Most of the lines we have worked with so far have been slanted, or oblique. In other words, they were neither horizontal nor vertical lines. The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as
[latex]x=c[/latex]
where [latex]c[/latex] is a constant. The slope of a vertical line is undefined, and regardless of the [latex]y[/latex]–value of any point on the line, the [latex]x[/latex]–coordinate of the point will be [latex]c[/latex].
vertical line
The equation of a vertical line is given as
[latex]x=c[/latex]
where [latex]c[/latex] is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be [latex]c[/latex].
Suppose that we want to find the equation of a line containing the following points: [latex]\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right)[/latex], and [latex]\left(-3,5\right)[/latex]. First, we will find the slope.
Zero in the denominator means that the slope is undefined and, therefore, we cannot use point-slope form. However, we can plot the points. Notice that all of the [latex]x[/latex]–coordinates are the same and we find a vertical line through [latex]x=-3[/latex].
The equation of a horizontal line is given as
[latex]y=c[/latex]
where [latex]c[/latex]is a constant. The slope of a horizontal line is zero, and for any [latex]x[/latex]–value of a point on the line, the [latex]y[/latex]–coordinate will be [latex]c[/latex].
horizontal line
The equation of a horizontal line is given as
[latex]y=c[/latex]
where [latex]c[/latex] is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be [latex]c[/latex].
Suppose we want to find the equation of a line that contains the following set of points: [latex]\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right)[/latex], and [latex]\left(5,-2\right)[/latex]. We can use point-slope form. First, we find the slope using any two points on the line.
The graph is a horizontal line through [latex]y=-2[/latex].
Find the equation of a line containing the following points:
[latex]\left(-3,-5\right),\left(-3,1\right),\left(-3,3\right)[/latex], and [latex]\left(-3,5\right)[/latex].
[latex]\left(-2,-2\right),\left(0,-2\right),\left(3,-2\right)[/latex], and [latex]\left(5,-2\right)[/latex].
Let’s use the point-slope form: [latex]y-y_1 = m(x-x_1)[/latex].
First, we need to find the slope:
[latex]m=\frac{5 - 3}{-3-\left(-3\right)}=\frac{2}{0}[/latex]Zero in the denominator means that the slope is undefined and, therefore, we cannot use point-slope form. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through [latex]x=-3[/latex].
Let’s use the point-slope form again: [latex]y-y_1 = m(x-x_1)[/latex].
First, we need to find the slope:
[latex]\begin{array}{l}m=\frac{-2-\left(-2\right)}{0-\left(-2\right)}\hfill \\ =\frac{0}{2}\hfill \\ =0\hfill \end{array}[/latex]Use any point for [latex]\left({x}_{1},{y}_{1}\right)[/latex] in the formula, or use the y-intercept.
[latex]\begin{array}{l}y-\left(-2\right)=0\left(x - 3\right)\hfill \\ y+2=0\hfill \\ y=-2\hfill \end{array}[/latex]The graph is a horizontal line through [latex]y=-2[/latex]. Notice that all of the y-coordinates are the same.
The line x = −3 is a vertical line. The line y = −2 is a horizontal line.
