{"id":1206,"date":"2021-12-02T19:36:58","date_gmt":"2021-12-03T00:36:58","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/3-2-midpoint-and-distance-between-points\/"},"modified":"2023-08-30T12:26:15","modified_gmt":"2023-08-30T16:26:15","slug":"midpoint-and-distance-between-points","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/midpoint-and-distance-between-points\/","title":{"raw":"3.2 Midpoint and Distance Between Points","rendered":"3.2 Midpoint and Distance Between Points"},"content":{"raw":"

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 [latex](a^2 + b^2 = c^2)[\/latex] to find the distance.\r\n\r\nTo do this for the two data points [latex](x_1, y_1)[\/latex] and [latex](x_2, y_2)[\/latex], the distance between these two points [latex](d)[\/latex] will be found using [latex]\\Delta x = x_2 - x_1[\/latex] and [latex]\\Delta y = y_2 - y_1.[\/latex]\r\n\r\nUsing the Pythagorean theorem, this will end up looking like:\r\n

[latex]d^2 = \\Delta x^2 + \\Delta y^2[\/latex]<\/p>\r\n\r\n\r\nor, in expanded form:\r\n

[latex]d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2[\/latex]<\/p>\r\n \r\n\r\n\r\n\r\nOn graph paper, this looks like the following. For this illustration, both [latex]\\Delta x[\/latex] and [latex]\\Delta y[\/latex] are 7 units long, making the distance [latex]d^2 = 7^2 + 7^2[\/latex] or [latex]d^2 = 98[\/latex].\r\n\r\n\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 [latex](-6,-4)[\/latex] and [latex](6, 5)[\/latex].\r\n\r\nStart by identifying which are the two data points [latex](x_1, y_1)[\/latex] and [latex](x_2, y_2)[\/latex]. Let [latex](x_1, y_1)[\/latex] be [latex](-6,-4)[\/latex] and [latex](x_2, y_2)[\/latex] be [latex](6, 5)[\/latex].\r\n\r\nNow:\r\n

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

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

or<\/p>\r\n

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

[latex]d^2 = 144 + 81[\/latex]<\/p>\r\n

or<\/p>\r\n

[latex]d^2 = 225[\/latex]<\/p>\r\nTaking the square root, the result is [latex]d = 15[\/latex].\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 [latex](x_1, y_1)[\/latex] and [latex](x_2, y_2)[\/latex] on a graph involves finding the average values of the [latex]x[\/latex] data points [latex](x_1, x_2)[\/latex] and the of the [latex]y[\/latex] data points [latex](y_1, y_2)[\/latex]. The averages are found by adding both data points together and dividing them by [latex]2[\/latex].\r\n\r\nIn an equation, this looks like:\r\n

[latex]x_{\\text{mid}}=\\dfrac{x_2+x_1}{2}[\/latex] and [latex]y_{\\text{mid}}=\\dfrac{y_2+y_1}{2}[\/latex]<\/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 [latex](-2, 3)[\/latex] and [latex](6, 9)[\/latex].\r\n\r\n\r\n\r\nWe start by adding the two [latex]x[\/latex] data points [latex](x_1 + x_2)[\/latex] and then dividing this result by 2.\r\n

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

or<\/p>\r\n

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

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

or<\/p>\r\n

[latex]\\dfrac{12}{2} = 6[\/latex]<\/p>\r\nThe midpoint between the points [latex](-2, 3)[\/latex] and [latex](6, 9)[\/latex] is at the data point [latex](2, 6)[\/latex].\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 [latex](a^2 + b^2 = c^2)[\/latex] to find the distance.<\/p>\n

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

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

      [latex]d^2 = \\Delta x^2 + \\Delta y^2[\/latex]<\/p>\n

      \"image\"<\/p>\n

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

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

       <\/p>\n

      \"image\"<\/p>\n

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

      \"image\"<\/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 [latex](-6,-4)[\/latex] and [latex](6, 5)[\/latex].<\/p>\n

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

      Now:<\/p>\n

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

      This means that<\/p>\n

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

      or<\/p>\n

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

      which reduces to<\/p>\n

      [latex]d^2 = 144 + 81[\/latex]<\/p>\n

      or<\/p>\n

      [latex]d^2 = 225[\/latex]<\/p>\n

      Taking the square root, the result is [latex]d = 15[\/latex].<\/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 [latex](x_1, y_1)[\/latex] and [latex](x_2, y_2)[\/latex] on a graph involves finding the average values of the [latex]x[\/latex] data points [latex](x_1, x_2)[\/latex] and the of the [latex]y[\/latex] data points [latex](y_1, y_2)[\/latex]. The averages are found by adding both data points together and dividing them by [latex]2[\/latex].<\/p>\n

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

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

      \n
      \n

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

      \n

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

      \"image\"<\/p>\n

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

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

      or<\/p>\n

      [latex]\\dfrac{4}{2} = 2[\/latex]<\/p>\n

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

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

      or<\/p>\n

      [latex]\\dfrac{12}{2} = 6[\/latex]<\/p>\n

      The midpoint between the points [latex](-2, 3)[\/latex] and [latex](6, 9)[\/latex] is at the data point [latex](2, 6)[\/latex].<\/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":90,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1206","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1189,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1206","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1206\/revisions"}],"predecessor-version":[{"id":2082,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1206\/revisions\/2082"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1189"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1206\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1206"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1206"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1206"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}