# What is the distance,in units, between (3, -5) and (8, 7) in the coordinate plane?

Jul 19, 2016

$13 u n i t$.

#### Explanation:

The Distance $A B$, btwn. pts $A \left({x}_{1} , {y}_{1}\right) \mathmr{and} B \left({x}_{2} , {y}_{2}\right)$ is

$A B = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

Hence, reqd. dist.

$= \sqrt{{\left(3 - 8\right)}^{2} + {\left(- 5 - 7\right)}^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 u n i t$.