Parallel and Perpendicular Lines Cont.

Writing Equations of Parallel Lines

Suppose we are given the following equation:

[latex]y=3x+1[/latex]

We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line with a slope of 3 will be parallel to [latex]y=3x+1[/latex]. So all of the following lines will be parallel to the given line.

[latex]\begin{array}{lll}y=3x+6\hfill & \\ y=3x+1\hfill & \\ y=3x+\frac{2}{3}\hfill \end{array}[/latex]

Suppose then we want to write the equation of a line that is parallel to [latex]y=3x+6[/latex] and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for b will give the correct line. We can begin with point-slope form of a line and then rewrite it in slope-intercept form.

[latex]\begin{array}{llll}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]

So [latex]y=3x+4[/latex] is parallel to [latex]y=3x+1[/latex] and passes through the point (1, 7).

Writing Equations of Perpendicular Lines

We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following line:

[latex]y=2x+4[/latex]

The slope of the line is 2 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 [latex]y=2x+4[/latex]. So all of the following lines will be perpendicular to [latex]y=2x+4[/latex].

[latex]\begin{array}{lll}y=-\frac{1}{2}x+4\hfill & \\ y=-\frac{1}{2}x+2\hfill & \\ y=-\frac{1}{2}x-\frac{1}{2}\hfill \end{array}[/latex]

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

[latex]\begin{array}{lllll}y=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]

The equation for the function with a slope of [latex]-\frac{1}{2}[/latex] and a y-intercept of 2 is [latex]y=-\frac{1}{2}x+2[/latex].

So [latex]y=-\frac{1}{2}x+2[/latex] is perpendicular to [latex]y=2x+4[/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.