# How do you find the distance and midpoint between the two points. (4, -6) (-2, 8)?

Oct 16, 2015

$d = 2 \sqrt{58}$
$M = \left(1 , 1\right)$

#### Explanation:

To find the distance we just apply Pythagoras. Think of it this way:

The difference between the $x$ points causes a straight horizontal line, the difference between the $y$ points causes a straight vertical line, so the distance between the two points is the hypotenuse, or

${d}^{2} = \Delta {x}^{2} + \Delta {y}^{2}$

$d = \sqrt{\Delta {x}^{2} + \Delta {y}^{2}}$

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

$d = \sqrt{36 + 196}$

$d = \sqrt{232}$

$d = \sqrt{58 \cdot 4} = 2 \sqrt{58}$

The midpoint between two points, $M$, is literally just the average between the $x$ values and the average between the $y$ values, or

$M = \left(\overline{x} , \overline{y}\right)$

We have that

$\overline{x} = \frac{4 - 2}{2} = \frac{2}{2} = 1$

And that

$\overline{y} = \frac{8 - 6}{2} = \frac{2}{2} = 1$