We can utilize the Pythagorean Theorem to find the distance between any two points in the plane. Consider the figure below. The line segment contains the endpoints \(\left( {{x_1},{y_1}} \right)\) and \(({x_2},{y_2})\). We can construct a right triangle from this and draw some conclusions.

We can determine that the third coordinate of the triangle, that is, the coordinate located at the vertex of the right angle, is \(\left( {{x_2},{y_1}} \right)\). Then the length of the horizontal leg of the triangle is \({x_2} - {x_1}\) and the length of the vertical leg of the triangle is \({y_2} - {y_1}\).

Then if we call the hypotenuse *d* (for distance), we can determine the distance *d* between the two given coordinates. We have

\({d^2} = {\left( {{x_2} - {x_1}} \right)^2} + {\left( {{y_2} - {y_1}} \right)^2}\)

So that

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

The formula above is known as the *distance formula*. We can use it to find the distance* d* between *any* two points in the plane. Let’s look at an example.

Find the distance between the points \(\left( { - 8,6} \right)\) and \(\left( { - 5, - 4} \right)\). I label my coordinates and plug them into the distance formula. Here \({x_1} = - 8\), \({x_2} = - 5\), \({y_1} = 6\) and \({y_2} = - 4\). Then the distance *d *between the two points is

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

\( = \sqrt {{{\left( {\left( { - 5} \right) - \left( { - 8} \right)} \right)}^2} + {{\left( {\left( { - 4} \right) - 6} \right)}^2}} \)

\( = \sqrt {{3^2} + {{\left( { - 10} \right)}^2}} \)

\( = \sqrt {9 + 100} \)

\( = \sqrt {109} \)

So the distance between \(\left( { - 8,6} \right)\) and \(\left( { - 5, - 4} \right)\) is \(\sqrt {109} \) units.

Let’s do one more example. We want to find the distance between the points \(\left( {6, - 6} \right)\) and \(\left( { - 3, - 3} \right)\).

We will do the same technique as we did in the last example. Here \({x_1} = 6\), \({x_2} = - 3\), \({y_1} = - 6\), and \({y_2} = - 3\).

Then, by the distance formula, we have the distance *d* between the two points is

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

\( = \sqrt {{{\left( { - 3 - 6} \right)}^2} + {{\left( { - 3 - \left( { - 6} \right)} \right)}^2}} \)

\( = \sqrt {{{\left( { - 9} \right)}^2} + {{\left( 3 \right)}^2}} \)

\( = \sqrt {81 + 9} \)

\( = \sqrt {90} \)

\( = \sqrt {9*10} \)

\( = 3\sqrt {10} \)

Then the distance between \(\left( {6, - 6} \right)\) and \(\left( { - 3, - 3} \right)\) is \(3\sqrt {10} \) units.

Below you can **download** some** free** math worksheets and practice.

Find the distance between each pair of points.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch below how to solve this example:**

Find the distance between each pair of points.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch below how to solve this example:**

Find the distance between each pair of points.

This free worksheet contains 10 assignments each with 24 questions with answers.**Example of one question:**

**Watch below how to solve this example:**