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Math Lesson 14.4.6 - Finding the Distance between Parallel Lines
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Welcome to our Math lesson on Finding the Distance between Parallel Lines, this is the sixth lesson of our suite of math lessons covering the topic of Parallel, Perpendicular and Intersecting Graphs, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.
Finding the Distance between Parallel Lines
Now that we have covered parallel and perpendicular lines it is the right moment to explain how to find the distance between two parallel lines with known equations. First, it is necessary to point out that the distance between two parallel lines represents the shortest path that connects them, which corresponds to the distance between the two intercepts with the perpendicular line that intersects the two original parallel lines. In this way, we need three lines: the two parallel ones L1 and L2 and another line L3 that is perpendicular to the other two, as shown in the figure below.
We can use the intercepts A and B to find the equation of the missing line if two out of three lines have known equations. Thus, if the equations of the parallel lines L1 and L2 are known, you can use any of point A or B to find the equation of the perpendicular line L3. On the other hand, if the equation of L1 and L3 are known, we need the coordinates of point B to find the equation of L2, and so on.
However, the most important thing to find this situation regards the distance between two parallel lines if both of them have known equations. We cannot use the distance between x- or y-intercepts for this goal, as usually linear graphs are not perpendicular to the X- or Y-axes. Therefore, we must find other ways to calculate the distance between two parallel lines. The procedure used for this purpose is as follows:
Step 1: Choose an x-coordinate for any of the parallel lines and find the corresponding y-coordinate. In this way, you identify a point of that line.
Step 2: Find the gradient of the perpendicular line using the relationship between gradients in perpendicular lines.
Step 3: Find the equation of the perpendicular line given that the point identified in step 1 belongs to the perpendicular line as well.
Step 4: Make a system of equations with the equation of the perpendicular line you found in the previous step and the other parallel line, not the one of step 1.
Step 5: Solve the linear system to find the intercept of the perpendicular line to the other parallel line. This is the second point identified, after that identified in step 1.
Step 6: Calculate the distance d between the two lines by using the equation
where x1 and y1 are the coordinates of the leftmost point from those identified in steps 1 and 5 while x2 and y2 the coordinates of the rightmost one.
We explain the formula of distance in chapters 18 with particular focus in chapter 23. For now, it is sufficient to take it as true a-priori, without proof.
Let's consider an example to make this point clearer.
Example 6
What is the distance between the lines y = 2x - 4 and y = 2x + 6?
Solution 6
Let's denote the line y = 2x - 4 by L1 and y = 2x + 6 by L2. Since the gradients of the two lines L1 and L2 are equal (m1 = m2 = 2) they are parallel. Thus, if we choose randomly an x-coordinate from L1, for example, x = 2, we obtain the corresponding y-coordinate for that point (say point A). Thus,
= 2 ∙ 2 - 4
= 4 - 4
= 0
Hence, the coordinates of this point are A(2, 0).
Now, let's find the equation of the perpendicular line L3 that intercepts L1 at point A. Given the formula of gradients in perpendicular lines
we obtain for m3 after substitutions
m3 = -1/2
Therefore, the perpendicular line L3 will have the equation
To find the value of n, we substitute the coordinates of point A (since it belongs to L3 as well) in the above equation. Thus, for x = 2 and y = 0, we have
0 = -1 + n
n = 1
Therefore, the perpendicular line L3 has the equation
Now, we have to find the point B, which is the intercept of L2 and L3 by solving the system of the corresponding linear equations
Multiplying the second equation of the above system by 4 to eliminate x, yields
Thus, adding the two equations to each other yields
5y = 10
y = 2
Substituting this value in the first equation of the system (because it is easier to handle, as it does not contain fractions) yields
2x = 2 - 6
2x = -4
x = -2
Therefore, point B (which is the intercept of L2 and L3) has the coordinates B(-2, 2).
Now, let's calculate the distance between L1 and L2 using the formula of distance (B comes first, as it is the leftmost point)
= √(2 - (-2))2 + (0 - 2)2
= √(2 + 2)2 + (0 - 2)2
= √42 + (-2)2
= √16 + 4
= √20
≈ 4.5 units
Therefore, since A and B are points that belong to L1 and L2 respectively, these two parallel lines are approximately 4.5 units away from each other.
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