Explanation
In this page, we practice using the distance formula in \(1D\).
Remember:
In \(1D\), points lie on a straight line.
The distance between two points is how far apart they are.
Distance is always positive.
Formula / Rule
The distance formula in \(1D is:
\(d = \sqrt{(P_2 − P_1)^2}\)
You can also think of it as:
\(d = \left| P_2 − P_1 \right|\)
Both forms give the same distance.
Example 1
Find the distance between:
\(P_1 = 2\) and \(P_2 = 5\)
Use the formula:
\(d = \sqrt{(P_2 − P_1)^2}\)
Substitute:
\(d = \sqrt{(5 − 2)^2}\)
\(d = \sqrt{3^2}\)
\(d = 3\)
So, the distance is:
\(3\) units
Example 2
Find the distance between:
\(P_1 = 5\) and \(P_2 = 2\)
Use the formula:
\(d = \sqrt{(P_2 − P_1)^2}\)
Substitute:
\(d = \sqrt{(2 − 5)^2}\)
\(d = \sqrt{(-3)^2}\)
\(d = 3\)
So, the distance is:
\(3\) units
This shows that distance is always positive, even if we subtract in the opposite order.
Example 3
Find the distance between:
\(P_1 = −1\) and \(P_2 = −6\)
Use the formula:
\(d = \sqrt{(P_2 − P_1)^2}\)
Substitute:
\(d = \sqrt{(-6 − (−1))^2}\)
\(d = \sqrt{(-6 + 1)^2}\)
\(d = \sqrt{(-5)^2}\)
\(d = 5\)
So, the distance is:
\(5\) units
Example 4
Find the distance between:
\(P_1 = 2\) and \(P_2 = −2\)
Use the formula:
\(d = \sqrt{(P_2 − P_1)^2}\)
Substitute:
\(d = \sqrt{(−2 − 2)^2}\)
\(d = \sqrt{(-4)^2}\)
\(d = 4\)
So, the distance is:
\(4\) units
Example 5
Find the distance between:
\(P_1 = 1.5\) and \(P_2 = 4.5\)
Use the formula:
\(d = \sqrt{(P_2 − P_1)^2}\)
Substitute:
\(d = \sqrt{(4.5 − 1.5)^2}\)
\(d = \sqrt{(3^2}\)
\(d = 3\)
So, the distance is:
\(3\) units
Practice
- More Examples (Current page)
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