{"id":2627,"date":"2019-08-01T14:20:41","date_gmt":"2019-08-01T18:20:41","guid":{"rendered":"https:\/\/pressbooks.library.ryerson.ca\/ohsmath\/?post_type=chapter&p=2627"},"modified":"2020-02-07T11:55:31","modified_gmt":"2020-02-07T16:55:31","slug":"2-2-slope-of-a-line","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/chapter\/2-2-slope-of-a-line\/","title":{"raw":"2.2. Slope of a Line","rendered":"2.2. Slope of a Line"},"content":{"raw":"[Latexpage]\r\n

Slope of a Line<\/h1>\r\nIn this section, you will learn to:\r\n
    \r\n \t
  1. Find the slope of a line if two points are given.<\/li>\r\n \t
  2. Graph the line if a point and the slope are given.<\/li>\r\n \t
  3. Find the slope of the line that is written in the form y<\/em> = mx<\/em> + b<\/em>.<\/li>\r\n \t
  4. Find the slope of the line that is written in the form Ax<\/em>+ By<\/em> = c<\/em>.<\/li>\r\n<\/ol>\r\n \r\n\r\nIn the last section, we learned to graph a line by choosing two points on the line. A graph of a line can also be determined if one point and the \"steepness\" of the line is known. The precise number that refers to the steepness or inclination of a line is called the slope<\/strong> of the line.\r\n\r\nFrom previous math courses, many of you remember slope as the \"rise over run,\" or \"the vertical change over the horizontal change\" and have often seen it expressed as:\r\n

    $\\frac{rise}{run},\\frac{vertical\\;change}{horizontal\\;change},\\frac{\\Delta y}{\\Delta x}etc.$<\/p>\r\nWe give a precise definition.\r\n\r\nDefinition: <\/strong>\r\n\r\nIf ( x<\/em>1<\/sub>, y<\/em>1<\/sub>) and ( x<\/em>2<\/sub>, y<\/em>2<\/sub>) are two different points on a line, then the slope of the line is $Slope=m=\\frac{y_2-y_1}{x_2-x_1}$\r\n\r\n \r\n

    \r\n

    Example 2.2.1<\/p>\r\n\r\n<\/header>\r\n

    Find the slope of the line that passes through the points (-2, 3) and (4, -1), and graph the line.<\/div>\r\n
    Solution<\/strong><\/div>\r\n
    Let (x<\/em>1<\/sub>, y<\/em>1<\/sub>) = (-2, 3) and (x<\/em>2<\/sub>, y<\/em>2<\/sub>) = (4, \u22121) then the slope<\/div>\r\n
    $m=\\frac{-1-3}{4-(-2)}=-\\frac{4}{6}=-\\frac{2}{3}$<\/div>\r\n
    \r\n
    \r\n
    \"\"<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n\r\nTo give the reader a better understanding, both the vertical change, -4, and the horizontal change, 6, are shown in the above figure.\r\n\r\nWhen two points are given, it does not matter which point is denoted as (x<\/em>1<\/sub>, y<\/em>1<\/sub>) and which (x<\/em>2<\/sub>, y<\/em>2<\/sub>). The value for the slope will be the same. For example, if we choose (x<\/em>1<\/sub>, y<\/em>2<\/sub>) = (4, -1) and (x<\/em>2<\/sub>, y<\/em>2<\/sub>) = (-2, 3), we will get the same value for the slope as we obtained earlier. The steps involved are as follows.\r\n

    $m=\\frac{-3-(-1)}{-2-4}=\\frac{4}{-6}=-\\frac{2}{3}$<\/p>\r\nThe student should further observe that if a line rises when going from left to right, then it has a positive slope; and if it falls going from left to right, it has a negative slope.\r\n\r\n \r\n

    \r\n

    Example 2.2.2<\/p>\r\n\r\n<\/header>\r\n

    Find the slope of the line that passes through the points (2, 3) and (2, -1), and graph.<\/div>\r\n
    Solution<\/strong><\/div>\r\n
    Let (x<\/em>1, y<\/em>1) = (2, 3) and (x<\/em>2, y<\/em>2) = (2, -1) then the slope<\/div>\r\n
    $m=\\frac{-1-3}{2-2}=-\\frac{4}{0}=$ undefined<\/div>\r\n
    \r\n
    \r\n
    \"\"<\/div>\r\n<\/div>\r\n<\/div>\r\n

    Note: <\/strong>The slope of a vertical line is undefined.<\/p>\r\n\r\n<\/div>\r\n \r\n

    \r\n

    Example 2.2.3<\/p>\r\n\r\n<\/header>\r\n

    \r\n
    Graph the line that passes through the point (1, 2) and has slope $-\\frac{3}{4}$.<\/div>\r\n
    Solution <\/strong><\/div>\r\n
    Slope equals $\\frac{rise}{run}$. The fact that the slope is $\\frac{-3}{4}$ , means that for every rise of -3 units (fall of 3 units) there is a run of 4. So if from the given point (1, 2) we go down 3 units and go right 4 units, we reach the point (5, -1). The following graph is obtained by connecting these two points.<\/div>\r\n
    \r\n
    \r\n
    \"\"<\/div>\r\n<\/div>\r\nAlternatively, since $\\frac{3}{-4}$ represents the same number, the line can be drawn by starting at the point (1, 2) and choosing a rise of 3 units followed by a run of -4 units. So from the point (1, 2), we go up 3 units, and to the left 4, thus reaching the point (-3, 5) which is also on the same line. See the figure below.\r\n
    \r\n

    \"\"<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n

    \r\n

    Example 2.2.4<\/p>\r\n\r\n<\/header>\r\n

    \r\n
    Find the slope of the line 2x<\/em> + 3y<\/em> = 6.<\/div>\r\n
    Solution<\/strong><\/div>\r\n
    In order to find the slope of this line, we will choose any two points on this line.<\/div>\r\n
    Again, the selection of x<\/em> and y<\/em> intercepts seems to be a good choice. The x-intercept is (3, 0), and the y-intercept is (0, 2). Therefore, the slope is<\/div>\r\n
    $m=\\frac{2-0}{0-3}=-\\frac{2}{3}$<\/div>\r\n
    The graph below shows the line and the intercepts: x<\/em> and y.<\/em><\/div>\r\n
    \r\n
    \r\n
    \"\"<\/em><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n
    \r\n

    Example 2.2.5<\/p>\r\n\r\n<\/header>\r\n

    \r\n
    Find the slope of the line y<\/em> = 3x<\/em> + 2.<\/div>\r\n
    Solution <\/strong><\/div>\r\n
    We again find two points on the line. Say (0, 2) and (1, 5).<\/div>\r\n
    Therefore, the slope is $m=\\frac{5-2}{1-0}=\\frac{3}{1}=3$.<\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n\r\nLook at the slopes and the y-intercepts of the following lines.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    The line<\/strong><\/td>\r\nSlope<\/strong><\/td>\r\ny-intercept<\/strong><\/td>\r\n<\/tr>\r\n
    y<\/em> = 3x<\/em> + 2<\/td>\r\n3<\/td>\r\n2<\/td>\r\n<\/tr>\r\n
    y<\/em> = -2x<\/em> + 5<\/td>\r\n-2<\/td>\r\n5<\/td>\r\n<\/tr>\r\n
    y<\/em> = 3\/2x<\/em> \u2013 4<\/td>\r\n3\/2<\/td>\r\n-4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIt is no coincidence that when an equation of the line is solved for y<\/em>, the coefficient of the x<\/em> term represents the slope, and the constant term represents the y-intercept.\r\n\r\nIn other words, for the line y<\/em> = mx<\/em> + b<\/em>, m<\/em> is the slope, and b<\/em> is the y-intercept.\r\n\r\n \r\n
    \r\n

    Example 2.2.6<\/p>\r\n\r\n<\/header>\r\n

    \r\n
    \r\n\r\nDetermine the slope and y-intercept of the line 2x<\/em> + 3y<\/em> = 6.\r\n\r\nSolution<\/strong>\r\n\r\n<\/div>\r\n
    We solve for y<\/em>.<\/div>\r\n
    $2x + 3y = 6$<\/div>\r\n
    $3y = -2x + 6$<\/div>\r\n
    $y = -2\/3x + 2$<\/div>\r\n
    The slope = the coefficient of the x term = - 2\/3<\/div>\r\n
    The y-intercept = the constant term = 2.<\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n

    Practice questions<\/h1>\r\n1.<\/strong> Find the slope of the line passing through the following pair of points:\r\n

    a.<\/strong> (2, 3) and (5, 9)<\/p>\r\n

    b.<\/strong> (6, -5) and (4, -1)<\/p>\r\n

    c.<\/strong> (-3, -5) and (-1, -7)<\/p>\r\n2.<\/strong> Determine the slope of the line from the given equation of the line:\r\n

    a.<\/strong> $y = -2x + 1$<\/p>\r\n

    b.<\/strong> $3x - 4y = 12$<\/p>\r\n

    c.<\/strong> $2x - y = 6$<\/p>\r\n3.<\/strong> Graph the line that passes through the given point and has the given slope.\r\n

    a. <\/strong>(1, 2) and m<\/em> = -3\/4<\/p>\r\n

    b. <\/strong>(0, 2) and m<\/em> = -2<\/p>\r\n ","rendered":"

    Slope of a Line<\/h1>\n

    In this section, you will learn to:<\/p>\n

      \n
    1. Find the slope of a line if two points are given.<\/li>\n
    2. Graph the line if a point and the slope are given.<\/li>\n
    3. Find the slope of the line that is written in the form y<\/em> = mx<\/em> + b<\/em>.<\/li>\n
    4. Find the slope of the line that is written in the form Ax<\/em>+ By<\/em> = c<\/em>.<\/li>\n<\/ol>\n

       <\/p>\n

      In the last section, we learned to graph a line by choosing two points on the line. A graph of a line can also be determined if one point and the “steepness” of the line is known. The precise number that refers to the steepness or inclination of a line is called the slope<\/strong> of the line.<\/p>\n

      From previous math courses, many of you remember slope as the “rise over run,” or “the vertical change over the horizontal change” and have often seen it expressed as:<\/p>\n

      \"\frac{rise}{run},\frac{vertical\;change}{horizontal\;change},\frac{\Delta y}{\Delta x}etc.\"<\/p>\n

      We give a precise definition.<\/p>\n

      Definition: <\/strong><\/p>\n

      If ( x<\/em>1<\/sub>, y<\/em>1<\/sub>) and ( x<\/em>2<\/sub>, y<\/em>2<\/sub>) are two different points on a line, then the slope of the line is \"Slope=m=\frac{y_2-y_1}{x_2-x_1}\"<\/p>\n

       <\/p>\n

      \n
      \n

      Example 2.2.1<\/p>\n<\/header>\n

      Find the slope of the line that passes through the points (-2, 3) and (4, -1), and graph the line.<\/div>\n
      Solution<\/strong><\/div>\n
      Let (x<\/em>1<\/sub>, y<\/em>1<\/sub>) = (-2, 3) and (x<\/em>2<\/sub>, y<\/em>2<\/sub>) = (4, \u22121) then the slope<\/div>\n
      \"m=\frac{-1-3}{4-(-2)}=-\frac{4}{6}=-\frac{2}{3}\"<\/div>\n
      \n
      \n
      \"\"<\/div>\n<\/div>\n<\/div>\n<\/div>\n

       <\/p>\n

      To give the reader a better understanding, both the vertical change, -4, and the horizontal change, 6, are shown in the above figure.<\/p>\n

      When two points are given, it does not matter which point is denoted as (x<\/em>1<\/sub>, y<\/em>1<\/sub>) and which (x<\/em>2<\/sub>, y<\/em>2<\/sub>). The value for the slope will be the same. For example, if we choose (x<\/em>1<\/sub>, y<\/em>2<\/sub>) = (4, -1) and (x<\/em>2<\/sub>, y<\/em>2<\/sub>) = (-2, 3), we will get the same value for the slope as we obtained earlier. The steps involved are as follows.<\/p>\n

      \"m=\frac{-3-(-1)}{-2-4}=\frac{4}{-6}=-\frac{2}{3}\"<\/p>\n

      The student should further observe that if a line rises when going from left to right, then it has a positive slope; and if it falls going from left to right, it has a negative slope.<\/p>\n

       <\/p>\n

      \n
      \n

      Example 2.2.2<\/p>\n<\/header>\n

      Find the slope of the line that passes through the points (2, 3) and (2, -1), and graph.<\/div>\n
      Solution<\/strong><\/div>\n
      Let (x<\/em>1, y<\/em>1) = (2, 3) and (x<\/em>2, y<\/em>2) = (2, -1) then the slope<\/div>\n
      \"m=\frac{-1-3}{2-2}=-\frac{4}{0}=\" undefined<\/div>\n
      \n
      \n
      \"\"<\/div>\n<\/div>\n<\/div>\n

      Note: <\/strong>The slope of a vertical line is undefined.<\/p>\n<\/div>\n

       <\/p>\n

      \n
      \n

      Example 2.2.3<\/p>\n<\/header>\n

      \n
      Graph the line that passes through the point (1, 2) and has slope \"-\frac{3}{4}\".<\/div>\n
      Solution <\/strong><\/div>\n
      Slope equals \"\frac{rise}{run}\". The fact that the slope is \"\frac{-3}{4}\" , means that for every rise of -3 units (fall of 3 units) there is a run of 4. So if from the given point (1, 2) we go down 3 units and go right 4 units, we reach the point (5, -1). The following graph is obtained by connecting these two points.<\/div>\n
      \n
      \n
      \"\"<\/div>\n<\/div>\n

      Alternatively, since \"\frac{3}{-4}\" represents the same number, the line can be drawn by starting at the point (1, 2) and choosing a rise of 3 units followed by a run of -4 units. So from the point (1, 2), we go up 3 units, and to the left 4, thus reaching the point (-3, 5) which is also on the same line. See the figure below.<\/p>\n

      \n

      \"\"<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

       <\/p>\n

      \n
      \n

      Example 2.2.4<\/p>\n<\/header>\n

      \n
      Find the slope of the line 2x<\/em> + 3y<\/em> = 6.<\/div>\n
      Solution<\/strong><\/div>\n
      In order to find the slope of this line, we will choose any two points on this line.<\/div>\n
      Again, the selection of x<\/em> and y<\/em> intercepts seems to be a good choice. The x-intercept is (3, 0), and the y-intercept is (0, 2). Therefore, the slope is<\/div>\n
      \"m=\frac{2-0}{0-3}=-\frac{2}{3}\"<\/div>\n
      The graph below shows the line and the intercepts: x<\/em> and y.<\/em><\/div>\n
      \n
      \n
      \"\"<\/em><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

       <\/p>\n

      \n
      \n

      Example 2.2.5<\/p>\n<\/header>\n

      \n
      Find the slope of the line y<\/em> = 3x<\/em> + 2.<\/div>\n
      Solution <\/strong><\/div>\n
      We again find two points on the line. Say (0, 2) and (1, 5).<\/div>\n
      Therefore, the slope is \"m=\frac{5-2}{1-0}=\frac{3}{1}=3\".<\/div>\n<\/div>\n<\/div>\n

       <\/p>\n

      Look at the slopes and the y-intercepts of the following lines.<\/p>\n\n\n\n\n\n\n
      The line<\/strong><\/td>\nSlope<\/strong><\/td>\ny-intercept<\/strong><\/td>\n<\/tr>\n
      y<\/em> = 3x<\/em> + 2<\/td>\n3<\/td>\n2<\/td>\n<\/tr>\n
      y<\/em> = -2x<\/em> + 5<\/td>\n-2<\/td>\n5<\/td>\n<\/tr>\n
      y<\/em> = 3\/2x<\/em> \u2013 4<\/td>\n3\/2<\/td>\n-4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      It is no coincidence that when an equation of the line is solved for y<\/em>, the coefficient of the x<\/em> term represents the slope, and the constant term represents the y-intercept.<\/p>\n

      In other words, for the line y<\/em> = mx<\/em> + b<\/em>, m<\/em> is the slope, and b<\/em> is the y-intercept.<\/p>\n

       <\/p>\n

      \n
      \n

      Example 2.2.6<\/p>\n<\/header>\n

      \n
      \n

      Determine the slope and y-intercept of the line 2x<\/em> + 3y<\/em> = 6.<\/p>\n

      Solution<\/strong><\/p>\n<\/div>\n

      We solve for y<\/em>.<\/div>\n
      \"2x + 3y = 6\"<\/div>\n
      \"3y = -2x + 6\"<\/div>\n
      \"y = -2/3x + 2\"<\/div>\n
      The slope = the coefficient of the x term = – 2\/3<\/div>\n
      The y-intercept = the constant term = 2.<\/div>\n<\/div>\n<\/div>\n

       <\/p>\n

      Practice questions<\/h1>\n

      1.<\/strong> Find the slope of the line passing through the following pair of points:<\/p>\n

      a.<\/strong> (2, 3) and (5, 9)<\/p>\n

      b.<\/strong> (6, -5) and (4, -1)<\/p>\n

      c.<\/strong> (-3, -5) and (-1, -7)<\/p>\n

      2.<\/strong> Determine the slope of the line from the given equation of the line:<\/p>\n

      a.<\/strong> \"y = -2x + 1\"<\/p>\n

      b.<\/strong> \"3x - 4y = 12\"<\/p>\n

      c.<\/strong> \"2x - y = 6\"<\/p>\n

      3.<\/strong> Graph the line that passes through the given point and has the given slope.<\/p>\n

      a. <\/strong>(1, 2) and m<\/em> = -3\/4<\/p>\n

      b. <\/strong>(0, 2) and m<\/em> = -2<\/p>\n

       <\/p>\n","protected":false},"author":130,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[48],"contributor":[],"license":[],"class_list":["post-2627","chapter","type-chapter","status-publish","hentry","chapter-type-numberless"],"part":956,"_links":{"self":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/pressbooks\/v2\/chapters\/2627","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/wp\/v2\/users\/130"}],"version-history":[{"count":9,"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/pressbooks\/v2\/chapters\/2627\/revisions"}],"predecessor-version":[{"id":3121,"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/pressbooks\/v2\/chapters\/2627\/revisions\/3121"}],"part":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/pressbooks\/v2\/parts\/956"}],"metadata":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/pressbooks\/v2\/chapters\/2627\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/wp\/v2\/media?parent=2627"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/pressbooks\/v2\/chapter-type?post=2627"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/wp\/v2\/contributor?post=2627"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.torontomu.ca\/ohsmath\/wp-json\/wp\/v2\/license?post=2627"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}