{"id":347,"date":"2015-09-18T22:41:01","date_gmt":"2015-09-18T22:41:01","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=347"},"modified":"2015-11-04T22:52:59","modified_gmt":"2015-11-04T22:52:59","slug":"using-the-distance-formula","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/using-the-distance-formula\/","title":{"raw":"Using the Distance Formula","rendered":"Using the Distance Formula"},"content":{"raw":"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 a <\/em>and b<\/em> are the lengths of the legs adjacent to the right angle, and c<\/em> is the length of the hypotenuse.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This Figure 14<\/b>[\/caption]\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\nWe do not have to use the absolute value symbols in this definition because any number squared is positive.\r\n
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A General Note: The Distance Formula<\/h3>\r\nGiven endpoints [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex], the distance between two points is given by\r\n
[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n
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Example 5: Finding the Distance between Two Points<\/h3>\r\nFind the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex].\r\n\r\n<\/div>\r\n
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Solution<\/h3>\r\nLet us first look at the graph of the two points. Connect the points to form a right triangle as in\u00a0Figure 15.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"This Figure 15<\/b>[\/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<\/div>\r\n
\r\n\r\nFind the distance between two points: [latex]\\left(1,4\\right)[\/latex] and [latex]\\left(11,9\\right)[\/latex].\r\n\r\n<\/div>\r\n
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[latex]\\sqrt{125}=5\\sqrt{5}[\/latex]<\/p>\r\n\r\n<\/div>\r\n

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Example 6: Finding the Distance between Two Locations<\/h3>\r\nLet\u2019s return to the situation introduced at the beginning of this section.\r\n\r\nTracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions.\r\n\r\n<\/div>\r\n
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Solution<\/h3>\r\nThe first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at [latex]\\left(1,1\\right)[\/latex]. The next stop is 5 blocks to the east, so it is at [latex]\\left(5,1\\right)[\/latex]. After that, she traveled 3 blocks east and 2 blocks north to [latex]\\left(8,3\\right)[\/latex]. Lastly, she traveled 4 blocks north to [latex]\\left(8,7\\right)[\/latex]. We can label these points on the grid as in Figure 16.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"This Figure 16<\/b>[\/caption]\r\n\r\nNext, we can calculate the distance. Note that each grid unit represents 1,000 feet.\r\n