# How do you find the distance between (–3, –2) and (1, 4)?

Jul 1, 2015

$D = 2. \sqrt{13}$

#### Explanation:

The distance between two points A(x;y) and B(x';y') can be calculate with the formula :

$D = \sqrt{{\left(x ' - x\right)}^{2} + {\left(y ' - y\right)}^{2}}$

Then for : A(-3;-2) and B(1;4) we have :

$D = \sqrt{{\left(1 - \left(- 3\right)\right)}^{2} + {\left(4 - \left(- 2\right)\right)}^{2}}$

$D = \sqrt{{4}^{2} + {6}^{2}}$

$D = \sqrt{16 + 36} = \sqrt{52} = 2. \sqrt{13}$

The distance between A(-3;-2) and B(1;4) is exactly $2. \sqrt{13}$

Why this formula work ? In fact, we only calculate the length of vector(BA), and we implicitly use the Pythagorean theorem on it.