On this page, we practice finding the distance between different pairs of points on the coordinate plane.

We practice:

• points with positive coordinates
• points with negative coordinates
• points in different quadrants
• horizontal and vertical distances
• answers involving square roots

Remember

For two points:

$$P_1=(x_1,y_1)$$

and

$$P_2=(x_2,y_2)$$

use:

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Example 1

Find the distance between:

$$P_1=(2,3)$$

and

$$P_2=(5,7)$$

Solution:

\begin{align}
d &= \sqrt{(5-2)^2+(7-3)^2}\\
\\
&= \sqrt{3^2+4^2}\\
\\
&= \sqrt{9+16}\\
\\
&= \sqrt{25}\\
\\
&= 5
\end{align}

So, the distance is \(5\) units.

Example 2

Find the distance between:

$$P_1=(-4,1)$$

and

$$P_2=(2,9)$$

Solution:

\begin{align}
d &= \sqrt{(2-(-4))^2+(9-1)^2}\\
\\
&= \sqrt{6^2+8^2}\\
\\
&= \sqrt{36+64}\\
\\
&= \sqrt{100}\\
\\
&= 10
\end{align}

So, the distance is \(10\) units.

Example 3

Find the distance between:

$$P_1=(-2,-3)$$

and

$$P_2=(1,1)$$

Solution:

\begin{align}
d &= \sqrt{(1-(-2))^2+(1-(-3))^2}\\
\\
&= \sqrt{3^2+4^2}\\
\\
&= \sqrt{9+16}\\
\\
&= \sqrt{25}\\
\\
&= 5
\end{align}

So, the distance is \(5\) units.

Example 4 – Vertical Distance

Find the distance between:

$$P_1=(4,-2)$$

and

$$P_2=(4,5)$$

Solution:

\begin{align}
d &= \sqrt{(4-4)^2+(5-(-2))^2}\\
\\
&= \sqrt{0^2+7^2}\\
\\
&= \sqrt{49}\\
\\
&= 7
\end{align}

So, the distance is \(7\) units.

Because the two points have the same \(x\)-coordinate, the distance is vertical.

Example 5 – Horizontal Distance

Find the distance between:

$$P_1=(-3,2)$$

and

$$P_2=(2,2)$$

Solution:

\begin{align}
d &= \sqrt{(2-(-3))^2+(2-2)^2}\\
\\
&= \sqrt{5^2+0^2}\\
\\
&= \sqrt{25}\\
\\
&= 5\\
\end{align}

So, the distance is \(5\) units.

Because the two points have the same \(y\)-coordinate, the distance is horizontal.

Example 6 – Answer in Square-Root Form

Find the distance between:

$$P_1=(0,0)$$

and

$$P_2=(2,3)$$

Solution:

\begin{align}
d &= \sqrt{(2-0)^2+(3-0)^2}\\
\\
&= \sqrt{2^2+3^2}\\
\\
&= \sqrt{4+9}\\
\\
&= \sqrt{13}\\
\end{align}

As a decimal:

$$d\approx3.61$$

So, the distance is:

$$\sqrt{13}\approx3.61\text{ units}$$

Practice

• More Examples (Current page)
• Take the Quiz

Continue Learning

1. Plotting Points on the Coordinate Plane (\(2D\))
2. Which Quadrant or Axis? Points on the Coordinate Plane \(2D\)
3. Distance Formula in \(2D\)
4. Distance between Two Points in \(2D\)
5. Distance Formula: Radius and Area of a Circle

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