{"id":1111,"date":"2015-11-12T18:35:31","date_gmt":"2015-11-12T18:35:31","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1111"},"modified":"2015-11-12T18:35:31","modified_gmt":"2015-11-12T18:35:31","slug":"write-the-equation-of-a-line-parallel-or-perpendicular-to-a-given-line","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/chapter\/write-the-equation-of-a-line-parallel-or-perpendicular-to-a-given-line\/","title":{"raw":"Write the equation of a line parallel or perpendicular to a given line","rendered":"Write the equation of a line parallel or perpendicular to a given line"},"content":{"raw":"

If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.<\/p>\n\n

Writing Equations of Parallel Lines<\/h2>\n

Suppose for example, we are given the following equation.<\/p>\n\n

[latex]f\\left(x\\right)=3x+1\\\\[\/latex]<\/div>\n

We know that the slope of the line formed by the function is 3. We also know that the y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to f<\/em>(x<\/em>). So the lines formed by all of the following functions will be parallel to f<\/em>(x<\/em>).<\/p>\n\n

[latex]\\begin{cases}g\\left(x\\right)=3x+6\\hfill \\\\ h\\left(x\\right)=3x+1\\hfill \\\\ p\\left(x\\right)=3x+\\frac{2}{3}\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

Suppose then we want to write the equation of a line that is parallel to f<\/em>\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for b<\/em>\u00a0will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.<\/p>\n\n

[latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 7=3\\left(x - 1\\right)\\hfill \\\\ y - 7=3x - 3\\hfill \\\\ \\text{ }y=3x+4\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

So [latex]g\\left(x\\right)=3x+4\\\\[\/latex] is parallel to [latex]f\\left(x\\right)=3x+1\\\\[\/latex] and passes through the point (1, 7).<\/p>\n\n

\n

How To: Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.<\/h3>\n
  1. Find the slope of the function.<\/li>\n\t
  2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.<\/li>\n\t
  3. Simplify.<\/li>\n<\/ol><\/div>\n
    \n
    \n
    \n

    Example 9: Finding a Line Parallel to a Given Line<\/h3>\n

    Find a line parallel to the graph of [latex]f\\left(x\\right)=3x+6\\\\[\/latex] that passes through the point (3, 0).<\/p>\n\n<\/div>\n

    \n

    Solutions<\/h3>\n

    The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m\u00a0<\/em>= 3, x\u00a0<\/em>= 3, and f<\/em>(x<\/em>) = 0 into the slope-intercept form to find the y-<\/em>intercept.<\/p>\n\n

    [latex]\\begin{cases}g\\left(x\\right)=3x+b\\hfill \\\\ \\text{ }0=3\\left(3\\right)+b\\hfill \\\\ \\text{ }b=-9\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

    The line parallel to\u00a0f<\/em>(x<\/em>) that passes through (3, 0) is [latex]g\\left(x\\right)=3x - 9\\\\[\/latex].<\/p>\n\n<\/div>\n

    \n

    Analysis of the Solution<\/h3>\n

    We can confirm that the two lines are parallel by graphing them. Figure 22\u00a0shows that the two lines will never intersect.\n<\/span><\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]\"GraphFigure 22<\/b>[\/caption]\n\n<\/div>\n<\/div>\n<\/div>\n<\/section>

    Writing Equations of Perpendicular Lines<\/h2>\n

    We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:<\/p>\n\n

    [latex]f\\left(x\\right)=2x+4\\\\[\/latex]<\/div>\n

    The slope of the line is 2, and its negative reciprocal is [latex]-\\frac{1}{2}\\\\[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}\\\\[\/latex] will be perpendicular to\u00a0f<\/em>(x<\/em>). So the lines formed by all of the following functions will be perpendicular to\u00a0f<\/em>(x<\/em>).<\/p>\n\n

    [latex]\\begin{cases}g\\left(x\\right)=-\\frac{1}{2}x+4\\hfill \\\\ h\\left(x\\right)=-\\frac{1}{2}x+2\\hfill \\\\ p\\left(x\\right)=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

    As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to\u00a0f<\/em>(x<\/em>) and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}\\\\[\/latex]. Now we can use the point to find the y<\/em>-intercept by substituting the given values into the slope-intercept form of a line and solving for b<\/em>.<\/p>\n\n

    [latex]\\begin{cases}g\\left(x\\right)=mx+b\\hfill \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill \\\\ b=2\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

    The equation for the function with a slope of [latex]-\\frac{1}{2}\\\\[\/latex] and a y-<\/em>intercept of 2 is<\/p>\n\n

    [latex]g\\left(x\\right)=-\\frac{1}{2}x+2\\\\[\/latex].<\/div>\n

    So [latex]g\\left(x\\right)=-\\frac{1}{2}x+2\\\\[\/latex] is perpendicular to [latex]f\\left(x\\right)=2x+4\\\\[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/p>\n\n

    \n

    Q & A<\/h3>\n

    A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong><\/p>\n

    No. For two perpendicular linear functions, the product of their slopes is \u20131. However, a vertical line is not a function so the definition is not contradicted.<\/em><\/p>\n\n<\/div>\n

    \n

    How To: Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.<\/h3>\n
    1. Find the slope of the function.<\/li>\n\t
    2. Determine the negative reciprocal of the slope.<\/li>\n\t
    3. Substitute the new slope and the values for x<\/em>\u00a0and y<\/em>\u00a0from the coordinate pair provided into [latex]g\\left(x\\right)=mx+b\\\\[\/latex].<\/li>\n\t
    4. Solve for b<\/em>.<\/li>\n\t
    5. Write the equation for the line.<\/li>\n<\/ol><\/div>\n
      \n
      \n
      \n

      Example 10: Finding the Equation of a Perpendicular Line<\/h3>\n

      Find the equation of a line perpendicular to [latex]f\\left(x\\right)=3x+3\\\\[\/latex] that passes through the point (3, 0).<\/p>\n\n<\/div>\n

      \n

      Solution<\/h3>\n

      The original line has slope m\u00a0<\/em>= 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\\frac{1}{3}\\\\[\/latex]. Using this slope and the given point, we can find the equation for the line.<\/p>\n\n

      [latex]\\begin{cases}g\\left(x\\right)=-\\frac{1}{3}x+b\\hfill \\\\ \\text{ }0=-\\frac{1}{3}\\left(3\\right)+b\\hfill \\\\ \\text{ }1=b\\hfill \\\\ \\text{ }b=1\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

      The line perpendicular to\u00a0f<\/em>(x<\/em>)\u00a0that passes through (3, 0) is [latex]g\\left(x\\right)=-\\frac{1}{3}x+1\\\\[\/latex].<\/p>\n\n<\/div>\n

      \n

      Analysis of the Solution<\/h3>\n

      A graph of the two lines is shown in Figure 23.\n<\/span><\/p>\n\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"GraphFigure 23<\/b>[\/caption]\n\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Try It 5<\/h3>\n

      Given the function [latex]h\\left(x\\right)=2x - 4\\\\[\/latex], write an equation for the line passing through (0, 0) that is<\/p>\n

      a. parallel to h<\/em>(x<\/em>)\nb. perpendicular to h<\/em>(x<\/em>)<\/p>\nSolution<\/a>\n\n<\/div>\n

      \n

      How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/h3>\n
      1. Determine the slope of the line passing through the points.<\/li>\n\t
      2. Find the negative reciprocal of the slope.<\/li>\n\t
      3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\n\t
      4. Simplify.<\/li>\n<\/ol><\/div>\n
        \n
        \n
        \n

        Example 11: Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point<\/h3>\n

        A line passes through the points (\u20132, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).<\/p>\n\n<\/div>\n

        \n

        Solution<\/h3>\nFrom the two points of the given line, we can calculate the slope of that line.\n
        [latex]\\begin{cases}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill \\\\ =\\frac{-1}{6}\\hfill \\\\ =-\\frac{1}{6}\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

        Find the negative reciprocal of the slope.<\/p>\n\n

        [latex]\\begin{cases}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill \\\\ =-1\\left(-\\frac{6}{1}\\right)\\hfill \\\\ =6\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

        We can then solve for the y-<\/em>intercept of the line passing through the point (4, 5).<\/p>\n\n

        [latex]\\begin{cases}g\\left(x\\right)=6x+b\\hfill \\\\ 5=6\\left(4\\right)+b\\hfill \\\\ 5=24+b\\hfill \\\\ -19=b\\hfill \\\\ b=-19\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

        The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is<\/p>\n\n

        [latex]y=6x - 19\\\\[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
        \n

        Try It 6<\/h3>\n

        A line passes through the points, (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point, (6, 4).<\/p>\nSolution<\/a>\n\n<\/div>\n<\/section>","rendered":"

        If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.<\/p>\n

        \n

        Writing Equations of Parallel Lines<\/h2>\n

        Suppose for example, we are given the following equation.<\/p>\n

        [latex]f\\left(x\\right)=3x+1\\\\[\/latex]<\/div>\n

        We know that the slope of the line formed by the function is 3. We also know that the y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to f<\/em>(x<\/em>). So the lines formed by all of the following functions will be parallel to f<\/em>(x<\/em>).<\/p>\n

        [latex]\\begin{cases}g\\left(x\\right)=3x+6\\hfill \\\\ h\\left(x\\right)=3x+1\\hfill \\\\ p\\left(x\\right)=3x+\\frac{2}{3}\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

        Suppose then we want to write the equation of a line that is parallel to f<\/em>\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for b<\/em>\u00a0will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.<\/p>\n

        [latex]\\begin{cases}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 7=3\\left(x - 1\\right)\\hfill \\\\ y - 7=3x - 3\\hfill \\\\ \\text{ }y=3x+4\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

        So [latex]g\\left(x\\right)=3x+4\\\\[\/latex] is parallel to [latex]f\\left(x\\right)=3x+1\\\\[\/latex] and passes through the point (1, 7).<\/p>\n

        \n

        How To: Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.<\/h3>\n
          \n
        1. Find the slope of the function.<\/li>\n
        2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.<\/li>\n
        3. Simplify.<\/li>\n<\/ol>\n<\/div>\n
          \n
          \n
          \n

          Example 9: Finding a Line Parallel to a Given Line<\/h3>\n

          Find a line parallel to the graph of [latex]f\\left(x\\right)=3x+6\\\\[\/latex] that passes through the point (3, 0).<\/p>\n<\/div>\n

          \n

          Solutions<\/h3>\n

          The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m\u00a0<\/em>= 3, x\u00a0<\/em>= 3, and f<\/em>(x<\/em>) = 0 into the slope-intercept form to find the y-<\/em>intercept.<\/p>\n

          [latex]\\begin{cases}g\\left(x\\right)=3x+b\\hfill \\\\ \\text{ }0=3\\left(3\\right)+b\\hfill \\\\ \\text{ }b=-9\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

          The line parallel to\u00a0f<\/em>(x<\/em>) that passes through (3, 0) is [latex]g\\left(x\\right)=3x - 9\\\\[\/latex].<\/p>\n<\/div>\n

          \n

          Analysis of the Solution<\/h3>\n

          We can confirm that the two lines are parallel by graphing them. Figure 22\u00a0shows that the two lines will never intersect.
          \n<\/span><\/p>\n

          \"Graph<\/p>\n

          Figure 22<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n

          \n

          Writing Equations of Perpendicular Lines<\/h2>\n

          We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:<\/p>\n

          [latex]f\\left(x\\right)=2x+4\\\\[\/latex]<\/div>\n

          The slope of the line is 2, and its negative reciprocal is [latex]-\\frac{1}{2}\\\\[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}\\\\[\/latex] will be perpendicular to\u00a0f<\/em>(x<\/em>). So the lines formed by all of the following functions will be perpendicular to\u00a0f<\/em>(x<\/em>).<\/p>\n

          [latex]\\begin{cases}g\\left(x\\right)=-\\frac{1}{2}x+4\\hfill \\\\ h\\left(x\\right)=-\\frac{1}{2}x+2\\hfill \\\\ p\\left(x\\right)=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

          As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to\u00a0f<\/em>(x<\/em>) and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}\\\\[\/latex]. Now we can use the point to find the y<\/em>-intercept by substituting the given values into the slope-intercept form of a line and solving for b<\/em>.<\/p>\n

          [latex]\\begin{cases}g\\left(x\\right)=mx+b\\hfill \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill \\\\ b=2\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

          The equation for the function with a slope of [latex]-\\frac{1}{2}\\\\[\/latex] and a y-<\/em>intercept of 2 is<\/p>\n

          [latex]g\\left(x\\right)=-\\frac{1}{2}x+2\\\\[\/latex].<\/div>\n

          So [latex]g\\left(x\\right)=-\\frac{1}{2}x+2\\\\[\/latex] is perpendicular to [latex]f\\left(x\\right)=2x+4\\\\[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/p>\n

          \n

          Q & A<\/h3>\n

          A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong><\/p>\n

          No. For two perpendicular linear functions, the product of their slopes is \u20131. However, a vertical line is not a function so the definition is not contradicted.<\/em><\/p>\n<\/div>\n

          \n

          How To: Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.<\/h3>\n
            \n
          1. Find the slope of the function.<\/li>\n
          2. Determine the negative reciprocal of the slope.<\/li>\n
          3. Substitute the new slope and the values for x<\/em>\u00a0and y<\/em>\u00a0from the coordinate pair provided into [latex]g\\left(x\\right)=mx+b\\\\[\/latex].<\/li>\n
          4. Solve for b<\/em>.<\/li>\n
          5. Write the equation for the line.<\/li>\n<\/ol>\n<\/div>\n
            \n
            \n
            \n

            Example 10: Finding the Equation of a Perpendicular Line<\/h3>\n

            Find the equation of a line perpendicular to [latex]f\\left(x\\right)=3x+3\\\\[\/latex] that passes through the point (3, 0).<\/p>\n<\/div>\n

            \n

            Solution<\/h3>\n

            The original line has slope m\u00a0<\/em>= 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\\frac{1}{3}\\\\[\/latex]. Using this slope and the given point, we can find the equation for the line.<\/p>\n

            [latex]\\begin{cases}g\\left(x\\right)=-\\frac{1}{3}x+b\\hfill \\\\ \\text{ }0=-\\frac{1}{3}\\left(3\\right)+b\\hfill \\\\ \\text{ }1=b\\hfill \\\\ \\text{ }b=1\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

            The line perpendicular to\u00a0f<\/em>(x<\/em>)\u00a0that passes through (3, 0) is [latex]g\\left(x\\right)=-\\frac{1}{3}x+1\\\\[\/latex].<\/p>\n<\/div>\n

            \n

            Analysis of the Solution<\/h3>\n

            A graph of the two lines is shown in Figure 23.
            \n<\/span><\/p>\n

            \"Graph<\/p>\n

            Figure 23<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

            \n

            Try It 5<\/h3>\n

            Given the function [latex]h\\left(x\\right)=2x - 4\\\\[\/latex], write an equation for the line passing through (0, 0) that is<\/p>\n

            a. parallel to h<\/em>(x<\/em>)
            \nb. perpendicular to h<\/em>(x<\/em>)<\/p>\n

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

            \n

            How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/h3>\n
              \n
            1. Determine the slope of the line passing through the points.<\/li>\n
            2. Find the negative reciprocal of the slope.<\/li>\n
            3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\n
            4. Simplify.<\/li>\n<\/ol>\n<\/div>\n
              \n
              \n
              \n

              Example 11: Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point<\/h3>\n

              A line passes through the points (\u20132, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).<\/p>\n<\/div>\n

              \n

              Solution<\/h3>\n

              From the two points of the given line, we can calculate the slope of that line.<\/p>\n

              [latex]\\begin{cases}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill \\\\ =\\frac{-1}{6}\\hfill \\\\ =-\\frac{1}{6}\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

              Find the negative reciprocal of the slope.<\/p>\n

              [latex]\\begin{cases}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill \\\\ =-1\\left(-\\frac{6}{1}\\right)\\hfill \\\\ =6\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

              We can then solve for the y-<\/em>intercept of the line passing through the point (4, 5).<\/p>\n

              [latex]\\begin{cases}g\\left(x\\right)=6x+b\\hfill \\\\ 5=6\\left(4\\right)+b\\hfill \\\\ 5=24+b\\hfill \\\\ -19=b\\hfill \\\\ b=-19\\hfill \\end{cases}\\\\[\/latex]<\/div>\n

              The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is<\/p>\n

              [latex]y=6x - 19\\\\[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
              \n

              Try It 6<\/h3>\n

              A line passes through the points, (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point, (6, 4).<\/p>\n

              Solution<\/a><\/p>\n<\/div>\n<\/section>\n\n\t\t\t

              \n\t\t\t

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