{"id":4638,"date":"2017-06-07T18:53:57","date_gmt":"2017-06-07T18:53:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-find-the-slope-of-horizontal-and-vertical-lines\/"},"modified":"2017-08-16T03:04:07","modified_gmt":"2017-08-16T03:04:07","slug":"read-find-the-slope-of-horizontal-and-vertical-lines","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-find-the-slope-of-horizontal-and-vertical-lines\/","title":{"raw":"Find the Slope of Horizontal and Vertical Lines","rendered":"Find the Slope of Horizontal and Vertical Lines"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Find the Slope of Horizontal and Vertical Lines\r\n<ul>\r\n \t<li>Find the slope of the lines [latex]x=a[\/latex] and [latex]y=b[\/latex]<\/li>\r\n \t<li>Recognize that horizontal lines have slope = 0<\/li>\r\n \t<li>Recognize that vertical lines have slopes that are undefined<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Finding the Slopes of Horizontal and Vertical Lines<\/h2>\r\nSo far you\u2019ve considered lines that run \u201cuphill\u201d or \u201cdownhill.\u201d Their slopes may be steep or gradual, but they are always positive or negative numbers. But there are two other kinds of lines, horizontal and vertical. What is the slope of a flat line or level ground? Of a wall or a vertical line?\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185352\/image040.jpg\" alt=\"The line y=3 crosses through the point (-3,3); the point (0,3); the point (2,3); and the point (5,3).\" width=\"335\" height=\"324\" \/>\r\n\r\nUsing the form [latex]y=0x+3[\/latex], you can see that the slope is 0. You can also use the slope formula with two points on this horizontal line to calculate the slope of this horizontal line. Using [latex](\u22123,3)[\/latex] as Point 1 and (2, 3) as Point 2, you get:\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{3-3}{2-\\left(-3\\right)}=\\frac{0}{5}=0\\end{array}[\/latex]<\/p>\r\nThe slope of this horizontal line is 0.\r\n\r\nLet\u2019s consider any horizontal line. No matter which two points you choose on the line, they will always have the same <i>y<\/i>-coordinate. So, when you apply the slope formula, the numerator will always be 0. Zero divided by any non-zero number is 0, so the slope of any horizontal line is always 0.\r\n\r\nThe equation for the horizontal line [latex]y=3[\/latex]\u00a0is telling you that no matter which two points you choose on this line, the <i>y-<\/i>coordinate will always be 3.\r\n\r\nHow about vertical lines? In their case, no matter which two points you choose, they will always have the same <i>x<\/i>-coordinate. The equation for this line is [latex]x=2[\/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\/07185354\/image041.jpg\" alt=\"The line x=2 runs through the point (2,-2), the point (2,1), the point (2,3), and the point (2,4).\" width=\"387\" height=\"374\" \/>\r\n\r\nThere is no way that this equation can be put in the slope-point form, as the coefficient of <i>y<\/i> is [latex]0\\left(x=0y+2\\right)[\/latex].\r\n\r\nSo, what happens when you use the slope formula with two points on this vertical line to calculate the slope? Using [latex](2,1)[\/latex] as Point 1 and [latex](2,3)[\/latex] as Point 2, you get:\r\n<p style=\"text-align: center\">[latex] \\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{3-1}{2-2}=\\frac{2}{0}\\end{array}[\/latex]<\/p>\r\nBut division by zero has no meaning for the set of real numbers. Because of this fact, it is said that the slope of this vertical line is undefined. This is true for all vertical lines\u2014they all have a slope that is undefined.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nWhat is the slope of the line that contains the points [latex](3,2)[\/latex] and [latex](\u22128,2)[\/latex]?\r\n\r\n[reveal-answer q=\"111566\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"111566\"]\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}{{x}_{1}}=3\\\\{{y}_{1}}=2\\end{array}[\/latex]<\/td>\r\n<td>[latex](3,2)=\\text{Point }1[\/latex], [latex] \\displaystyle \\left(x_{1},x_{2}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}{{x}_{2}}=-8\\\\\\\\{{y}_{2}}=2\\end{array}[\/latex]<\/td>\r\n<td>[latex](\u22128,2)=\\text{Point }2[\/latex], [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex] \\displaystyle \\begin{array}{l}\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\\\frac{(2)-(2)}{(-8)-(3)}=\\frac{0}{-11}=0\\\\\\\\m=0\\end{array}[\/latex]<\/td>\r\n<td>Substitute the values into the slope formula and simplify.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h4>Answer<\/h4>\r\nThe slope is 0, so the line is horizontal.[\/hidden-answer]\r\n\r\n<\/div>\r\nIn this video, you will see more examples of how to find the slope of horizontal and vertical lines:\r\n\r\nhttps:\/\/youtu.be\/zoLM3rxzndo\r\n<h2>Summary<\/h2>\r\nSlope describes the steepness of a line. The slope of any line remains constant along the line. The slope can also tell you information about the direction of the line on the coordinate plane. Slope can be calculated either by looking at the graph of a line or by using the coordinates of any two points on a line. There are two common formulas for slope: [latex] \\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex] and [latex] \\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex] where [latex]m=\\text{slope}[\/latex]\u00a0and [latex] \\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex] and [latex] \\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex] are two points on the line.\r\n\r\nThe images below summarize the slopes of different types of lines.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185356\/image042.gif\" alt=\"Uphill line with positive slope has a line that starts at the bottom-left and goes into the top-right of the graph. Downhill line with negative slope starts in the top-left and ends in the bottom-right part of the graph. Horizontal lines have a slope of 0. Vertical lines have an undefined slope.\" width=\"456\" height=\"183\" \/>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Find the Slope of Horizontal and Vertical Lines\n<ul>\n<li>Find the slope of the lines [latex]x=a[\/latex] and [latex]y=b[\/latex]<\/li>\n<li>Recognize that horizontal lines have slope = 0<\/li>\n<li>Recognize that vertical lines have slopes that are undefined<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2>Finding the Slopes of Horizontal and Vertical Lines<\/h2>\n<p>So far you\u2019ve considered lines that run \u201cuphill\u201d or \u201cdownhill.\u201d Their slopes may be steep or gradual, but they are always positive or negative numbers. But there are two other kinds of lines, horizontal and vertical. What is the slope of a flat line or level ground? Of a wall or a vertical line?<\/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\/07185352\/image040.jpg\" alt=\"The line y=3 crosses through the point (-3,3); the point (0,3); the point (2,3); and the point (5,3).\" width=\"335\" height=\"324\" \/><\/p>\n<p>Using the form [latex]y=0x+3[\/latex], you can see that the slope is 0. You can also use the slope formula with two points on this horizontal line to calculate the slope of this horizontal line. Using [latex](\u22123,3)[\/latex] as Point 1 and (2, 3) as Point 2, you get:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{3-3}{2-\\left(-3\\right)}=\\frac{0}{5}=0\\end{array}[\/latex]<\/p>\n<p>The slope of this horizontal line is 0.<\/p>\n<p>Let\u2019s consider any horizontal line. No matter which two points you choose on the line, they will always have the same <i>y<\/i>-coordinate. So, when you apply the slope formula, the numerator will always be 0. Zero divided by any non-zero number is 0, so the slope of any horizontal line is always 0.<\/p>\n<p>The equation for the horizontal line [latex]y=3[\/latex]\u00a0is telling you that no matter which two points you choose on this line, the <i>y-<\/i>coordinate will always be 3.<\/p>\n<p>How about vertical lines? In their case, no matter which two points you choose, they will always have the same <i>x<\/i>-coordinate. The equation for this line is [latex]x=2[\/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\/07185354\/image041.jpg\" alt=\"The line x=2 runs through the point (2,-2), the point (2,1), the point (2,3), and the point (2,4).\" width=\"387\" height=\"374\" \/><\/p>\n<p>There is no way that this equation can be put in the slope-point form, as the coefficient of <i>y<\/i> is [latex]0\\left(x=0y+2\\right)[\/latex].<\/p>\n<p>So, what happens when you use the slope formula with two points on this vertical line to calculate the slope? Using [latex](2,1)[\/latex] as Point 1 and [latex](2,3)[\/latex] as Point 2, you get:<\/p>\n<p style=\"text-align: center\">[latex]\\displaystyle \\begin{array}{l}m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\m=\\frac{3-1}{2-2}=\\frac{2}{0}\\end{array}[\/latex]<\/p>\n<p>But division by zero has no meaning for the set of real numbers. Because of this fact, it is said that the slope of this vertical line is undefined. This is true for all vertical lines\u2014they all have a slope that is undefined.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>What is the slope of the line that contains the points [latex](3,2)[\/latex] and [latex](\u22128,2)[\/latex]?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q111566\">Show Solution<\/span><\/p>\n<div id=\"q111566\" class=\"hidden-answer\" style=\"display: none\">\n<table>\n<tbody>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}{{x}_{1}}=3\\\\{{y}_{1}}=2\\end{array}[\/latex]<\/td>\n<td>[latex](3,2)=\\text{Point }1[\/latex], [latex]\\displaystyle \\left(x_{1},x_{2}\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}{{x}_{2}}=-8\\\\\\\\{{y}_{2}}=2\\end{array}[\/latex]<\/td>\n<td>[latex](\u22128,2)=\\text{Point }2[\/latex], [latex]\\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\displaystyle \\begin{array}{l}\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\\\\\\\\\\frac{(2)-(2)}{(-8)-(3)}=\\frac{0}{-11}=0\\\\\\\\m=0\\end{array}[\/latex]<\/td>\n<td>Substitute the values into the slope formula and simplify.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h4>Answer<\/h4>\n<p>The slope is 0, so the line is horizontal.<\/p><\/div>\n<\/div>\n<\/div>\n<p>In this video, you will see more examples of how to find the slope of horizontal and vertical lines:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  Determine the Slope a Line Given Two Points on a Horizontal and Vertical Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zoLM3rxzndo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>Slope describes the steepness of a line. The slope of any line remains constant along the line. The slope can also tell you information about the direction of the line on the coordinate plane. Slope can be calculated either by looking at the graph of a line or by using the coordinates of any two points on a line. There are two common formulas for slope: [latex]\\displaystyle \\text{Slope }=\\frac{\\text{rise}}{\\text{run}}[\/latex] and [latex]\\displaystyle m=\\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}[\/latex] where [latex]m=\\text{slope}[\/latex]\u00a0and [latex]\\displaystyle ({{x}_{1}},{{y}_{1}})[\/latex] and [latex]\\displaystyle ({{x}_{2}},{{y}_{2}})[\/latex] are two points on the line.<\/p>\n<p>The images below summarize the slopes of different types of lines.<\/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\/07185356\/image042.gif\" alt=\"Uphill line with positive slope has a line that starts at the bottom-left and goes into the top-right of the graph. Downhill line with negative slope starts in the top-left and ends in the bottom-right part of the graph. Horizontal lines have a slope of 0. Vertical lines have an undefined slope.\" width=\"456\" height=\"183\" \/><\/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-4638\">\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 the Slope a Line Given Two Points on a Horizontal and Vertical Line. <strong>Authored by<\/strong>: mathispower4u. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/zoLM3rxzndo\">https:\/\/youtu.be\/zoLM3rxzndo<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/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 <\/section>","protected":false},"author":17533,"menu_order":9,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Ex: Determine the Slope a Line Given Two Points on a Horizontal and Vertical Line\",\"author\":\"mathispower4u\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/zoLM3rxzndo\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"}]","CANDELA_OUTCOMES_GUID":"a7ffbc48-4f97-4bde-9cbc-53a0ae834823","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4638","chapter","type-chapter","status-publish","hentry"],"part":4587,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4638","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\/4638\/revisions"}],"predecessor-version":[{"id":5081,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapters\/4638\/revisions\/5081"}],"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\/4638\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/media?parent=4638"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/pressbooks\/v2\/chapter-type?post=4638"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/contributor?post=4638"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/wp-json\/wp\/v2\/license?post=4638"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}