{"id":76,"date":"2019-02-06T17:31:12","date_gmt":"2019-02-06T17:31:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/coursedemo-collegealgebracoreq\/chapter\/distance-in-the-plane\/"},"modified":"2019-02-27T16:22:56","modified_gmt":"2019-02-27T16:22:56","slug":"distance-in-the-plane","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/chapter\/distance-in-the-plane\/","title":{"raw":"Distance in the Coordinate Plane","rendered":"Distance in the Coordinate Plane"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the distance formula to find the distance between two points in the plane.<\/li>\r\n \t<li>Use the midpoint formula to find the midpoint between two points.<\/li>\r\n<\/ul>\r\n<\/div>\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 <em>a <\/em>and <em>b<\/em> are the lengths of the legs adjacent to the right angle, and <em>c<\/em> is the length of the hypotenuse.\r\n\r\n<img class=\"aligncenter\" 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=\"487\" height=\"331\" \/>\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\nWe do not have to use the absolute value symbols in this definition because any number squared is positive.\r\n<div class=\"textbox\">\r\n<h3>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<div style=\"text-align: center;\">[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: 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[reveal-answer q=\"737169\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"737169\"]\r\n\r\nLet us first look at the graph of the two points. Connect the points to form a right triangle.\r\n\r\n<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=\"487\" height=\"289\" \/>\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<\/div>\r\n<div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<div>\r\n\r\nFind the distance between two points: [latex]\\left(1,4\\right)[\/latex] and [latex]\\left(11,9\\right)[\/latex].\r\n[reveal-answer q=\"934526\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934526\"]\r\n\r\n[latex]\\sqrt{125}=5\\sqrt{5}[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nIn the following video, we present more worked examples of how to use the distance formula to find the distance between two points in the coordinate plane.\r\n\r\nhttps:\/\/youtu.be\/Vj7twkiUgf0\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: 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. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions.\r\n\r\n[reveal-answer q=\"411250\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"411250\"]\r\n\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.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042433\/CNX_CAT_Figure_02_01_017.jpg\" alt=\"This is an image of a road map of a city. The point (1, 1) is on North Avenue and Bertau Avenue. The point (5, 1) is on North Avenue and Wolf Road. The point (8, 3) is on Mannheim Road and McLean Street. The point (8, 7) is on Mannheim Road and Schiller Avenue.\" width=\"731\" height=\"480\" \/>\r\n\r\nNext, we can calculate the distance. Note that each grid unit represents 1,000 feet.\r\n<ul>\r\n \t<li>From her starting location to her first stop at [latex]\\left(1,1\\right)[\/latex], Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop.<\/li>\r\n \t<li>Her second stop is at [latex]\\left(5,1\\right)[\/latex]. So from [latex]\\left(1,1\\right)[\/latex] to [latex]\\left(5,1\\right)[\/latex], Tracie drove east 4,000 feet.<\/li>\r\n \t<li>Her third stop is at [latex]\\left(8,3\\right)[\/latex]. There are a number of routes from [latex]\\left(5,1\\right)[\/latex] to [latex]\\left(8,3\\right)[\/latex]. Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let\u2019s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet.<\/li>\r\n \t<li>Tracie\u2019s final stop is at [latex]\\left(8,7\\right)[\/latex]. This is a straight drive north from [latex]\\left(8,3\\right)[\/latex] for a total of 4,000 feet.<\/li>\r\n<\/ul>\r\nNext, we will add the distances listed in the table.\r\n<table summary=\"A table with 6 rows and 2 columns. The entries in the first row are: From\/To and Number of Feet Driven. The entries in the second row are: (0, 0) to (1, 1) and 2,000. The entries in the third row are: (1, 1) to (5, 1) and 4,000. The entries in the fourth row are: (5, 1) to (8, 3) and 5,000. The entries in the fourth row are: (8, 3) to (8, 7) and 4,000. The entries in the sixth row are: Total and 15,000.\">\r\n<thead>\r\n<tr>\r\n<th>From\/To<\/th>\r\n<th>Number of Feet Driven<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]\\left(0,0\\right)[\/latex] to [latex]\\left(1,1\\right)[\/latex]<\/td>\r\n<td>2,000<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\left(1,1\\right)[\/latex] to [latex]\\left(5,1\\right)[\/latex]<\/td>\r\n<td>4,000<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\left(5,1\\right)[\/latex] to [latex]\\left(8,3\\right)[\/latex]<\/td>\r\n<td>5,000<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\\left(8,3\\right)[\/latex] to [latex]\\left(8,7\\right)[\/latex]<\/td>\r\n<td>4,000<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Total<\/td>\r\n<td>15,000<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe total distance Tracie drove is 15,000 feet or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points [latex]\\left(0,0\\right)[\/latex] and [latex]\\left(8,7\\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left(8 - 0\\right)}^{2}+{\\left(7 - 0\\right)}^{2}}\\hfill \\\\ =\\sqrt{64+49}\\hfill \\\\ =\\sqrt{113}\\hfill \\\\ =10.63\\text{ units}\\hfill \\end{array}[\/latex]<\/div>\r\nAt 1,000 feet per grid unit, the distance between Elmhurst, IL to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point [latex]\\left(8,7\\right)[\/latex]. Perhaps you have heard the saying \"as the crow flies,\" which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using the 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 <strong>midpoint formula<\/strong>. Given 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 formula states how to find the coordinates of the midpoint [latex]M[\/latex].\r\n<div style=\"text-align: center;\">[latex]M=\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex]<\/div>\r\nA graphical view of a midpoint is shown below. Notice that the line segments on either side of the midpoint are congruent.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042436\/CNX_CAT_Figure_02_01_018.jpg\" alt=\"This is a line graph on an x, y coordinate plane with the x and y axes ranging from 0 to 6. The points (x sub 1, y sub 1), (x sub 2, y sub 2), and (x sub 1 plus x sub 2 all over 2, y sub 1 plus y sub 2 all over 2) are plotted. A straight line runs through these three points. Pairs of short parallel lines bisect the two sections of the line to note that they are equivalent.\" width=\"487\" height=\"290\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Midpoint of the Line Segment<\/h3>\r\nFind the midpoint of the line segment with the endpoints [latex]\\left(7,-2\\right)[\/latex] and [latex]\\left(9,5\\right)[\/latex].\r\n\r\n[reveal-answer q=\"788934\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"788934\"]\r\n\r\nUse the formula to find the midpoint of the line segment.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)\\hfill&amp;=\\left(\\frac{7+9}{2},\\frac{-2+5}{2}\\right)\\hfill \\\\ \\hfill&amp;=\\left(8,\\frac{3}{2}\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the midpoint of the line segment with endpoints [latex]\\left(-2,-1\\right)[\/latex] and [latex]\\left(-8,6\\right)[\/latex].\r\n\r\n[reveal-answer q=\"964077\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"964077\"]\r\n\r\n[latex]\\left(-5,\\frac{5}{2}\\right)[\/latex][\/hidden-answer]\r\n\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=2308&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Center of a Circle<\/h3>\r\nThe 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.\r\n\r\n[reveal-answer q=\"418175\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"418175\"]\r\n\r\nThe center of a circle is the center or midpoint of its diameter. Thus, the midpoint formula will yield the center point.\r\n<div style=\"text-align: center;\">[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]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110938&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the distance formula to find the distance between two points in the plane.<\/li>\n<li>Use the midpoint formula to find the midpoint between two points.<\/li>\n<\/ul>\n<\/div>\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 <em>a <\/em>and <em>b<\/em> are the lengths of the legs adjacent to the right angle, and <em>c<\/em> is the length of the hypotenuse.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" 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=\"487\" height=\"331\" \/><\/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<p>We do not have to use the absolute value symbols in this definition because any number squared is positive.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Distance Formula<\/h3>\n<p>Given 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<\/p>\n<div style=\"text-align: center;\">[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Distance between Two Points<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737169\">Show Solution<\/span><\/p>\n<div id=\"q737169\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let us first look at the graph of the two points. Connect the points to form a right triangle.<\/p>\n<p><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=\"487\" height=\"289\" \/><\/p>\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<\/div>\n<div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<div>\n<p>Find the distance between two points: [latex]\\left(1,4\\right)[\/latex] and [latex]\\left(11,9\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934526\">Show Solution<\/span><\/p>\n<div id=\"q934526\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\sqrt{125}=5\\sqrt{5}[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we present more worked examples of how to use the distance formula to find the distance between two points in the coordinate plane.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Example:  Determine the Distance Between Two Points\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Vj7twkiUgf0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Distance between Two Locations<\/h3>\n<p>Let\u2019s return to the situation introduced at the beginning of this section.<\/p>\n<p>Tracie 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. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q411250\">Show Solution<\/span><\/p>\n<div id=\"q411250\" class=\"hidden-answer\" style=\"display: none\">\n<p>The 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.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042433\/CNX_CAT_Figure_02_01_017.jpg\" alt=\"This is an image of a road map of a city. The point (1, 1) is on North Avenue and Bertau Avenue. The point (5, 1) is on North Avenue and Wolf Road. The point (8, 3) is on Mannheim Road and McLean Street. The point (8, 7) is on Mannheim Road and Schiller Avenue.\" width=\"731\" height=\"480\" \/><\/p>\n<p>Next, we can calculate the distance. Note that each grid unit represents 1,000 feet.<\/p>\n<ul>\n<li>From her starting location to her first stop at [latex]\\left(1,1\\right)[\/latex], Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop.<\/li>\n<li>Her second stop is at [latex]\\left(5,1\\right)[\/latex]. So from [latex]\\left(1,1\\right)[\/latex] to [latex]\\left(5,1\\right)[\/latex], Tracie drove east 4,000 feet.<\/li>\n<li>Her third stop is at [latex]\\left(8,3\\right)[\/latex]. There are a number of routes from [latex]\\left(5,1\\right)[\/latex] to [latex]\\left(8,3\\right)[\/latex]. Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let\u2019s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet.<\/li>\n<li>Tracie\u2019s final stop is at [latex]\\left(8,7\\right)[\/latex]. This is a straight drive north from [latex]\\left(8,3\\right)[\/latex] for a total of 4,000 feet.<\/li>\n<\/ul>\n<p>Next, we will add the distances listed in the table.<\/p>\n<table summary=\"A table with 6 rows and 2 columns. The entries in the first row are: From\/To and Number of Feet Driven. The entries in the second row are: (0, 0) to (1, 1) and 2,000. The entries in the third row are: (1, 1) to (5, 1) and 4,000. The entries in the fourth row are: (5, 1) to (8, 3) and 5,000. The entries in the fourth row are: (8, 3) to (8, 7) and 4,000. The entries in the sixth row are: Total and 15,000.\">\n<thead>\n<tr>\n<th>From\/To<\/th>\n<th>Number of Feet Driven<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]\\left(0,0\\right)[\/latex] to [latex]\\left(1,1\\right)[\/latex]<\/td>\n<td>2,000<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\left(1,1\\right)[\/latex] to [latex]\\left(5,1\\right)[\/latex]<\/td>\n<td>4,000<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\left(5,1\\right)[\/latex] to [latex]\\left(8,3\\right)[\/latex]<\/td>\n<td>5,000<\/td>\n<\/tr>\n<tr>\n<td>[latex]\\left(8,3\\right)[\/latex] to [latex]\\left(8,7\\right)[\/latex]<\/td>\n<td>4,000<\/td>\n<\/tr>\n<tr>\n<td>Total<\/td>\n<td>15,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>The total distance Tracie drove is 15,000 feet or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points [latex]\\left(0,0\\right)[\/latex] and [latex]\\left(8,7\\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}d=\\sqrt{{\\left(8 - 0\\right)}^{2}+{\\left(7 - 0\\right)}^{2}}\\hfill \\\\ =\\sqrt{64+49}\\hfill \\\\ =\\sqrt{113}\\hfill \\\\ =10.63\\text{ units}\\hfill \\end{array}[\/latex]<\/div>\n<p>At 1,000 feet per grid unit, the distance between Elmhurst, IL to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point [latex]\\left(8,7\\right)[\/latex]. Perhaps you have heard the saying &#8220;as the crow flies,&#8221; which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using the Midpoint Formula<\/h2>\n<p>When 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 <strong>midpoint formula<\/strong>. Given 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 formula states how to find the coordinates of the midpoint [latex]M[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]M=\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex]<\/div>\n<p>A graphical view of a midpoint is shown below. Notice that the line segments on either side of the midpoint are congruent.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/12042436\/CNX_CAT_Figure_02_01_018.jpg\" alt=\"This is a line graph on an x, y coordinate plane with the x and y axes ranging from 0 to 6. The points (x sub 1, y sub 1), (x sub 2, y sub 2), and (x sub 1 plus x sub 2 all over 2, y sub 1 plus y sub 2 all over 2) are plotted. A straight line runs through these three points. Pairs of short parallel lines bisect the two sections of the line to note that they are equivalent.\" width=\"487\" height=\"290\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Midpoint of the Line Segment<\/h3>\n<p>Find the midpoint of the line segment with the endpoints [latex]\\left(7,-2\\right)[\/latex] and [latex]\\left(9,5\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q788934\">Show Solution<\/span><\/p>\n<div id=\"q788934\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the formula to find the midpoint of the line segment.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(\\frac{{x}_{1}+{x}_{2}}{2},\\frac{{y}_{1}+{y}_{2}}{2}\\right)\\hfill&=\\left(\\frac{7+9}{2},\\frac{-2+5}{2}\\right)\\hfill \\\\ \\hfill&=\\left(8,\\frac{3}{2}\\right)\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the midpoint of the line segment with endpoints [latex]\\left(-2,-1\\right)[\/latex] and [latex]\\left(-8,6\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q964077\">Show Solution<\/span><\/p>\n<div id=\"q964077\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(-5,\\frac{5}{2}\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=2308&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Center of a Circle<\/h3>\n<p>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.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q418175\">Show Solution<\/span><\/p>\n<div id=\"q418175\" class=\"hidden-answer\" style=\"display: none\">\n<p>The center of a circle is the center or midpoint of its diameter. Thus, the midpoint formula will yield the center point.<\/p>\n<div style=\"text-align: center;\">[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]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=110938&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\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-76\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Midpoint Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/j2blmpiid7\">https:\/\/www.desmos.com\/calculator\/j2blmpiid7<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Distance in the Plane Interactive. <strong>Authored by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/nrtjcgmy69\">https:\/\/www.desmos.com\/calculator\/nrtjcgmy69<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 19140. <strong>Authored by<\/strong>: Amy Lambert. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 2308. <strong>Authored by<\/strong>: David Lippman. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 110938. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Example: Determine the Distance Between Two Points. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Vj7twkiUgf0\">https:\/\/youtu.be\/Vj7twkiUgf0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: OpenStax College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/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\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"OpenStax College Algebra\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"original\",\"description\":\"Midpoint Interactive\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/www.desmos.com\/calculator\/j2blmpiid7\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Distance in the Plane Interactive\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/www.desmos.com\/calculator\/nrtjcgmy69\",\"project\":\"\",\"license\":\"pd\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 19140\",\"author\":\"Amy Lambert\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC- BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 2308\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC- BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 110938\",\"author\":\"Lumen 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