{"id":455,"date":"2019-04-29T13:20:53","date_gmt":"2019-04-29T17:20:53","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=455"},"modified":"2020-01-04T15:04:27","modified_gmt":"2020-01-04T20:04:27","slug":"3-2-midpoint-and-distance-between-points","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/3-2-midpoint-and-distance-between-points\/","title":{"raw":"3.2 Midpoint and Distance Between Points","rendered":"3.2 Midpoint and Distance Between Points"},"content":{"raw":"[latexpage]\r\n

Finding the Distance Between Two Points<\/h1>\r\nThe logic used to find the distance between two data points on a graph involves the construction of a right triangle using the two data points and the Pythagorean theorem \\((a^2 + b^2 = c^2)\\) to find the distance.\r\n\r\nTo do this for the two data points \\((x_1, y_1)\\) and \\((x_2, y_2)\\), the distance between these two points \\((d)\\) will be found using \\(\\Delta x = x_2 - x_1\\) and \\(\\Delta y = y_2 - y_1.\\)\r\n\r\nUsing the Pythagorean theorem, this will end up looking like:\r\n

\\(d^2 = \\Delta x^2 + \\Delta y^2\\)<\/p>\r\n\"The\r\n\r\nor, in expanded form:\r\n

\\(d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\\)<\/p>\r\n \r\n\r\n\"Delta\r\n\r\nOn graph paper, this looks like the following. For this illustration, both \\(\\Delta x\\) and \\(\\Delta y\\) are 7 units long, making the distance \\(d^2 = 7^2 + 7^2\\) or \\(d^2 = 98\\).\r\n\r\n\"Triangle\r\n\r\nThe square root of 98 is approximately 9.899 units long.\r\n

\r\n

Example 3.2.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the distance between the points \\((-6,-4)\\) and \\((6, 5)\\).\r\n\r\nStart by identifying which are the two data points \\((x_1, y_1)\\) and \\((x_2, y_2)\\). Let \\((x_1, y_1)\\) be \\((-6,-4)\\) and \\((x_2, y_2)\\) be \\((6, 5)\\).\r\n\r\nNow:\r\n

\\(\\Delta x^2 = (x_2 - x_1)^2\\) or \\([6 - (-6)]^2\\) and \\(\\Delta y^2 = (y_2 - y_1)^2\\) or \\([5 - (-4)]^2\\).<\/p>\r\nThis means that\r\n

\\(d^2 = [6 - (-6)]^2 + [5 - (-4)]^2\\)<\/p>\r\n

or<\/p>\r\n

\\(d^2 = [12]^2 + [9]^2\\)<\/p>\r\nwhich reduces to\r\n

\\(d^2 = 144 + 81\\)<\/p>\r\n

or<\/p>\r\n

\\(d^2 = 225\\)<\/p>\r\nTaking the square root, the result is \\(d = 15\\).\r\n\r\n<\/div>\r\n<\/div>\r\n

Finding the Midway Between Two Points (Midpoint)<\/h1>\r\nThe logic used to find the midpoint between two data points \\((x_1, y_1)\\) and \\((x_2, y_2)\\) on a graph involves finding the average values of the \\(x\\) data points \\((x_1, x_2)\\) and the of the \\(y\\) data points \\((y_1, y_2)\\). The averages are found by adding both data points together and dividing them by \\(2\\).\r\n\r\nIn an equation, this looks like:\r\n

\\(x_{\\text{mid}}=\\dfrac{x_2+x_1}{2}\\) and \\(y_{\\text{mid}}=\\dfrac{y_2+y_1}{2}\\)<\/p>\r\n\r\n

\r\n

Example 3.2.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the midpoint between the\u00a0points \\((-2, 3)\\) and \\((6, 9)\\).\r\n\r\n\"Triangle\r\n\r\nWe start by adding the two \\(x\\) data points \\((x_1 + x_2)\\) and then dividing this result by 2.\r\n

\\(x_{\\text{mid}} = \\dfrac{(-2 + 6)}{2}\\)<\/p>\r\n

or<\/p>\r\n

\\(\\dfrac{4}{2} = 2\\)<\/p>\r\nThe midpoint's \\(y\\)-coordinate is found by adding the two \\(y\\) data points \\((y_1 + y_2)\\) and then dividing this result by 2.\r\n

\\(y_{\\text{mid}} = \\dfrac{(9 + 3)}{2}\\)<\/p>\r\n

or<\/p>\r\n

\\(\\dfrac{12}{2} = 6\\)<\/p>\r\nThe midpoint between the points \\((-2, 3)\\) and \\((6, 9)\\) is at the data point \\((2, 6)\\).\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nFor questions 1 to 8, find the distance between the points.\r\n
    \r\n \t
  1. \u00a0(\u22126, \u22121) and (6, 4)<\/li>\r\n \t
  2. (1, \u22124) and (5, \u22121)<\/li>\r\n \t
  3. (\u22125, \u22121) and (3, 5)<\/li>\r\n \t
  4. (6, \u22124) and (12, 4)<\/li>\r\n \t
  5. (\u22128, \u22122) and (4, 3)<\/li>\r\n \t
  6. (3, \u22122) and (7, 1)<\/li>\r\n \t
  7. (\u221210, \u22126) and (\u22122, 0)<\/li>\r\n \t
  8. (8, \u22122) and (14, 6)<\/li>\r\n<\/ol>\r\nFor questions 9 to 16, find the midpoint between the points.\r\n
      \r\n \t
    1. (\u22126, \u22121) and (6, 5)<\/li>\r\n \t
    2. (1, \u22124) and (5, \u22122)<\/li>\r\n \t
    3. (\u22125, \u22121) and (3, 5)<\/li>\r\n \t
    4. (6, \u22124) and (12, 4)<\/li>\r\n \t
    5. (\u22128, \u22121) and (6, 7)<\/li>\r\n \t
    6. (1, \u22126) and (3, \u22122)<\/li>\r\n \t
    7. (\u22127, \u22121) and (3, 9)<\/li>\r\n \t
    8. (2, \u22122) and (12, 4)<\/li>\r\n<\/ol>\r\nAnswer Key 3.2<\/a>","rendered":"

      Finding the Distance Between Two Points<\/h1>\n

      The logic used to find the distance between two data points on a graph involves the construction of a right triangle using the two data points and the Pythagorean theorem \"(a^2 + b^2 = c^2)\" to find the distance.<\/p>\n

      To do this for the two data points \"(x_1, y_1)\" and \"(x_2, y_2)\", the distance between these two points \"(d)\" will be found using \"\Delta x = x_2 - x_1\" and \"\Delta y = y_2 - y_1.\"<\/p>\n

      Using the Pythagorean theorem, this will end up looking like:<\/p>\n

      \"d^2 = \Delta x^2 + \Delta y^2\"<\/p>\n

      \"The<\/p>\n

      or, in expanded form:<\/p>\n

      \"d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\"<\/p>\n

       <\/p>\n

      \"Delta<\/p>\n

      On graph paper, this looks like the following. For this illustration, both \"\Delta x\" and \"\Delta y\" are 7 units long, making the distance \"d^2 = 7^2 + 7^2\" or \"d^2 = 98\".<\/p>\n

      \"Triangle<\/p>\n

      The square root of 98 is approximately 9.899 units long.<\/p>\n

      \n
      \n

      Example 3.2.1<\/p>\n<\/header>\n

      \n

      Find the distance between the points \"(-6,-4)\" and \"(6, 5)\".<\/p>\n

      Start by identifying which are the two data points \"(x_1, y_1)\" and \"(x_2, y_2)\". Let \"(x_1, y_1)\" be \"(-6,-4)\" and \"(x_2, y_2)\" be \"(6, 5)\".<\/p>\n

      Now:<\/p>\n

      \"\Delta x^2 = (x_2 - x_1)^2\" or \"[6 - (-6)]^2\" and \"\Delta y^2 = (y_2 - y_1)^2\" or \"[5 - (-4)]^2\".<\/p>\n

      This means that<\/p>\n

      \"d^2 = [6 - (-6)]^2 + [5 - (-4)]^2\"<\/p>\n

      or<\/p>\n

      \"d^2 = [12]^2 + [9]^2\"<\/p>\n

      which reduces to<\/p>\n

      \"d^2 = 144 + 81\"<\/p>\n

      or<\/p>\n

      \"d^2 = 225\"<\/p>\n

      Taking the square root, the result is \"d = 15\".<\/p>\n<\/div>\n<\/div>\n

      Finding the Midway Between Two Points (Midpoint)<\/h1>\n

      The logic used to find the midpoint between two data points \"(x_1, y_1)\" and \"(x_2, y_2)\" on a graph involves finding the average values of the \"x\" data points \"(x_1, x_2)\" and the of the \"y\" data points \"(y_1, y_2)\". The averages are found by adding both data points together and dividing them by \"2\".<\/p>\n

      In an equation, this looks like:<\/p>\n

      \"x_{\text{mid}}=\dfrac{x_2+x_1}{2}\" and \"y_{\text{mid}}=\dfrac{y_2+y_1}{2}\"<\/p>\n

      \n
      \n

      Example 3.2.2<\/p>\n<\/header>\n

      \n

      Find the midpoint between the\u00a0points \"(-2, 3)\" and \"(6, 9)\".<\/p>\n

      \"Triangle<\/p>\n

      We start by adding the two \"x\" data points \"(x_1 + x_2)\" and then dividing this result by 2.<\/p>\n

      \"x_{\text{mid}} = \dfrac{(-2 + 6)}{2}\"<\/p>\n

      or<\/p>\n

      \"\dfrac{4}{2} = 2\"<\/p>\n

      The midpoint’s \"y\"-coordinate is found by adding the two \"y\" data points \"(y_1 + y_2)\" and then dividing this result by 2.<\/p>\n

      \"y_{\text{mid}} = \dfrac{(9 + 3)}{2}\"<\/p>\n

      or<\/p>\n

      \"\dfrac{12}{2} = 6\"<\/p>\n

      The midpoint between the points \"(-2, 3)\" and \"(6, 9)\" is at the data point \"(2, 6)\".<\/p>\n<\/div>\n<\/div>\n

      Questions<\/h1>\n

      For questions 1 to 8, find the distance between the points.<\/p>\n

        \n
      1. \u00a0(\u22126, \u22121) and (6, 4)<\/li>\n
      2. (1, \u22124) and (5, \u22121)<\/li>\n
      3. (\u22125, \u22121) and (3, 5)<\/li>\n
      4. (6, \u22124) and (12, 4)<\/li>\n
      5. (\u22128, \u22122) and (4, 3)<\/li>\n
      6. (3, \u22122) and (7, 1)<\/li>\n
      7. (\u221210, \u22126) and (\u22122, 0)<\/li>\n
      8. (8, \u22122) and (14, 6)<\/li>\n<\/ol>\n

        For questions 9 to 16, find the midpoint between the points.<\/p>\n

          \n
        1. (\u22126, \u22121) and (6, 5)<\/li>\n
        2. (1, \u22124) and (5, \u22122)<\/li>\n
        3. (\u22125, \u22121) and (3, 5)<\/li>\n
        4. (6, \u22124) and (12, 4)<\/li>\n
        5. (\u22128, \u22121) and (6, 7)<\/li>\n
        6. (1, \u22126) and (3, \u22122)<\/li>\n
        7. (\u22127, \u22121) and (3, 9)<\/li>\n
        8. (2, \u22122) and (12, 4)<\/li>\n<\/ol>\n

          Answer Key 3.2<\/a><\/p>\n","protected":false},"author":540,"menu_order":11,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-455","chapter","type-chapter","status-publish","hentry"],"part":360,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/455","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/540"}],"version-history":[{"count":16,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/455\/revisions"}],"predecessor-version":[{"id":3766,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/455\/revisions\/3766"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/360"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/455\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=455"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=455"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=455"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=455"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}