3.11 Find the Equation of a Line

Since we are given the slope and [latex]y[/latex]-intercept of the line, we can substitute the needed values into the slope–intercept form, [latex]y=mx+b[/latex].

Step 1: Name the slope.

[latex]m=7[/latex]

Step 2: Name the [latex]y[/latex]-intercept.

[latex]y[/latex]-intercept = [latex]{(0,\;}{\color{blue}{-}}{}{\color{blue}{1}}{)}[/latex]

Step 3: Substitute the values into [latex]y=mx+b[/latex].

[latex]\begin{align*}y&={\color{red}{m}}x+{\color{blue}{b}}\\y&={\color{red}{-}}{\color{red}{7}}x+({\color{blue}{-}}{\color{blue}{1}})\\y&=-7x-1\end{align*}[/latex]

We need to find the slope and [latex]y[/latex]-intercept of the line from the graph so we can substitute the needed values into the slope–intercept form, [latex]y=mx+b[/latex].

To find the slope, we choose two points on the graph.

The [latex]y[/latex]-intercept is [latex](0,-4)[/latex] and the graph passes through [latex](3,-2)[/latex].

Step 1: Find the slope by counting the rise and run.

[latex]\begin{align*}m&=\frac{rise}{run}\\m&={\color{red}{\frac23}}\end{align*}[/latex]

Step 2: Find the [latex]y[/latex]-intercept.

[latex]\text{y-intercept}=(0,{\color{red}{-}}{\color{red}{4}})[/latex]

Step 3: Substitute the values into [latex]y=mx+b[/latex].

[latex]\begin{align*}y&={\color{red}{m}}x+{\color{blue}{b}}\\y&={\color{red}{\frac23}}x{\color{blue}{-}}{\color{blue}{\;}}{\color{blue}{4}}\end{align*}[/latex]

Since we are given a point and the slope of the line, we can substitute the needed values into the point–slope form, [latex]y-y_1[/latex]

Step 1: Identify the slope.

[latex]{m\;=\;}{\color{red}{-}}{}{\color{red}{\frac13}}{}[/latex]

Step 2: Identify the point.

[latex]\left(\overset{{\color{blue}{x}}_{\color{blue}{1}}}6,\;\overset{{\color{blue}{y}}_{\color{blue}{1}}}-4\right)[/latex]

Step 3: Substitute the values into [latex]y-y_1=m(x-x_1)[/latex]

[latex]\begin{align*}y-{\color{blue}{y}}_{\color{blue}{1}}&={\color{red}{m}}(x-{\color{blue}{x}}_{\color{blue}{1}})\\y-({\color{blue}{-}}{\color{blue}{4}})&={\color{red}{-}}{\color{red}{\frac13}}(x-{\color{blue}{6}})\end{align*}[/latex]

Step 4: Simplify.

[latex]{y\;-\;4\;=\;}{\color{black}{-}}{\color{black}{\frac13}}{x\;-\;6}[/latex]

Step 5: Write in slope–intercept form.

[latex]{y\;=\;}{\color{black}{-}}{\color{black}{\frac13}}{x\;-\;2}[/latex]

Every horizontal line has slope [latex]0[/latex]. We can substitute the slope and points into the point–slope form, [latex]y-y_1=m(x-x_1)[/latex].

Step 1: Identify the slope.

[latex]{m\;=\;}{\color{red}{0}}[/latex]

Step 2: Identify the point.

[latex]\left(\overset{{\color{blue}{x}}_{\color{blue}{1}}}{-1},\;\overset{{\color{blue}{y}}_{\color{blue}{1}}}2\right)[/latex]

Step 3: Substitute the values into [latex]y-y_1=m(x-x_1)[/latex].

[latex]\begin{align*} &\;&y-{\color{blue}{y}}_{\color{blue}{1}}&={\color{red}{m}}(x-{\color{blue}{x}}_{\color{blue}{1}})\\ &\;&y-{\color{blue}{2}}&={\color{red}{0}}(x-({\color{blue}{-}}{\color{blue}{1}}))\\ &\text{Simplify.}&y-2&=(x+1)\\ &\;&y-2&=0\\ &\;&y&=2 \end{align*}[/latex]

Step 4: Write in slope–intercept form.

It is in [latex]y[/latex]-form, but could be written [latex]y=0x+2[/latex].

Did we end up with the form of a horizontal line, [latex]y=a[/latex]?

Since we have two points, we will find an equation of the line using the point–slope form. The first step will be to find the slope.

Step 1: Find the slope of the line through [latex](−3, −1)[/latex] and [latex](2, −2)[/latex].

[latex]\begin{align*} m&=\frac{y_2-y_1}{x_2-x_1}\\ m&=\frac{-2-(-1)}{2-(-3)}\\ m&=\frac{-1}5\\ {\color{red}{m}}&={\color{blue}{\frac{-1}5}} \end{align*}[/latex]

Step 2: Choose either point.

[latex]\left(\overset{{\color{blue}{x}}_{\color{blue}{1}}}2,\;-\overset{{\color{blue}{y}}_{\color{blue}{1}}}2\right)[/latex]

Step 3: Substitute the values into [latex]y-y_1=m(x-x_1)[/latex].

[latex]\begin{align*}y-{\color{blue}{y}}_{\color{blue}{1}}&={\color{red}{m}}(x-{\color{blue}{x}}_{\color{blue}{1}})\\y-{\color{blue}{(}}{\color{blue}{-}}{\color{blue}{2}}{\color{blue}{)}}&={\color{red}{-}}{\color{red}{\frac15}}(x-{\color{blue}{2}})\\y+2&=-\frac15x+\frac25\end{align*}[/latex]

Step 4: Write in slope–intercept form.

[latex]y\;=\;-\frac15x\;-\;\frac85[/latex]

Again, the first step will be to find the slope.

Find the slope of the line through [latex](-2,4)[/latex] and [latex](-2,-3)[/latex]

[latex]\begin{align*} m&=\frac{y_2-y_1}{x_2-x_1}\\ m&=\frac{-3-4}{-2-(-2)}\\ m&=\frac{-7}{0} \end{align*}[/latex]

The slope is undefined.

This tells us it is a vertical line. Both of our points have an [latex]x[/latex]-coordinate of [latex]-2[/latex]. So our equation of the line is [latex]x=-2[/latex]. Since there is no [latex]y[/latex], we cannot write it in slope–intercept form.

You may want to sketch a graph using the two given points. Does the graph agree with our conclusion that this is a vertical line?

Again, since we know one point, the point–slope option seems more promising than the slope–intercept option. We need the slope to use this form, and we know the new line will be perpendicular to [latex]x=5[/latex]. This line is vertical, so its perpendicular will be horizontal. This tells us the [latex]m_\perp=0[/latex].

Step 1: Identify the point.

[latex](3,-2)[/latex]

Step 2: Identify the slope of the perpendicular line.

[latex]m_\perp=0[/latex]

Step 3: Substitute the values into [latex]y-y_1=m(x-x_1)[/latex].

[latex]\begin{align*} y-y_1&=m(x-x_1)\\ y-(-2)&=0(x-3)\\ y+2&=0\\ y&=-2 \end{align*}[/latex]

Step 4: Simplify.

[latex]y=-2[/latex]

Sketch the graph of both lines. Do they appear to be perpendicular?

The line [latex]y=-4[/latex] is a horizontal line. Any line perpendicular to it must be vertical, in the form [latex]x=a[/latex]. Since the perpendicular line is vertical and passes through [latex](-4,2)[/latex], every point on it has an [latex]x[/latex]-coordinate of [latex]-4[/latex]. The equation of the perpendicular line is [latex]x=-4[/latex]. You may want to sketch the lines. Do they appear perpendicular?