>Write the equation of a line given the slope and a point on the line.<\/li>\r\n \t<li>Identify which parts of a linear equation are given and which parts need to be solved for using algebra<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Find the Equation of a Line Given the Slope and a Point on the Line<\/h2>\r\nUsing the slope-intercept equation of a line is possible when you know both the slope (<i>m<\/i>) and the <i>y<\/i>-intercept (<i>b<\/i>), but what if you know the slope and just any point on the line, not specifically the <i>y<\/i>-intercept? Can you still write the equation? The answer is <i>yes<\/i>, but you will need to put in a little more thought and work than you did previously.\r\n\r\nRecall that a point is an (<i>x<\/i>, <i>y<\/i>) coordinate pair and that all points on the line will satisfy the linear equation. So, if you have a point on the line, it must be a solution to the equation. Although you don\u2019t know the exact equation yet, you know that you can express the line in slope-intercept form, [latex]y=mx+b[\/latex].\r\n\r\nYou do know the slope (<i>m<\/i>), but you just don\u2019t know the value of the <i>y<\/i>-intercept (<i>b<\/i>). Since point (<i>x<\/i>, <i>y<\/i>) is a solution to the equation, you can substitute its coordinates for <i>x<\/i> and <i>y<\/i> in [latex]y=mx+b[\/latex]\u00a0and solve to find <i>b<\/i>!\r\n\r\nThis may seem a bit confusing with all the variables, but an example with an actual slope and a point will help to clarify.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line that has a slope of 3 and contains the point [latex](1,4)[\/latex].\r\n\r\n[reveal-answer q=\"161353\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"161353\"]\r\n\r\nSubstitute the slope (<i>m<\/i>) into\u00a0[latex]y=mx+b[\/latex].\r\n<p style=\"text-align: center\">[latex]y=3x+b[\/latex]<\/p>\r\nSubstitute the point [latex](1,4)[\/latex] for <i>x <\/i>and <i>y.<\/i>\r\n<p style=\"text-align: center\">[latex]4=3\\left(1\\right)+b[\/latex]<\/p>\r\nSolve for <i>b.<\/i>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}4=3+b\\\\1=b\\end{array}[\/latex]<\/p>\r\nRewrite [latex]y=mx+b[\/latex]\u00a0with [latex]m=3[\/latex]\u00a0and [latex]b=1[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]y=3x+1[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nTo confirm our algebra, you can check by graphing the equation [latex]y=3x+1[\/latex]. The equation checks because when graphed it passes through the point [latex](1,4)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185403\/image045.jpg\" alt=\"An uphill line passes through the y-intercept of (0,1) and the point (1,4). The rise is 3 and the run is 1.\" width=\"348\" height=\"349\" \/>\r\n\r\nIf you know the slope of a line and a point on the line, you can draw a graph. Using an equation in the point-slope form allows you to identify the slope and a point. Consider the equation [latex] \\displaystyle y=-3x-1[\/latex]. The <em>y<\/em>-intercept is the point on the line where it passes through the <em>y<\/em>-axis. What is the value of <em>x<\/em> at this point?\r\n<div class=\"textbox shaded\"><span style=\"color: #993366\"><strong>Reminder<\/strong><\/span>: All <em>y<\/em>-intercepts are points in the form [latex](0,y)[\/latex]. \u00a0The <em>x<\/em> value of any <em>y<\/em>-intercept is <em>always<\/em>\u00a0zero.<\/div>\r\nTherefore, you can tell from this equation that the <i>y<\/i>-intercept is at [latex](0,\u22121)[\/latex], check this by replacing <em>x<\/em> with 0 and solving for <em>y<\/em>. To graph the line, start by plotting that point, [latex](0,\u22121)[\/latex], on a graph.\r\n\r\nYou can also tell from the equation that the slope of this line is [latex]\u22123[\/latex]. So start at [latex](0,\u22121)[\/latex] and count up 3 and over [latex]\u22121[\/latex] (1 unit in the negative direction, left) and plot a second point. (You could also have gone down 3 and over 1.) Then draw a line through both points, and there it is, the graph of [latex] \\displaystyle y=-3x-1[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185405\/image044.jpg\" alt=\"A downhill line passes through the point (-1,2) and the y-intercept (0,-1). The rise is 3 and the run is -1.\" width=\"325\" height=\"326\" \/>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example (Advanced)<\/h3>\r\nWrite the equation of the line that has a slope of [latex]\\frac{7}{8}[\/latex]\u00a0and contains the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex].\r\n\r\n[reveal-answer q=\"31452\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"31452\"]\r\n\r\nSubstitute the slope (<i>m<\/i>) into [latex]y=mx+b[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y=mx+b\\\\\\\\y=-\\frac{7}{8}x+b\\end{array}[\/latex]<\/p>\r\nSubstitute the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex]\u00a0for <i>x <\/i>and <i>y.<\/i>\r\n<p style=\"text-align: center\">[latex]\\frac{5}{4}=-\\frac{7}{8}\\left(4\\right)+b[\/latex]<\/p>\r\nSolve for <i>b.<\/i>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=-\\frac{28}{8}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=-\\frac{14}{4}+b\\\\\\\\\\frac{5}{4}+\\frac{14}{4}=-\\frac{14}{4}+\\frac{14}{4}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{19}{4}=b\\end{array}[\/latex]<\/p>\r\nRewrite [latex]y=mx+b[\/latex] with [latex] \\displaystyle m=-\\frac{7}{8}[\/latex] and [latex] \\displaystyle b=\\frac{19}{4}[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]y=-\\frac{7}{8}x+\\frac{19}{4}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nIn this video, you will see an additional example of how to find the equation of a line given the slope and a point.\r\n\r\nhttps:\/\/youtu.be\/URYnKqEctgc","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Write and solve equations of lines using slope and a point on the line\n<ul>\n<li>Write the equation of a line given the slope and a point on the line.<\/li>\n<li>Identify which parts of a linear equation are given and which parts need to be solved for using algebra<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2>Find the Equation of a Line Given the Slope and a Point on the Line<\/h2>\n<p>Using the slope-intercept equation of a line is possible when you know both the slope (<i>m<\/i>) and the <i>y<\/i>-intercept (<i>b<\/i>), but what if you know the slope and just any point on the line, not specifically the <i>y<\/i>-intercept? Can you still write the equation? The answer is <i>yes<\/i>, but you will need to put in a little more thought and work than you did previously.<\/p>\n<p>Recall that a point is an (<i>x<\/i>, <i>y<\/i>) coordinate pair and that all points on the line will satisfy the linear equation. So, if you have a point on the line, it must be a solution to the equation. Although you don\u2019t know the exact equation yet, you know that you can express the line in slope-intercept form, [latex]y=mx+b[\/latex].<\/p>\n<p>You do know the slope (<i>m<\/i>), but you just don\u2019t know the value of the <i>y<\/i>-intercept (<i>b<\/i>). Since point (<i>x<\/i>, <i>y<\/i>) is a solution to the equation, you can substitute its coordinates for <i>x<\/i> and <i>y<\/i> in [latex]y=mx+b[\/latex]\u00a0and solve to find <i>b<\/i>!<\/p>\n<p>This may seem a bit confusing with all the variables, but an example with an actual slope and a point will help to clarify.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Write the equation of the line that has a slope of 3 and contains the point [latex](1,4)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q161353\">Show Solution<\/span><\/p>\n<div id=\"q161353\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the slope (<i>m<\/i>) into\u00a0[latex]y=mx+b[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]y=3x+b[\/latex]<\/p>\n<p>Substitute the point [latex](1,4)[\/latex] for <i>x <\/i>and <i>y.<\/i><\/p>\n<p style=\"text-align: center\">[latex]4=3\\left(1\\right)+b[\/latex]<\/p>\n<p>Solve for <i>b.<\/i><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}4=3+b\\\\1=b\\end{array}[\/latex]<\/p>\n<p>Rewrite [latex]y=mx+b[\/latex]\u00a0with [latex]m=3[\/latex]\u00a0and [latex]b=1[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=3x+1[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>To confirm our algebra, you can check by graphing the equation [latex]y=3x+1[\/latex]. The equation checks because when graphed it passes through the point [latex](1,4)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185403\/image045.jpg\" alt=\"An uphill line passes through the y-intercept of (0,1) and the point (1,4). The rise is 3 and the run is 1.\" width=\"348\" height=\"349\" \/><\/p>\n<p>If you know the slope of a line and a point on the line, you can draw a graph. Using an equation in the point-slope form allows you to identify the slope and a point. Consider the equation [latex]\\displaystyle y=-3x-1[\/latex]. The <em>y<\/em>-intercept is the point on the line where it passes through the <em>y<\/em>-axis. What is the value of <em>x<\/em> at this point?<\/p>\n<div class=\"textbox shaded\"><span style=\"color: #993366\"><strong>Reminder<\/strong><\/span>: All <em>y<\/em>-intercepts are points in the form [latex](0,y)[\/latex]. \u00a0The <em>x<\/em> value of any <em>y<\/em>-intercept is <em>always<\/em>\u00a0zero.<\/div>\n<p>Therefore, you can tell from this equation that the <i>y<\/i>-intercept is at [latex](0,\u22121)[\/latex], check this by replacing <em>x<\/em> with 0 and solving for <em>y<\/em>. To graph the line, start by plotting that point, [latex](0,\u22121)[\/latex], on a graph.<\/p>\n<p>You can also tell from the equation that the slope of this line is [latex]\u22123[\/latex]. So start at [latex](0,\u22121)[\/latex] and count up 3 and over [latex]\u22121[\/latex] (1 unit in the negative direction, left) and plot a second point. (You could also have gone down 3 and over 1.) Then draw a line through both points, and there it is, the graph of [latex]\\displaystyle y=-3x-1[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185405\/image044.jpg\" alt=\"A downhill line passes through the point (-1,2) and the y-intercept (0,-1). The rise is 3 and the run is -1.\" width=\"325\" height=\"326\" \/><\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example (Advanced)<\/h3>\n<p>Write the equation of the line that has a slope of [latex]\\frac{7}{8}[\/latex]\u00a0and contains the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q31452\">Show Solution<\/span><\/p>\n<div id=\"q31452\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the slope (<i>m<\/i>) into [latex]y=mx+b[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y=mx+b\\\\\\\\y=-\\frac{7}{8}x+b\\end{array}[\/latex]<\/p>\n<p>Substitute the point [latex]\\left(4,\\frac{5}{4}\\right)[\/latex]\u00a0for <i>x <\/i>and <i>y.<\/i><\/p>\n<p style=\"text-align: center\">[latex]\\frac{5}{4}=-\\frac{7}{8}\\left(4\\right)+b[\/latex]<\/p>\n<p>Solve for <i>b.<\/i><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=-\\frac{28}{8}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{5}{4}=-\\frac{14}{4}+b\\\\\\\\\\frac{5}{4}+\\frac{14}{4}=-\\frac{14}{4}+\\frac{14}{4}+b\\\\\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\frac{19}{4}=b\\end{array}[\/latex]<\/p>\n<p>Rewrite [latex]y=mx+b[\/latex] with [latex]\\displaystyle m=-\\frac{7}{8}[\/latex] and [latex]\\displaystyle b=\\frac{19}{4}[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]y=-\\frac{7}{8}x+\\frac{19}{4}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<p>In this video, you will see an additional example of how to find the equation of a line given the slope and a point.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Determine a Linear Equation Given Slope and a Point (Slope-Intercept Form) (09x-32)\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/URYnKqEctgc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-4646\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Ex: Determine a Linear Equation Given Slope and a Point (Slope-Intercept Form) (09x-32) Mathispower4u. <strong>Authored by<\/strong>: mathispower4u. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/URYnKqEctgc\">https:\/\/youtu.be\/URYnKqEctgc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t 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License\"}]","CANDELA_OUTCOMES_GUID":"668c8d89-6409-45e1-b1b8-61e72ae22d9c","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4646","chapter","type-chapter","status-publish","hentry"],"part":4587,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4646","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4646\/revisions"}],"predecessor-version":[{"id":5085,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4646\/revisions\/5085"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/parts\/4587"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4646\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/media?parent=4646"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapter-type?post=4646"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/contributor?post=4646"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/license?post=4646"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}