# How do you find the distance between points (3,-3), (7,2)?

May 24, 2017

$d = \sqrt{41}$

#### Explanation:

Use the distance formula: $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
If we let $\left(3 , - 3\right) \to \left(\textcolor{b l u e}{{x}_{1}} , \textcolor{red}{{y}_{1}}\right)$ and $\left(7 , 2\right) \to \left(\textcolor{b l u e}{{x}_{2}} , \textcolor{red}{{y}_{2}}\right)$ then...

$d = \sqrt{{\left(\textcolor{b l u e}{7 - 3}\right)}^{2} + {\left(\textcolor{red}{2 - - 3}\right)}^{2}}$

$d = \sqrt{{\left(\textcolor{b l u e}{4}\right)}^{2} + {\left(\textcolor{red}{5}\right)}^{2}}$

d=sqrt((color(blue)(16))+(color(red)(25))

$d = \sqrt{41}$

May 24, 2017

The distance between points (3,-3), (7,2) is $d = \sqrt{41}$

That is $\approx 6.4 u n i t s$

#### Explanation:

Use the distance formula which is derived from the Pythagorean Theorem.

They did it here:
http://www.purplemath.com/modules/distform.htm

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

We have points (3,-3), (7,2):

Then: ${d}^{2} = {\left(3 - 7\right)}^{2} + {\left(- 3 - 2\right)}^{2}$

${d}^{2} = {\left(- 4\right)}^{2} + {\left(- 5\right)}^{2}$

See how squaring gets rid of those nasty negatives?

${d}^{2} = \left(16\right) + \left(25\right)$

Too bad we cannot take the roots of the two terms without adding.

${d}^{2} = 41$

$d = \sqrt{41} \approx 6.4$

To check, compare this $r i g h t \angle \triangle 4 , 5 , 6.4$ to a standard $r i g h t \angle \triangle 3 , 4 , 5$, and they appear to be similar.