# What is the distance between (31,-201) and (28,-209)?

Jul 23, 2017

See a solution process below:

#### Explanation:

The formula for calculating the distance between two points is:

$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2}}$

Substituting the values from the points in the problem gives:

$d = \sqrt{{\left(\textcolor{red}{28} - \textcolor{b l u e}{31}\right)}^{2} + {\left(\textcolor{red}{- 209} - \textcolor{b l u e}{- 201}\right)}^{2}}$

$d = \sqrt{{\left(\textcolor{red}{28} - \textcolor{b l u e}{31}\right)}^{2} + {\left(\textcolor{red}{- 209} + \textcolor{b l u e}{201}\right)}^{2}}$

$d = \sqrt{{\left(- 3\right)}^{2} + {\left(- 8\right)}^{2}}$

$d = \sqrt{9 + 64}$

$d = \sqrt{73}$

Or

$d = 8.544$ rounded to the nearest thousandth.

Jul 23, 2017

color(blue)(8.544

#### Explanation:

$\therefore y - y = \left(- 201\right) - \left(- 209\right) = 8 = o p p o s i t e$

$\therefore x - x = 31 - 28 = 3 = a \mathrm{dj} a c e n t$

:.8/3=tantheta=2.666666667=69°26'38''

hypotenuse=distance

Distance$\therefore = \sec \theta \times 3$

Distance:.=sec69°26'38'' xx 3

Distance$\therefore = 2.848001248 \times 3 = 8.544003745$

:.color(blue)(=8.544 to 3 decimals