{"id":169,"date":"2023-11-08T16:10:12","date_gmt":"2023-11-08T16:10:12","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-pythagorean-theorem\/"},"modified":"2026-02-05T11:34:50","modified_gmt":"2026-02-05T11:34:50","slug":"5-8-distance-and-midpoint-formulas-pythagorean-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/5-8-distance-and-midpoint-formulas-pythagorean-theorem\/","title":{"raw":"5.8 Distance and Midpoint Formulas, Pythagorean Theorem","rendered":"5.8 Distance and Midpoint Formulas, Pythagorean Theorem"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve applications using the Pythagorean Theorem<\/li>\r\n \t<li>Compute the distance between two points using the Distance Formula<\/li>\r\n \t<li>Find the midpoint between two points using the Midpoint Formula<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>The Pythagorean Theorem<\/h2>\r\nThe <b>Pythagorean theorem<\/b> is a statement about the sides of a right triangle.\u00a0One of the angles of a right triangle is always equal to\u00a0[latex]90[\/latex] degrees. This angle is the <strong>right angle<\/strong>. The two sides next to the right angle are called the <strong>legs<\/strong> and the other side is called the <strong>hypotenuse<\/strong>. The hypotenuse is the side opposite the right angle, and it is always the longest side.\r\n\r\n[caption id=\"attachment_4904\" align=\"alignleft\" width=\"132\"]<img class=\"wp-image-4904\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161955\/Screen-Shot-2016-06-14-at-8.32.56-PM-268x300.png\" alt=\"Right triangle with right angle in bottom left. Horizontal length of a. Vertical length of b and diagonal length of c across from right angle.\" width=\"132\" height=\"148\" \/> A right triangle with sides labeled.[\/caption]\r\n<p style=\"text-align: left;\">The Pythagorean theorem is often used to find unknown lengths of the sides of right triangles. If the hypotenuse is labeled [latex]c[\/latex], and the other two sides [latex]a[\/latex] and [latex]b[\/latex] as in the image on the left, \u00a0The Pythagorean Theorem states that<\/p>\r\n<p style=\"text-align: center;\">[latex]a^2+b^2=c^2[\/latex]<\/p>\r\n<p style=\"text-align: left;\">If we are given any two side lengths of a right triangle, we can use the Pythagorean Theorem to solve for the remaining side.<\/p>\r\n<p style=\"text-align: left;\">Consider an example where\u00a0[latex]a=5[\/latex] and\u00a0[latex]b=7.[\/latex] We can write<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}5^2+7^2&amp;=c^2\\\\25+49&amp;=c^2\\\\74&amp;=c^2\\end{align}[\/latex]<\/p>\r\nWe would normally solve a quadratic equation such as this by factoring, but\u00a0[latex]74[\/latex] is not a perfect square, so none of our factoring techniques will be successful. We need to introduce a new technique - the square root property.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>SQUARE ROOT PROPERTY<\/h3>\r\nIf [latex]x^2=n[\/latex] where\u00a0[latex]n[\/latex] is positive, then\u00a0[latex]x=\\sqrt{n}[\/latex] or\u00a0[latex]x=-\\sqrt{n}.[\/latex]\r\n\r\n<\/div>\r\nThe reason we get two solutions is because of the observation earlier in this chapter that every positive number has two square roots - the principle root and the negative root. Squaring either of them results in a positive quantity. In this section we will be working with lengths and distances, so only the positive answer makes sense. In the next chapter we will need both solutions.\r\n\r\nThe length of the hypotenuse is\u00a0[latex]c=\\sqrt{74}\\approx 8.60[\/latex] The final answer can either be stated as a simplified radical or approximated as a decimal, depending on the prompt of the problem.\r\n\r\nTry this example yourself.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nA triangle has one leg of length [latex]6[\/latex] and hypotenuse of length [latex]12[\/latex]. Find the length of the remaining leg.\u00a0Write the answer both in simplified exact form and as an approximation rounded to two decimal places.\r\n[reveal-answer q=\"613245\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"613245\"]\r\n\r\nSince the two legs are interchangeable in the Pythagorean Theorem, say that [latex]a=6,c=12,[\/latex] and the unknown leg is [latex]b.[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}a^2+b^2&amp;=c^2\\\\6^2+b^2&amp;=12^2\\\\36+b^2&amp;=144\\\\b^2&amp;=108&amp;&amp;\\color{blue}{\\textsf{subtract 36 from both sides}}\\\\b&amp;=\\sqrt{108}&amp;&amp;\\color{blue}{\\textsf{Square Root Property}}\\\\b&amp;=\\sqrt{4\\cdot9\\cdot3}\\\\&amp;=2\\cdot 3\\sqrt{3}&amp;&amp;\\color{blue}{\\textsf{simplify the radical}}\\\\b&amp;=6\\sqrt{3}\\end{align}[\/latex]<\/p>\r\nThe exact answer is [latex]6\\sqrt{3}.[\/latex] The approximate answer is [latex]10.39.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nMany application problems involve finding the lengths of sides in a right triangle.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nTV screens are typically measured by the length of their diagonal from one corner to the opposite corner. Suppose that a [latex]55[\/latex] inch TV has a width of [latex]48[\/latex] inches. How tall is the screen? Round your answer to the nearest tenth of an inch.\r\n[reveal-answer q=\"613247\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"613247\"]\r\n\r\n<img class=\"aligncenter wp-image-1588 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8TV-300x201.png\" alt=\"Rectangle divided from bottom left to upper right. Right triangle formed with hypotenuse 55, base 48, and height b.\" width=\"300\" height=\"201\" \/>\r\n\r\nIf we draw the diagonal across a TV screen, it creates a right triangle with legs the height and the width. In the Pythagorean Theorem, label [latex]a=48,c=55,[\/latex] and the unknown leg (the height of the TV) is [latex]b.[\/latex]\r\n<p style=\"text-align: center;\">[latex]\\begin{align}a^2+b^2&amp;=c^2\\\\48^2+b^2&amp;=55^2\\\\2304+b^2&amp;=3025\\\\b^2&amp;=721&amp;&amp;\\color{blue}{\\textsf{subtract 2304 from both sides}}\\\\b&amp;=\\sqrt{721}&amp;&amp;\\color{blue}{\\textsf{Square Root Property}}\\end{align}[\/latex]<\/p>\r\nThe approximate answer is [latex]\\sqrt{721}\\approx 26.9[\/latex] inches.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p style=\"text-align: left;\">Sometimes we can solve when multiple side lengths are unknown, if we know a relationship between the sides.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA\u00a0right triangle has one leg with length [latex]x[\/latex], another whose length is greater by two, and the length of the hypotenuse is greater by four. Find the lengths of the sides of the triangle. Use the image below.\r\n\r\n<img class=\"alignnone wp-image-4907 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161957\/Screen-Shot-2016-06-14-at-8.49.02-PM-264x300.png\" alt=\"Right triangle with horizontal leg labeled x, vertical leg labeled X plus 2 and hypotenuse labeled X plus 4.\" width=\"264\" height=\"300\" \/>\r\n[reveal-answer q=\"133740\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"133740\"]\r\n\r\nWe can use the three side lengths in the Pythagorean Theorem:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}a^2+b^2=c^2\\\\x^2+\\left(x+2\\right)^2=\\left(x+4\\right)^2\\end{array}[\/latex]<\/p>\r\nNext, we expand and combine like terms so we can determine what kind of equation we have and what technique will work best to solve it.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^2+\\left(x+2\\right)^2=\\left(x+4\\right)^2\\\\x^2+x^2+4x+4=x^2+8x+16\\\\2x^2+4x+4=x^2+8x+16\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Since this is a polynomial equation with multiple variable terms, move all the terms to one side and see if we can factor.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^2+4x+4=x^2+8x+16\\\\\\underline{-x^2}\\,\\,\\,\\underline{-8x}\\,\\,\\,\\underline{-16}\\,\\,\\,\\,\\,\\underline{-x^2}\\,\\,\\,\\underline{-8x}\\,\\,\\,\\underline{-16}\\\\x^2-4x-12=0\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">The left side is now factorable:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(x-6\\right)\\left(x+2\\right)=0[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Set each factor equal to zero:<\/p>\r\n<p style=\"text-align: center;\">[latex]x-6=0, x=6[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x+2=0, x=-2[\/latex]<\/p>\r\n<p style=\"text-align: left;\">It does not make sense to have a length equal to\u00a0[latex]-2[\/latex], so we can safely throw that solution out. \u00a0The lengths of the sides are as follows:<\/p>\r\n<p style=\"text-align: left;\">[latex]x=6[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[latex]x+2=6+2=8[\/latex]<\/p>\r\n<p style=\"text-align: left;\">[latex]x+4=6+4=10[\/latex]<\/p>\r\n<p style=\"text-align: left;\"><strong>Check:\u00a0<\/strong>Since we know the relationship between the sides of a right triangle, we can check that we are correct. Sometimes it helps to draw a picture<\/p>\r\n<p style=\"text-align: left;\">.<img class=\"alignnone wp-image-4908 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161959\/Screen-Shot-2016-06-14-at-9.20.06-PM-300x263.png\" alt=\"Right triangle with horizontal leg labeled X = 6 = a, vertical leg labeled X plus 2 = 8 = b and hypotenuse X plus 4 = 10 = c.\" width=\"300\" height=\"263\" \/><\/p>\r\n<p style=\"text-align: left;\">We know that [latex]a^2+b^2=c^2[\/latex], so we can substitute the values we found:<\/p>\r\n<p style=\"text-align: left;\">[latex]\\begin{array}{l}6^2+8^2=10^2\\\\36+64=100\\\\100=100\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Our solution checks out.<\/p>\r\nThe lengths of the sides of the right triangle are\u00a0[latex]6, 8[\/latex], and\u00a0[latex]10[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is a video illustrating the same concept.\r\n\r\nhttps:\/\/youtu.be\/xeP5pRBqsNs\r\n<h2>The Distance Formula<\/h2>\r\nDerived from the Pythagorean Theorem, the <strong>distance formula<\/strong> is used to find the distance between two points in the plane. Consider points [latex](x_1,y_1)[\/latex] and\u00a0[latex](x_2,y_2).[\/latex] The diagram below shows the desired distance [latex]d[\/latex]\u00a0between them, which is the hypotenuse of a right triangle with sides parallel to the axes. We can find the lengths of the legs by finding the horizontal distance\u00a0[latex]x_2-x_1[\/latex] and vertical distance\u00a0[latex]y_2-y_1[\/latex] between the points. We use absolute values to make sure that these differences are nonnegative.\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=\"Triangle on an x, y coordinate plane showing quadrant one. 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 right triangle. Base labeled: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse labeled: d = c. Vertical side labeled: the absolute value of y sub 2 minus y sub 1 equals b.\" width=\"487\" height=\"331\" \/>\r\n\r\nUsing these three side lengths in the Pythagorean Theorem, we obtain\r\n<p style=\"text-align: center;\">[latex]\\begin{align}d^2&amp;={\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}\\\\ d&amp;=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\end{align}[\/latex]<\/p>\r\nNote that we ended up ignoring the absolute values because squaring always results in a nonnegative quantity anyway. Here is the distance formula stated formally:\r\n<div class=\"textbox key-takeaways\">\r\n<h3>The distance formula<\/h3>\r\nThe distance between the two points [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] 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<\/h3>\r\nFind the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex]. Write the answer both in exact form and as an approximation rounded to 2 decimal places.\r\n\r\n[reveal-answer q=\"737169\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"737169\"]\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 class=\"alignnone\" 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=\"Triangle on an x, y coordinate plane. The points (negative 3, negative 1); (2, negative 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 the line segment connecting the points 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 \\\\ d=\\sqrt{{\\left(5\\right)}^{2}+{\\left(4\\right)}^{2}}\\hfill \\\\ d=\\sqrt{25+16}\\hfill \\\\ d=\\sqrt{41}\\hfill \\end{array}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>The exact answer is [latex]\\sqrt{41}.[\/latex] The approximate answer is [latex]6.40.[\/latex]<\/div>\r\n<div><\/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 YOURSELF<\/h3>\r\n<div>\r\n\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=213543&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\" data-mce-fragment=\"1\"><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/iframe>\r\n\r\n<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA subway takes passengers from a station at 9th street East, 3rd street North in a straight line to another station at 12th street East, 10th street North. If each city block (for example, 3rd East to 4th East) is one tenth of a mile, how far is it between the two stations to the nearest hundredth of a mile?\r\n\r\n[reveal-answer q=\"907635\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"907635\"]\r\n\r\nWe first determine the distance between the two stops in units of city blocks. Then we convert this to miles.\r\n\r\nLabel the locations of the stations as ordered pairs, [latex](9, 3)[\/latex] and [latex](12,10).[\/latex] Then use the distance formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}d&amp;=\\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\r\nd&amp;=\\sqrt{(12-9)^2+(10-3)^2}\\\\\r\nd&amp;=\\sqrt{(3)^2+(7)^2}\\\\\r\nd&amp;=\\sqrt{9+49}\\\\\r\nd&amp;=\\sqrt{58}\\end{align}[\/latex]<\/p>\r\nMake sure not to round to the nearest hundredth until the final answer! We convert city blocks to miles by multiplying [latex]\\sqrt{58}\\textsf{ blocks}\\cdot\\frac{0.1\\textsf{ miles}}{1\\textsf{ block}}\\approx 0.76\\textsf{ miles}[\/latex]\r\n\r\nIt is approximately [latex]0.76[\/latex] miles in between the two stations in a straight line.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>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 <strong>midpoint<\/strong> and the formula is known as the <strong>midpoint formula<\/strong>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>THE 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 <strong>midpoint<\/strong> is\r\n<p style=\"text-align: center;\">[latex]\\displaystyle M=\\left(\\dfrac{{x}_{1}+{x}_{2}}{2},\\dfrac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\nIt makes sense that since the midpoint is halfway from Point\u00a0[latex]1[\/latex] to Point [latex]2,[\/latex] it should be halfway in both the\u00a0[latex]x[\/latex] direction and the\u00a0[latex]y[\/latex] direction. The \"midpoint\" between two real numbers is called the <strong>average <\/strong>or<strong> mean<\/strong>. Thus, you can equivalently view the midpoint formula as\r\n<p style=\"text-align: center;\">[latex]\\displaystyle M=\\left(\\textsf{avg. of x coords, avg. of y coords}\\right)[\/latex]<\/p>\r\nA graphical view of a midpoint is shown below. Notice that the line segments on either side of the midpoint are congruent. If you like, you can verify that these two segments have equal length using the Distance Formula.\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=\"Line segment on an x, y coordinate plane. 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<\/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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"788934\"]\r\nUse the formula to find the midpoint of the line segment.\r\n<div style=\"text-align: center;\">[latex]\\large \\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><img class=\"aligncenter wp-image-1758 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8Midpoint-300x288.png\" alt=\"Coordinate plane grid with line segment from (7, negative 2) to (9,5) and midpoint (8,1.5).\" width=\"300\" height=\"288\" \/><\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It YOURSELF<\/h3>\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288858&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nThe Pythagorean Theorem [latex]a^2+b^2=c^2[\/latex] can be used to determine the side lengths of a right triangle with legs [latex]a, b[\/latex] and hypotenuse [latex]c.[\/latex] While solving the equation, we can use the Square Root Property: if [latex]x^2=n[\/latex], then [latex]x=\\sqrt{n}[\/latex] (assuming [latex]x[\/latex] is positive.)\r\n\r\nThe distance formula\u00a0[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex] can find the distance between any two points in the plane.\u00a0 The midpoint formula [latex]M=\\left(\\dfrac{{x}_{1}+{x}_{2}}{2},\\dfrac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex] finds the midpoint of the line segment connecting any two points.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve applications using the Pythagorean Theorem<\/li>\n<li>Compute the distance between two points using the Distance Formula<\/li>\n<li>Find the midpoint between two points using the Midpoint Formula<\/li>\n<\/ul>\n<\/div>\n<h2>The Pythagorean Theorem<\/h2>\n<p>The <b>Pythagorean theorem<\/b> is a statement about the sides of a right triangle.\u00a0One of the angles of a right triangle is always equal to\u00a0[latex]90[\/latex] degrees. This angle is the <strong>right angle<\/strong>. The two sides next to the right angle are called the <strong>legs<\/strong> and the other side is called the <strong>hypotenuse<\/strong>. The hypotenuse is the side opposite the right angle, and it is always the longest side.<\/p>\n<div id=\"attachment_4904\" style=\"width: 142px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4904\" class=\"wp-image-4904\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161955\/Screen-Shot-2016-06-14-at-8.32.56-PM-268x300.png\" alt=\"Right triangle with right angle in bottom left. Horizontal length of a. Vertical length of b and diagonal length of c across from right angle.\" width=\"132\" height=\"148\" \/><\/p>\n<p id=\"caption-attachment-4904\" class=\"wp-caption-text\">A right triangle with sides labeled.<\/p>\n<\/div>\n<p style=\"text-align: left;\">The Pythagorean theorem is often used to find unknown lengths of the sides of right triangles. If the hypotenuse is labeled [latex]c[\/latex], and the other two sides [latex]a[\/latex] and [latex]b[\/latex] as in the image on the left, \u00a0The Pythagorean Theorem states that<\/p>\n<p style=\"text-align: center;\">[latex]a^2+b^2=c^2[\/latex]<\/p>\n<p style=\"text-align: left;\">If we are given any two side lengths of a right triangle, we can use the Pythagorean Theorem to solve for the remaining side.<\/p>\n<p style=\"text-align: left;\">Consider an example where\u00a0[latex]a=5[\/latex] and\u00a0[latex]b=7.[\/latex] We can write<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}5^2+7^2&=c^2\\\\25+49&=c^2\\\\74&=c^2\\end{align}[\/latex]<\/p>\n<p>We would normally solve a quadratic equation such as this by factoring, but\u00a0[latex]74[\/latex] is not a perfect square, so none of our factoring techniques will be successful. We need to introduce a new technique &#8211; the square root property.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>SQUARE ROOT PROPERTY<\/h3>\n<p>If [latex]x^2=n[\/latex] where\u00a0[latex]n[\/latex] is positive, then\u00a0[latex]x=\\sqrt{n}[\/latex] or\u00a0[latex]x=-\\sqrt{n}.[\/latex]<\/p>\n<\/div>\n<p>The reason we get two solutions is because of the observation earlier in this chapter that every positive number has two square roots &#8211; the principle root and the negative root. Squaring either of them results in a positive quantity. In this section we will be working with lengths and distances, so only the positive answer makes sense. In the next chapter we will need both solutions.<\/p>\n<p>The length of the hypotenuse is\u00a0[latex]c=\\sqrt{74}\\approx 8.60[\/latex] The final answer can either be stated as a simplified radical or approximated as a decimal, depending on the prompt of the problem.<\/p>\n<p>Try this example yourself.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>A triangle has one leg of length [latex]6[\/latex] and hypotenuse of length [latex]12[\/latex]. Find the length of the remaining leg.\u00a0Write the answer both in simplified exact form and as an approximation rounded to two decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q613245\">Show Solution<\/span><\/p>\n<div id=\"q613245\" class=\"hidden-answer\" style=\"display: none\">\n<p>Since the two legs are interchangeable in the Pythagorean Theorem, say that [latex]a=6,c=12,[\/latex] and the unknown leg is [latex]b.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}a^2+b^2&=c^2\\\\6^2+b^2&=12^2\\\\36+b^2&=144\\\\b^2&=108&&\\color{blue}{\\textsf{subtract 36 from both sides}}\\\\b&=\\sqrt{108}&&\\color{blue}{\\textsf{Square Root Property}}\\\\b&=\\sqrt{4\\cdot9\\cdot3}\\\\&=2\\cdot 3\\sqrt{3}&&\\color{blue}{\\textsf{simplify the radical}}\\\\b&=6\\sqrt{3}\\end{align}[\/latex]<\/p>\n<p>The exact answer is [latex]6\\sqrt{3}.[\/latex] The approximate answer is [latex]10.39.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Many application problems involve finding the lengths of sides in a right triangle.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>TV screens are typically measured by the length of their diagonal from one corner to the opposite corner. Suppose that a [latex]55[\/latex] inch TV has a width of [latex]48[\/latex] inches. How tall is the screen? Round your answer to the nearest tenth of an inch.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q613247\">Show Solution<\/span><\/p>\n<div id=\"q613247\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1588 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8TV-300x201.png\" alt=\"Rectangle divided from bottom left to upper right. Right triangle formed with hypotenuse 55, base 48, and height b.\" width=\"300\" height=\"201\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8TV-300x201.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8TV-65x44.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8TV-225x151.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8TV-350x235.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8TV.png 638w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>If we draw the diagonal across a TV screen, it creates a right triangle with legs the height and the width. In the Pythagorean Theorem, label [latex]a=48,c=55,[\/latex] and the unknown leg (the height of the TV) is [latex]b.[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}a^2+b^2&=c^2\\\\48^2+b^2&=55^2\\\\2304+b^2&=3025\\\\b^2&=721&&\\color{blue}{\\textsf{subtract 2304 from both sides}}\\\\b&=\\sqrt{721}&&\\color{blue}{\\textsf{Square Root Property}}\\end{align}[\/latex]<\/p>\n<p>The approximate answer is [latex]\\sqrt{721}\\approx 26.9[\/latex] inches.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p style=\"text-align: left;\">Sometimes we can solve when multiple side lengths are unknown, if we know a relationship between the sides.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A\u00a0right triangle has one leg with length [latex]x[\/latex], another whose length is greater by two, and the length of the hypotenuse is greater by four. Find the lengths of the sides of the triangle. Use the image below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4907 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161957\/Screen-Shot-2016-06-14-at-8.49.02-PM-264x300.png\" alt=\"Right triangle with horizontal leg labeled x, vertical leg labeled X plus 2 and hypotenuse labeled X plus 4.\" width=\"264\" height=\"300\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q133740\">Show Solution<\/span><\/p>\n<div id=\"q133740\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can use the three side lengths in the Pythagorean Theorem:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}a^2+b^2=c^2\\\\x^2+\\left(x+2\\right)^2=\\left(x+4\\right)^2\\end{array}[\/latex]<\/p>\n<p>Next, we expand and combine like terms so we can determine what kind of equation we have and what technique will work best to solve it.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^2+\\left(x+2\\right)^2=\\left(x+4\\right)^2\\\\x^2+x^2+4x+4=x^2+8x+16\\\\2x^2+4x+4=x^2+8x+16\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Since this is a polynomial equation with multiple variable terms, move all the terms to one side and see if we can factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2x^2+4x+4=x^2+8x+16\\\\\\underline{-x^2}\\,\\,\\,\\underline{-8x}\\,\\,\\,\\underline{-16}\\,\\,\\,\\,\\,\\underline{-x^2}\\,\\,\\,\\underline{-8x}\\,\\,\\,\\underline{-16}\\\\x^2-4x-12=0\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">The left side is now factorable:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x-6\\right)\\left(x+2\\right)=0[\/latex]<\/p>\n<p style=\"text-align: left;\">Set each factor equal to zero:<\/p>\n<p style=\"text-align: center;\">[latex]x-6=0, x=6[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x+2=0, x=-2[\/latex]<\/p>\n<p style=\"text-align: left;\">It does not make sense to have a length equal to\u00a0[latex]-2[\/latex], so we can safely throw that solution out. \u00a0The lengths of the sides are as follows:<\/p>\n<p style=\"text-align: left;\">[latex]x=6[\/latex]<\/p>\n<p style=\"text-align: left;\">[latex]x+2=6+2=8[\/latex]<\/p>\n<p style=\"text-align: left;\">[latex]x+4=6+4=10[\/latex]<\/p>\n<p style=\"text-align: left;\"><strong>Check:\u00a0<\/strong>Since we know the relationship between the sides of a right triangle, we can check that we are correct. Sometimes it helps to draw a picture<\/p>\n<p style=\"text-align: left;\">.<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-4908 size-medium\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/20161959\/Screen-Shot-2016-06-14-at-9.20.06-PM-300x263.png\" alt=\"Right triangle with horizontal leg labeled X = 6 = a, vertical leg labeled X plus 2 = 8 = b and hypotenuse X plus 4 = 10 = c.\" width=\"300\" height=\"263\" \/><\/p>\n<p style=\"text-align: left;\">We know that [latex]a^2+b^2=c^2[\/latex], so we can substitute the values we found:<\/p>\n<p style=\"text-align: left;\">[latex]\\begin{array}{l}6^2+8^2=10^2\\\\36+64=100\\\\100=100\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left;\">Our solution checks out.<\/p>\n<p>The lengths of the sides of the right triangle are\u00a0[latex]6, 8[\/latex], and\u00a0[latex]10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is a video illustrating the same concept.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factoring Application - Find the Lengths of Three Sides of a Right Triangle (Pythagorean Theorem)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/xeP5pRBqsNs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>The Distance Formula<\/h2>\n<p>Derived from the Pythagorean Theorem, the <strong>distance formula<\/strong> is used to find the distance between two points in the plane. Consider points [latex](x_1,y_1)[\/latex] and\u00a0[latex](x_2,y_2).[\/latex] The diagram below shows the desired distance [latex]d[\/latex]\u00a0between them, which is the hypotenuse of a right triangle with sides parallel to the axes. We can find the lengths of the legs by finding the horizontal distance\u00a0[latex]x_2-x_1[\/latex] and vertical distance\u00a0[latex]y_2-y_1[\/latex] between the points. We use absolute values to make sure that these differences are nonnegative.<\/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=\"Triangle on an x, y coordinate plane showing quadrant one. 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 right triangle. Base labeled: the absolute value of x sub 2 minus x sub 1 equals a. The hypotenuse labeled: d = c. Vertical side labeled: the absolute value of y sub 2 minus y sub 1 equals b.\" width=\"487\" height=\"331\" \/><\/p>\n<p>Using these three side lengths in the Pythagorean Theorem, we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}d^2&={\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}\\\\ d&=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\end{align}[\/latex]<\/p>\n<p>Note that we ended up ignoring the absolute values because squaring always results in a nonnegative quantity anyway. Here is the distance formula stated formally:<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>The distance formula<\/h3>\n<p>The distance between the two points [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2},{y}_{2}\\right)[\/latex] 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<\/h3>\n<p>Find the distance between the points [latex]\\left(-3,-1\\right)[\/latex] and [latex]\\left(2,3\\right)[\/latex]. Write the answer both in exact form and as an approximation rounded to 2 decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737169\">Solution<\/span><\/p>\n<div id=\"q737169\" class=\"hidden-answer\" style=\"display: none\">\nLet 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\" class=\"alignnone\" 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=\"Triangle on an x, y coordinate plane. The points (negative 3, negative 1); (2, negative 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 the line segment connecting the points 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 \\\\ d=\\sqrt{{\\left(5\\right)}^{2}+{\\left(4\\right)}^{2}}\\hfill \\\\ d=\\sqrt{25+16}\\hfill \\\\ d=\\sqrt{41}\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<div>The exact answer is [latex]\\sqrt{41}.[\/latex] The approximate answer is [latex]6.40.[\/latex]<\/div>\n<div><\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It YOURSELF<\/h3>\n<div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=213543&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\" data-mce-fragment=\"1\"><span data-mce-type=\"bookmark\" style=\"display: inline-block; width: 0px; overflow: hidden; line-height: 0;\" class=\"mce_SELRES_start\">\ufeff<\/span><\/iframe><\/p>\n<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A subway takes passengers from a station at 9th street East, 3rd street North in a straight line to another station at 12th street East, 10th street North. If each city block (for example, 3rd East to 4th East) is one tenth of a mile, how far is it between the two stations to the nearest hundredth of a mile?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q907635\">Show Solution<\/span><\/p>\n<div id=\"q907635\" class=\"hidden-answer\" style=\"display: none\">\n<p>We first determine the distance between the two stops in units of city blocks. Then we convert this to miles.<\/p>\n<p>Label the locations of the stations as ordered pairs, [latex](9, 3)[\/latex] and [latex](12,10).[\/latex] Then use the distance formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}d&=\\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\  d&=\\sqrt{(12-9)^2+(10-3)^2}\\\\  d&=\\sqrt{(3)^2+(7)^2}\\\\  d&=\\sqrt{9+49}\\\\  d&=\\sqrt{58}\\end{align}[\/latex]<\/p>\n<p>Make sure not to round to the nearest hundredth until the final answer! We convert city blocks to miles by multiplying [latex]\\sqrt{58}\\textsf{ blocks}\\cdot\\frac{0.1\\textsf{ miles}}{1\\textsf{ block}}\\approx 0.76\\textsf{ miles}[\/latex]<\/p>\n<p>It is approximately [latex]0.76[\/latex] miles in between the two stations in a straight line.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>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 <strong>midpoint<\/strong> and the formula is known as the <strong>midpoint formula<\/strong>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>THE MIDPOINT FORMULA<\/h3>\n<p>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 <strong>midpoint<\/strong> is<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle M=\\left(\\dfrac{{x}_{1}+{x}_{2}}{2},\\dfrac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex]<\/p>\n<\/div>\n<p>It makes sense that since the midpoint is halfway from Point\u00a0[latex]1[\/latex] to Point [latex]2,[\/latex] it should be halfway in both the\u00a0[latex]x[\/latex] direction and the\u00a0[latex]y[\/latex] direction. The &#8220;midpoint&#8221; between two real numbers is called the <strong>average <\/strong>or<strong> mean<\/strong>. Thus, you can equivalently view the midpoint formula as<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle M=\\left(\\textsf{avg. of x coords, avg. of y coords}\\right)[\/latex]<\/p>\n<p>A graphical view of a midpoint is shown below. Notice that the line segments on either side of the midpoint are congruent. If you like, you can verify that these two segments have equal length using the Distance Formula.<\/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=\"Line segment on an x, y coordinate plane. 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<\/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\">Solution<\/span><\/p>\n<div id=\"q788934\" class=\"hidden-answer\" style=\"display: none\">\nUse the formula to find the midpoint of the line segment.<\/p>\n<div style=\"text-align: center;\">[latex]\\large \\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><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1758 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8Midpoint-300x288.png\" alt=\"Coordinate plane grid with line segment from (7, negative 2) to (9,5) and midpoint (8,1.5).\" width=\"300\" height=\"288\" srcset=\"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8Midpoint-300x288.png 300w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8Midpoint-65x62.png 65w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8Midpoint-225x216.png 225w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8Midpoint-350x336.png 350w, https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-content\/uploads\/sites\/5871\/2023\/11\/5.8Midpoint.png 685w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It YOURSELF<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288858&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<\/div>\n<h2>Summary<\/h2>\n<p>The Pythagorean Theorem [latex]a^2+b^2=c^2[\/latex] can be used to determine the side lengths of a right triangle with legs [latex]a, b[\/latex] and hypotenuse [latex]c.[\/latex] While solving the equation, we can use the Square Root Property: if [latex]x^2=n[\/latex], then [latex]x=\\sqrt{n}[\/latex] (assuming [latex]x[\/latex] is positive.)<\/p>\n<p>The distance formula\u00a0[latex]d=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}[\/latex] can find the distance between any two points in the plane.\u00a0 The midpoint formula [latex]M=\\left(\\dfrac{{x}_{1}+{x}_{2}}{2},\\dfrac{{y}_{1}+{y}_{2}}{2}\\right)[\/latex] finds the midpoint of the line segment connecting any two points.<\/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-169\">\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>Pythagorean Theorem, Description and Examples. <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>Factoring Application - Find the Lengths of Three Sides of a Right Triangle (Pythagorean Theorem). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/xeP5pRBqsNs\">https:\/\/youtu.be\/xeP5pRBqsNs<\/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, Shared previously<\/div><ul class=\"citation-list\"><li>Celestial Navigation Math. <strong>Authored by<\/strong>: TabletClass Math. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=XWLZKmPU17M\">https:\/\/www.youtube.com\/watch?v=XWLZKmPU17M<\/a>. <strong>License<\/strong>: <em>All Rights Reserved<\/em>. <strong>License Terms<\/strong>: Standard YouTube License<\/li><li>Pythagorean Theorem. <strong>Provided by<\/strong>: Wikipedia. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/en.wikipedia.org\/wiki\/Pythagorean_theorem\">https:\/\/en.wikipedia.org\/wiki\/Pythagorean_theorem<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/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":395986,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Celestial Navigation Math\",\"author\":\"TabletClass Math\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/watch?v=XWLZKmPU17M\",\"project\":\"\",\"license\":\"arr\",\"license_terms\":\"Standard YouTube License\"},{\"type\":\"original\",\"description\":\"Pythagorean Theorem, Description and Examples\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Pythagorean Theorem\",\"author\":\"\",\"organization\":\"Wikipedia\",\"url\":\"https:\/\/en.wikipedia.org\/wiki\/Pythagorean_theorem\",\"project\":\"\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Factoring Application - 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