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Section2.4Equation of a line: point-slope form

Click here to open the Desmos graph full screen.

Subsection2.4.1Activities and Problems

Use the equation \(y-y_1=m(x-x_1)\) to determine the equation of the line that passes through the point \((-3,2)\) with a slope of \(-\frac{5}{2}\text{.}\) Express the equation in slope-intercept form (\(y=mx+b\)). Use the Desmos graph to check your answer.

Solution

We have \(x_1=-3\text{,}\) \(y_1=2\text{,}\) and \(m=-\frac{5}{2}\text{.}\) Plugging these values into the point-slope equation gives us \(y-2=-\frac{5}{2}(x-(-3))\text{.}\) Solving for \(y\) we see that the slope-intercept equation of the line is \(y=-\frac{5}{2}x-\frac{11}{2}\text{.}\)

Use the equation \(y-y_1=m(x-x_1)\) to determine the equation of the line that passes through the points shown in Table 2.4.1. Express the equation in slope-intercept form, \(y=mx+b\text{.}\) Use the Desmos graph to check your answer.

\(x\) \(y\)
\(-1\) \(6\)
\(4\) \(1\)
Table2.4.1
Solution

We begin by using the formula \(\frac{y_2-y_1}{x_2-x_1}\) to determine that the slope of the line is \(-1\text{.}\) Then making the substitutions \(x_1=-1\text{,}\) \(y_1=6\text{,}\) and \(m=-1\) in the point-slope equation and solving for \(y\text{,}\) we determine that the slope-intercept equation of the line is \(y=-x+5\text{.}\) Note that we also could have found the equation making the substitutions \(x_1=4\) and \(y_1=1\text{.}\)

Use the equation \(y-y_1=m(x-x_1)\) to determine the equation of the line that passes through the points shown in Table 2.4.2. Express the equation in slope-intercept form, \(y=mx+b\text{.}\) Use the Desmos graph to check your answer.

\(x\) \(y\)
\(-2\) \(-7\)
\(\frac{1}{2}\) \(-2\)
Table2.4.2
Solution

We begin by using the formula \(\frac{y_2-y_1}{x_2-x_1}\) to determine that the slope of the line is \(2\text{.}\) Then making the substitutions \(x_1=-2\text{,}\) \(y_1=-7\text{,}\) and \(m=2\) in the point-slope equation and solving for \(y\text{,}\) we determine that the slope-intercept equation of the line is \(y=2x-3\text{.}\)

Explain why it would be somewhat inefficient to use point-slope to determine an equation of the line that passes through the points \((0,3)\) and \((-2,6)\text{.}\)

Solution

Since we're given the \(y\)-intercept, \((0,3)\text{,}\) once we calculate the slope to be \(-\frac{3}{2}\text{,}\) we can use slope-intercept to simply write down the equation \(y=-\frac{3}{2}x+3\text{.}\)