{"id":776,"date":"2025-07-15T16:08:46","date_gmt":"2025-07-15T16:08:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=776"},"modified":"2026-01-12T16:04:23","modified_gmt":"2026-01-12T16:04:23","slug":"linear-functions-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/linear-functions-learn-it-6\/","title":{"raw":"Linear Functions: Learn It 6","rendered":"Linear Functions: Learn It 6"},"content":{"raw":"

Writing Equations of Parallel Lines<\/h2>\r\nIf 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.\r\n\r\n
Suppose we are given the following function:\r\n

[latex]f\\left(x\\right)=3x+1[\/latex]<\/p>\r\nWe know that the slope of the line is [latex]3[\/latex]. We also know that the [latex]y[\/latex]-<\/em>intercept is [latex](0, 1)[\/latex]. Any other line with a slope of [latex]3[\/latex] will be parallel to [latex]f(x)[\/latex]. The lines formed by all of the following functions will be parallel to [latex]f(x)[\/latex].\r\n

[latex]\\begin{array}{l}g\\left(x\\right)=3x+6\\hfill \\\\ h\\left(x\\right)=3x+1\\hfill \\\\ p\\left(x\\right)=3x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nSuppose then we want to write the equation of a line that is parallel to [latex]f [\/latex] and passes through the point [latex](1, 7)[\/latex]. We already know that the slope is [latex]3[\/latex]. We just need to determine which value for [latex]b[\/latex]\u00a0will give the correct line. We can begin by using point-slope form of an equation for a line. We can then rewrite it in slope-intercept form.\r\n

[latex]\\begin{array}{l}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{array}[\/latex]<\/p>\r\nSo [latex]g\\left(x\\right)=3x+4[\/latex] is parallel to [latex]f\\left(x\\right)=3x+1[\/latex] and passes through the point [latex](1, 7)[\/latex].\r\n\r\n<\/section>

How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is parallel to the given line.<\/strong>\r\n
    \r\n \t
  1. Find the slope of the function.<\/li>\r\n \t
  2. Substitute the slope and given point into point-slope or slope-intercept form.<\/li>\r\n \t
  3. Simplify.<\/li>\r\n<\/ol>\r\n<\/section>
    Find a line parallel to the graph of [latex]f\\left(x\\right)=3x+6[\/latex] that passes through the point [latex](3, 0)[\/latex].[reveal-answer q=\"961163\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"961163\"]The slope of the given line is [latex]3[\/latex]. If we choose slope-intercept form, we can substitute [latex]m=3[\/latex],\u00a0[latex]x=3[\/latex], and [latex]f(x)=0[\/latex] into slope-intercept form to find the [latex]y-[\/latex]intercept.\r\n

    [latex]\\begin{array}{l}g\\left(x\\right)=3x+b\\hfill \\\\ \\text{}0=3\\left(3\\right)+b\\hfill \\\\ \\text{}b=-9\\hfill \\end{array}[\/latex]<\/p>\r\nThe line parallel to\u00a0[latex]f(x)[\/latex] that passes through [latex](3, 0)[\/latex] is [latex]g\\left(x\\right)=3x - 9[\/latex].\r\n\r\nAnalysis of the Solution<\/strong>\r\n\r\nWe can confirm that the two lines are parallel by graphing them. The figure below shows that the two lines will never intersect.\r\n\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

    Writing Equations of Perpendicular Lines<\/h2>\r\nWe can use a very similar process to write the equation of a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope.\r\n\r\n
    Suppose we are given the following function:\r\n

    [latex]f\\left(x\\right)=2x+4[\/latex]<\/p>\r\nThe slope of the line is [latex]2[\/latex], 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\u00a0[latex]f(x)[\/latex]. The lines formed by all of the following functions will be perpendicular to\u00a0[latex]f(x)[\/latex].\r\n

    [latex]\\begin{array}{l}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{array}[\/latex]<\/p>\r\nAs before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose that we want to write the equation of a line that is perpendicular to\u00a0[latex]f(x)[\/latex] and passes through the point [latex](4, 0)[\/latex]. We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the [latex]y[\/latex]-intercept by substituting the given values into the slope-intercept form of a line and solving for [latex]b[\/latex].\r\n

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

    [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex].<\/p>\r\nSo [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 [latex](4, 0)[\/latex].\r\n\r\n<\/section>

    Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/section>
    How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is Perpendicular to the given line.<\/strong>\r\n
      \r\n \t
    1. Find the slope of the given function.<\/li>\r\n \t
    2. Determine the negative reciprocal of the slope.<\/li>\r\n \t
    3. Substitute the new slope and the values for [latex]x[\/latex]\u00a0and [latex]y[\/latex]\u00a0from given point into [latex]g\\left(x\\right)=mx+b[\/latex].<\/li>\r\n \t
    4. Solve for [latex]b[\/latex].<\/li>\r\n \t
    5. Write the equation of the line.<\/li>\r\n<\/ol>\r\n<\/section>
      Find the equation of a line perpendicular to [latex]f\\left(x\\right)=3x+3[\/latex] that passes through the point [latex](3, 0)[\/latex].[reveal-answer q=\"668867\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"668867\"]The original line has slope [latex]m=3[\/latex], 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.\r\n

      [latex]\\begin{array}{l}g\\left(x\\right)=-\\frac{1}{3}x+b\\hfill \\\\ 0=-\\frac{1}{3}\\left(3\\right)+b\\hfill \\\\ \\text{ }1=b\\hfill \\\\ b=1\\hfill \\end{array}[\/latex]<\/p>\r\nThe line perpendicular to\u00a0[latex]f(x)[\/latex]\u00a0that passes through [latex](3, 0)[\/latex] is [latex]g\\left(x\\right)=-\\frac{1}{3}x+1[\/latex].\r\n\r\nAnalysis of the Solution<\/strong>\r\n\r\nA graph of the two lines is shown below.\r\n\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      [ohm_question hide_question_numbers=1]318749[\/ohm_question]<\/section>
      How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/strong>\r\n
        \r\n \t
      1. Determine the slope of the line passing through the points.<\/li>\r\n \t
      2. Find the negative reciprocal of the slope.<\/li>\r\n \t
      3. Use slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\r\n \t
      4. Simplify.<\/li>\r\n<\/ol>\r\n<\/section>
        A line passes through the points [latex](\u20132, 6)[\/latex] and [latex](4, 5)[\/latex]. Find the equation of a line that is perpendicular and passes through the point [latex](4, 5)[\/latex].[reveal-answer q=\"289053\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"289053\"]From the two points of the given line, we can calculate the slope of that line.\r\n

        [latex]\\begin{array}{l}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill \\\\ {m}_{1}=\\frac{-1}{6}\\hfill \\\\ {m}_{1}=-\\frac{1}{6}\\hfill \\end{array}[\/latex]<\/p>\r\nFind the negative reciprocal of the slope.\r\n

        [latex]\\begin{array}{l}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill \\\\ {m}_{2}=-1\\left(-\\frac{6}{1}\\right)\\hfill \\\\ {m}_{2}=6\\hfill \\end{array}[\/latex]<\/p>\r\nWe can then solve for the y-<\/em>intercept of the line passing through the point [latex](4, 5)[\/latex].\r\n

        [latex]\\begin{array}{l}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{array}[\/latex]<\/p>\r\nThe equation of the line that passes through the point [latex](4, 5)[\/latex] and is perpendicular to the line passing through the two given points is [latex]y=6x - 19[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

        \r\n

        equations of parallel and perpendicular lines<\/h3>\r\n
          \r\n \t
        • Parallel lines will have the\u00a0same slope<\/em> but different [latex]y[\/latex]-intercepts.<\/li>\r\n \t
        • Perpendicular lines will have\u00a0opposite reciprocal\u00a0<\/em>slopes.<\/li>\r\n<\/ul>\r\n<\/section> ","rendered":"

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

          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

          Suppose we are given the following function:<\/p>\n

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

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

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

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

          [latex]\\begin{array}{l}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{array}[\/latex]<\/p>\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 [latex](1, 7)[\/latex].<\/p>\n<\/section>\n

          How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is parallel to the given line.<\/strong><\/p>\n
            \n
          1. Find the slope of the function.<\/li>\n
          2. Substitute the slope and given point into point-slope or slope-intercept form.<\/li>\n
          3. Simplify.<\/li>\n<\/ol>\n<\/section>\n
            Find a line parallel to the graph of [latex]f\\left(x\\right)=3x+6[\/latex] that passes through the point [latex](3, 0)[\/latex].<\/p>\n