Distance Formula

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The distance formula to find the distance between two points (x₁, y₁) and (x₂, y₂) is derived from the Pythagorean Theorem.

Let A and B be two points on the coordinate plane, with coordinates (x₁, y₁) and (x₂, y₂) respectively. See the figure below:

Then by Pythagorean Theorem, AB² = $\sqrt{ \left ( x_{2} -x_{1}\right )^{2}+ \left ( y_{2} -y_{1}\right )^{2}}$

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Example: Find the distance between the points (8, -4) and (8, -7).

Let us assume (x₁, y₁) = (8, -4) and (x₂, y₂) = (8, -7)

Then distance between the given points = $\sqrt{ \left ( x_{2} -x_{1}\right )^{2}+ \left ( y_{2} -y_{1}\right )^{2}}$

= $\sqrt{ \left (8 -8\right )^{2}+ \left ( -7 +4\right )^{2}}$ = $\sqrt{ \left (0 )^{2}+ \left ( -3\right )^{2}}$ = $\sqrt{9}$ = 3 units

Example: What is the distance between the points (-6, -1) and (2, 5)?

Let us assume (x₁, y₁) = (-6, -1) and (x₂, y₂) = (2, 5)

Then distance between the given points = $\sqrt{\left ( x_{2} -x_{1}\right )^{2}+ \left ( y_{2} -y_{1}\right )^{2}}$

= $\sqrt{\left ( 2 -(-6)\right )^{2} + \left ( 5 -(-1)\right )^{2}}$ = $\sqrt{\left ( 2 + 6\right )^{2} + \left ( 5 +1\right )^{2}}$

= $\sqrt{ 8^{2}+ 6^{2}}$ = $\sqrt{ 64+ 36}$ = $\sqrt{ 100}$ = 10 units

Check Point

Find the distance between each pair of points.

(-2, -2), (-5, -6)
(-4, 0), (-7, 0)
(-1, 6), (7, 0)
(2, -3), (3, -5)
(-8, 1), (7, 1)

Answer key

5 units
3 units
10 units
$\sqrt{ 5}$ units
15 units

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