{"id":1006,"date":"2024-05-01T21:27:13","date_gmt":"2024-05-01T21:27:13","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=1006"},"modified":"2025-08-21T23:10:41","modified_gmt":"2025-08-21T23:10:41","slug":"graphing-and-analyzing-linear-equations-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/graphing-and-analyzing-linear-equations-learn-it-5\/","title":{"raw":"Graphing and Analyzing Linear Equations: Learn It 5","rendered":"Graphing and Analyzing Linear Equations: Learn It 5"},"content":{"raw":"

Distance Formula<\/h2>\r\nBuilding on our understanding of how to measure changes on a graph, let\u2019s now shift our focus to the distance formula, a powerful tool for calculating the precise distance between any two points on a graph.\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"400\"]\"This Triangle with sides labeled on an x,y coordinate plane[\/caption]\r\n\r\nDerived from the Pythagorean Theorem<\/strong>, the 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>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 d<\/em> is the same as that of sides a <\/em>and b <\/em>to side 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 c<\/em>, take the square root of both sides of the Pythagorean Theorem.\r\n
[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
[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
\r\n

distance formula<\/h3>\r\nThe 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

[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>

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\"]\"This Triangle with points labeled on an x,y, coordinate plane[\/caption]\r\n\r\nThen, calculate the length of d <\/em>using the distance formula.\r\n
[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
[\/hidden-answer]<\/div>\r\n<\/section>
[ohm2_question hide_question_numbers=1]18922[\/ohm2_question]<\/section>\r\n

Midpoint Formula<\/h2>\r\nWhen the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula<\/strong>.\r\n\r\n
\r\n

midpoint formula<\/h3>\r\nGiven the endpoints of a line segment, [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the midpoint [latex]M[\/latex] can be calculated using the formula:\r\n
[latex]M=\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex]<\/div>\r\n
<\/div>\r\n
\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"386\"]\"This Line graph on an x,y coordinate plane[\/caption]\r\n\r\n<\/div>\r\n<\/section>
Find the midpoint between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].[reveal-answer q=\"788934\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788934\"]Use the formula to find the midpoint of the line segment.\r\n
[latex]M = \\left( \\frac{-3 + 2}{2}, \\frac{-1 + 3}{2} \\right) = \\left( -0.5, 1 \\right)[\/latex]<\/div>\r\n
<\/div>\r\n
\r\n\r\n[caption id=\"attachment_1010\" align=\"aligncenter\" width=\"487\"]\"\" Triangle in x,y coordinate plane with midpoint labeled[\/caption]\r\n\r\n<\/div>\r\n
[\/hidden-answer]<\/div>\r\n<\/section>\r\n
[ohm2_question hide_question_numbers=1]18923[\/ohm2_question]<\/section><\/div>\r\n
The diameter of a circle has endpoints [latex]\\left(-1,-4\\right)[\/latex] and [latex]\\left(5,-4\\right)[\/latex]. Find the center of the circle.[reveal-answer q=\"418175\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"418175\"]The center of a circle is the center or midpoint of its diameter. Thus, the midpoint formula will yield the center point.\r\n[latex]\\begin{array}{l}\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)\\\\ \\left(\\frac{-1+5}{2},\\frac{-4 - 4}{2}\\right)=\\left(\\frac{4}{2},-\\frac{8}{2}\\right)=\\left(2,-4\\right)\\end{array}[\/latex]\r\n[\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]110938[\/ohm_question]<\/section>","rendered":"

Distance Formula<\/h2>\n

Building on our understanding of how to measure changes on a graph, let\u2019s now shift our focus to the distance formula, a powerful tool for calculating the precise distance between any two points on a graph.<\/p>\n

\"This
Triangle with sides labeled on an x,y coordinate plane<\/figcaption><\/figure>\n

Derived from the Pythagorean Theorem<\/strong>, the 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>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

The relationship of sides [latex]|{x}_{2}-{x}_{1}|[\/latex] and [latex]|{y}_{2}-{y}_{1}|[\/latex] to side d<\/em> is the same as that of sides a <\/em>and b <\/em>to side 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 c<\/em>, take the square root of both sides of the Pythagorean Theorem.<\/p>\n

[latex]{c}^{2}={a}^{2}+{b}^{2}\\rightarrow c=\\sqrt{{a}^{2}+{b}^{2}}[\/latex]<\/div>\n

It follows that the distance formula is given as<\/p>\n

[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
\n

distance formula<\/h3>\n

The 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

[latex]\\text{Distance}: d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/p>\n<\/section>\n

Find the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].<\/p>\n