第4章グラフ

Now, let’s consider perpendicular lines. Suppose we need to find a line passing through a specific point and which is perpendicular to a given line. We can use the fact that perpendicular lines have slopes that are negative reciprocals. We will again use the point–slope equation, like we did with parallel lines.

The graph shows the graph of $y=2x-3$. Now, we want to graph a line perpendicular to this line and passing through $\left(\mathrm{-2},1\right)$.

We know that perpendicular lines have slopes that are negative reciprocals. We’ll use the notation $m_{\text{⊥}}$ to represent the slope of a line perpendicular to a line with slope $m$. (Notice that the subscript ⊥ looks like the right angles made by two perpendicular lines.)

We now know the perpendicular line will pass through $\left(\mathrm{-2},1\right)$ with $m_{\text{⊥}}=-\frac{1}{2}$.

To graph the line, we will start at $\left(\mathrm{-2},1\right)$ and count out the rise $\mathrm{-1}$ and the run 2. Then we draw the line.

Do the lines appear perpendicular? Does the second line pass through $\left(\mathrm{-2},1\right)$?

Now, let’s see how to do this algebraically. We can use either the slope–intercept form or the point–slope form to find an equation of a line. In this example we know one point, and can find the slope, so we will use the point–slope form.

In Example 4.67, we used the point–slope form to find the equation. We could have looked at this in a different way.

We want to find a line that is perpendicular to $x=5$ that contains the point $\left(3,\mathrm{-2}\right)$. The graph shows us the line$x=5$ and the point $\left(3,\mathrm{-2}\right)$.

We know every line perpendicular to a vetical line is horizontal, so we will sketch the horizontal line through $\left(3,\mathrm{-2}\right)$.

Do the lines appear perpendicular?

If we look at a few points on this horizontal line, we notice they all have y-coordinates of $\mathrm{-2}$. So, the equation of the line perpendicular to the vertical line $x=5$ is $y=\mathrm{-2}$.