{"id":768,"date":"2025-07-15T16:07:45","date_gmt":"2025-07-15T16:07:45","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=768"},"modified":"2026-01-12T15:51:27","modified_gmt":"2026-01-12T15:51:27","slug":"linear-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/linear-functions-learn-it-2\/","title":{"raw":"Linear Functions: Learn It 2","rendered":"Linear Functions: Learn It 2"},"content":{"raw":"

Finding a Linear Equation<\/h2>\r\n

The Point-Slope Formula<\/h3>\r\nGiven the slope and one point on a line, we can find the equation of the line using point-slope form.\r\n
[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\n
\r\n

point-slope form<\/h3>\r\nThe point-slope form of the equation of a line is:\r\n

[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\nwhere:\r\n

    \r\n \t
  • [latex]m[\/latex] is the slope of the line<\/li>\r\n \t
  • [latex](x_1, y_1)[\/latex] are the coordinates of any point on the line.<\/li>\r\n<\/ul>\r\n<\/section>This is an important formula, as it will be used in other areas of Precalculus and often in Calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and re-write it in slope-intercept form [latex]y = mx+b[\/latex].\r\n\r\n
    Write the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex] in point-slope form and in slope-intercept form.\r\n<\/strong>\r\n
      \r\n \t
    • Point-Slope Form: <\/strong>Substituting the given point [latex]\\left(4,8\\right)[\/latex] and slope [latex]m=-3[\/latex], we have:<\/li>\r\n<\/ul>\r\n

      [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\n

      [latex]y-8=3\\left(x-4\\right)[\/latex]<\/p>\r\n\r\n

        \r\n \t
      • Slope-Intercept Form: <\/strong>To convert the point-slope form to slope-intercept form, we can expand and simplify out point-slope form.<\/li>\r\n<\/ul>\r\n

        [latex]\\begin{align*} \\text{Point-Slope Form:} & \\quad y - 8 = -3(x - 4) \\\\ \\text{Expand and Simplify:} & \\quad y - 8 = -3x + 12 \\\\ \\text{Slope-Intercept Form:} & \\quad y = -3x + 20 \\end{align*}[\/latex]<\/p>\r\n\r\n<\/section>

        Find the equation of the line passing through the points [latex]\\left(3,4\\right)[\/latex] and [latex]\\left(0,-3\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n[reveal-answer q=\"975043\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"975043\"]First, we calculate the slope using the slope formula and two points.\r\n
        [latex]\\begin{array}{l}\\hfill \\\\ m=\\frac{-3 - 4}{0 - 3}\\hfill \\\\ =\\frac{-7}{-3}\\hfill \\\\ =\\frac{7}{3}\\hfill \\end{array}[\/latex]<\/div>\r\nNext, we use point-slope form with the slope of [latex]\\frac{7}{3}[\/latex] and either point. Let\u2019s pick the point [latex]\\left(3,4\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n
        [latex]\\begin{array}{l}y - 4=\\frac{7}{3}\\left(x - 3\\right)\\hfill \\\\ y - 4=\\frac{7}{3}x - 7\\hfill& \\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/div>\r\nIn slope-intercept form, the equation is written as [latex]y=\\frac{7}{3}x - 3[\/latex].\r\n\r\nAnalysis of the Solution<\/strong>\r\n
        \r\n\r\nTo prove that either point can be used, let us use the second point [latex]\\left(0,-3\\right)[\/latex] and see if we get the same equation.\r\n
        [latex]\\begin{array}{l}y-\\left(-3\\right)=\\frac{7}{3}\\left(x - 0\\right)\\hfill \\\\ y+3=\\frac{7}{3}x\\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/div>\r\nWe see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
        [ohm_question hide_question_numbers=1]318743[\/ohm_question]<\/section>
        [ohm_question hide_question_numbers=1]318744[\/ohm_question]<\/section>\r\n

        Standard Form of a Line<\/h3>\r\nAnother way that we can represent the equation of a line is in standard form<\/strong>. Standard form is given as\r\n
        [latex]Ax+By=C[\/latex]<\/div>\r\nwhere [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] are integers. The x\u00a0<\/em>and y-<\/em>terms are on one side of the equal sign and the constant term is on the other side.\r\n\r\n
        \r\n

        standard form<\/h3>\r\nThe standard form of a line equation is written as:\r\n

        [latex]Ax+By=C[\/latex]<\/p>\r\nwhere:\r\n

          \r\n \t
        • [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] are integers,<\/li>\r\n \t
        • [latex]A[\/latex] and [latex]B[\/latex] not both zero.<\/li>\r\n<\/ul>\r\n<\/section>
          Write the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex] in standard form.To find the standard form of a line, first, you should find the slope-intercept form [latex]y =mx+b[\/latex].\r\n
            \r\n \t
          • In the example above, we have found the the slope-intercept form of this line is: [latex]y = -3x + 20[\/latex].<\/li>\r\n<\/ul>\r\nOnce the slope-intercept form is established, rearrange the equation by moving all variable terms to one side and the constant term to the other to achieve the standard form:\r\n

            [latex]\\begin{align*} \\text{Slope-Intercept Form:} & \\quad y = -3x + 20 \\\\ \\text{Add } 3x \\text{ to both sides:} & \\quad 3x + y = 3x - 3x + 20 \\\\ \\text{Simplify:} & \\quad 3x + y = 20 \\end{align*}[\/latex]<\/p>\r\n\r\n<\/section>

            Find the equation of the line with [latex]m=-6[\/latex] and passing through the point [latex]\\left(\\frac{1}{4},-2\\right)[\/latex]. Write the equation in standard form.\r\n[reveal-answer q=\"111657\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"111657\"]We begin by using point-slope form.\r\n
            [latex]\\begin{array}{l}y-\\left(-2\\right)=-6\\left(x-\\frac{1}{4}\\right)\\hfill \\\\ y+2=-6x+\\frac{3}{2}\\hfill \\end{array}[\/latex]<\/div>\r\nFrom here, we multiply through by 2 as no fractions are permitted in standard form. Then we move both variables to the left aside of the equal sign and move the constants to the right.\r\n
            [latex]\\begin{array}{l}2\\left(y+2\\right)=\\left(-6x+\\frac{3}{2}\\right)2\\hfill \\\\ 2y+4=-12x+3\\hfill \\\\ 12x+2y=-1\\hfill \\end{array}[\/latex]<\/div>\r\nThis equation is now written in standard form.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
            [ohm_question hide_question_numbers=1]318745[\/ohm_question]<\/section>
            [ohm_question hide_question_numbers=1]318746[\/ohm_question]<\/section>","rendered":"

            Finding a Linear Equation<\/h2>\n

            The Point-Slope Formula<\/h3>\n

            Given the slope and one point on a line, we can find the equation of the line using point-slope form.<\/p>\n

            [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n
            \n

            point-slope form<\/h3>\n

            The point-slope form of the equation of a line is:<\/p>\n

            [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n

            where:<\/p>\n

              \n
            • [latex]m[\/latex] is the slope of the line<\/li>\n
            • [latex](x_1, y_1)[\/latex] are the coordinates of any point on the line.<\/li>\n<\/ul>\n<\/section>\n

              This is an important formula, as it will be used in other areas of Precalculus and often in Calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and re-write it in slope-intercept form [latex]y = mx+b[\/latex].<\/p>\n

              Write the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex] in point-slope form and in slope-intercept form.
              \n<\/strong><\/p>\n
                \n
              • Point-Slope Form: <\/strong>Substituting the given point [latex]\\left(4,8\\right)[\/latex] and slope [latex]m=-3[\/latex], we have:<\/li>\n<\/ul>\n

                [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n

                [latex]y-8=3\\left(x-4\\right)[\/latex]<\/p>\n

                  \n
                • Slope-Intercept Form: <\/strong>To convert the point-slope form to slope-intercept form, we can expand and simplify out point-slope form.<\/li>\n<\/ul>\n

                  [latex]\\begin{align*} \\text{Point-Slope Form:} & \\quad y - 8 = -3(x - 4) \\\\ \\text{Expand and Simplify:} & \\quad y - 8 = -3x + 12 \\\\ \\text{Slope-Intercept Form:} & \\quad y = -3x + 20 \\end{align*}[\/latex]<\/p>\n<\/section>\n

                  Find the equation of the line passing through the points [latex]\\left(3,4\\right)[\/latex] and [latex]\\left(0,-3\\right)[\/latex]. Write the final equation in slope-intercept form.<\/p>\n