{"id":2530,"date":"2024-08-01T22:38:04","date_gmt":"2024-08-01T22:38:04","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=2530"},"modified":"2025-08-15T16:33:22","modified_gmt":"2025-08-15T16:33:22","slug":"module-15-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/module-15-background-youll-need-2\/","title":{"raw":"Conic Sections: Background You'll Need 2","rendered":"Conic Sections: Background You&#8217;ll Need 2"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Use the distance formula<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Distance Formula<\/h2>\r\n[caption id=\"\" align=\"alignright\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042428\/CNX_CAT_Figure_02_01_015.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x and y axes range from 0 to 7. The points (x sub 1, y sub 1); (x sub 2, y sub 1); and (x sub 2, y sub 2) are labeled and connected to form a triangle. Along the base of the triangle, the following equation is displayed: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse of the triangle is labeled: d = c. The remaining side is labeled: the absolute value of y sub 2 minus y sub 1 equals b.\" width=\"400\" height=\"272\" \/> Triangle on a coordinate plane with all sides and vertices labeled[\/caption]\r\n\r\nDerived from the <strong>Pythagorean Theorem<\/strong>, the <strong>distance formula<\/strong> is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], is based on a right triangle where [latex]a[\/latex]<i> <\/i>and [latex]b[\/latex] are the lengths of the legs adjacent to the right angle, and [latex]c[\/latex]\u00a0is the length of the hypotenuse.\r\n\r\nThe relationship of sides [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] to side <em>d<\/em> is the same as that of sides <em>a <\/em>and <em>b <\/em>to side <em>c.<\/em> We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, [latex]|-3|=3[\/latex]. ) The symbols [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] indicate that the lengths of the sides of the triangle are positive. To find the length <em>c<\/em>, take the square root of both sides of the Pythagorean Theorem.\r\n<div style=\"text-align: center;\">[latex]{c}^{2}={a}^{2}+{b}^{2}\\rightarrow c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]<\/div>\r\nIt follows that the distance formula is given as\r\n<div style=\"text-align: center;\">[latex]{d}^{2}={\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}\\to d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>Distance Formula<\/h3>\r\nThe <strong>distance formula<\/strong> is a mathematical equation used to determine the exact distance between two points ([latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex]) on a coordinate plane.\r\n<p style=\"text-align: center;\">[latex]\\text{Distance}: d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the distance formula to find the distance between the points [latex](-5, -3)[\/latex] and [latex](7,2)[\/latex].[latex]\\begin{align*} d &amp;= \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\\\ d &amp;= \\sqrt{(72 - (-5))^2 + (2 - (-3))^2} \\\\ d &amp;= \\sqrt{12^2 + 5^2} = \\sqrt{144 + 25} = \\sqrt{169} \\\\ d &amp;= 13 \\end{align*}[\/latex]<\/section><section class=\"textbox example\">Find the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].[reveal-answer q=\"891596\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"891596\"]Let us first look at the graph of the two points. Connect the points to form a right triangle.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"324\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042430\/CNX_CAT_Figure_02_01_016.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x-axis ranges from negative 4 to 4. The y-axis ranges from negative 2 to 4. The points (-3, -1); (2, -1); and (2, 3) are plotted and labeled on the graph. The points are connected to form a triangle\" width=\"324\" height=\"192\" \/> Triangle on a coordinate plane with vertices labeled[\/caption]\r\n\r\nThen, calculate the length of <em>d <\/em>using the distance formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\hfill \\\\ d=\\sqrt{{\\left(2-\\left(-3\\right)\\right)}^{2}+{\\left(3-\\left(-1\\right)\\right)}^{2}}\\hfill \\\\ =\\sqrt{{\\left(5\\right)}^{2}+{\\left(4\\right)}^{2}}\\hfill \\\\ =\\sqrt{25+16}\\hfill \\\\ =\\sqrt{41}\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]18922[\/ohm2_question]<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm2_question hide_question_numbers=1]24919[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Use the distance formula<\/li>\n<\/ul>\n<\/section>\n<h2>Distance Formula<\/h2>\n<figure style=\"width: 400px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042428\/CNX_CAT_Figure_02_01_015.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x and y axes range from 0 to 7. The points (x sub 1, y sub 1); (x sub 2, y sub 1); and (x sub 2, y sub 2) are labeled and connected to form a triangle. Along the base of the triangle, the following equation is displayed: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse of the triangle is labeled: d = c. The remaining side is labeled: the absolute value of y sub 2 minus y sub 1 equals b.\" width=\"400\" height=\"272\" \/><figcaption class=\"wp-caption-text\">Triangle on a coordinate plane with all sides and vertices labeled<\/figcaption><\/figure>\n<p>Derived from the <strong>Pythagorean Theorem<\/strong>, the <strong>distance formula<\/strong> is used to find the distance between two points in the plane. The Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], is based on a right triangle where [latex]a[\/latex]<i> <\/i>and [latex]b[\/latex] are the lengths of the legs adjacent to the right angle, and [latex]c[\/latex]\u00a0is the length of the hypotenuse.<\/p>\n<p>The relationship of sides [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] to side <em>d<\/em> is the same as that of sides <em>a <\/em>and <em>b <\/em>to side <em>c.<\/em> We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, [latex]|-3|=3[\/latex]. ) The symbols [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] indicate that the lengths of the sides of the triangle are positive. To find the length <em>c<\/em>, take the square root of both sides of the Pythagorean Theorem.<\/p>\n<div style=\"text-align: center;\">[latex]{c}^{2}={a}^{2}+{b}^{2}\\rightarrow c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]<\/div>\n<p>It follows that the distance formula is given as<\/p>\n<div style=\"text-align: center;\">[latex]{d}^{2}={\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}\\to d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\n<section class=\"textbox keyTakeaway\">\n<h3>Distance Formula<\/h3>\n<p>The <strong>distance formula<\/strong> is a mathematical equation used to determine the exact distance between two points ([latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex]) on a coordinate plane.<\/p>\n<p style=\"text-align: center;\">[latex]\\text{Distance}: d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use the distance formula to find the distance between the points [latex](-5, -3)[\/latex] and [latex](7,2)[\/latex].[latex]\\begin{align*} d &= \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \\\\ d &= \\sqrt{(72 - (-5))^2 + (2 - (-3))^2} \\\\ d &= \\sqrt{12^2 + 5^2} = \\sqrt{144 + 25} = \\sqrt{169} \\\\ d &= 13 \\end{align*}[\/latex]<\/section>\n<section class=\"textbox example\">Find the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q891596\">Show Answer<\/button><\/p>\n<div id=\"q891596\" class=\"hidden-answer\" style=\"display: none\">Let us first look at the graph of the two points. Connect the points to form a right triangle.<\/p>\n<figure style=\"width: 324px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042430\/CNX_CAT_Figure_02_01_016.jpg\" alt=\"This is an image of a triangle on an x, y coordinate plane. The x-axis ranges from negative 4 to 4. The y-axis ranges from negative 2 to 4. The points (-3, -1); (2, -1); and (2, 3) are plotted and labeled on the graph. The points are connected to form a triangle\" width=\"324\" height=\"192\" \/><figcaption class=\"wp-caption-text\">Triangle on a coordinate plane with vertices labeled<\/figcaption><\/figure>\n<p>Then, calculate the length of <em>d <\/em>using the distance formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\hfill \\\\ d=\\sqrt{{\\left(2-\\left(-3\\right)\\right)}^{2}+{\\left(3-\\left(-1\\right)\\right)}^{2}}\\hfill \\\\ =\\sqrt{{\\left(5\\right)}^{2}+{\\left(4\\right)}^{2}}\\hfill \\\\ =\\sqrt{25+16}\\hfill \\\\ =\\sqrt{41}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm18922\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18922&theme=lumen&iframe_resize_id=ohm18922&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm24919\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24919&theme=lumen&iframe_resize_id=ohm24919&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":345,"module-header":"background_you_need","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2530"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2530\/revisions"}],"predecessor-version":[{"id":7877,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2530\/revisions\/7877"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/345"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2530\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=2530"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2530"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=2530"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=2530"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}