# How do you find the distance between (-sqrt2,-sqrt2), (sqrt2, 6sqrt2)?

Jun 13, 2017

abs((-sqrt2,-sqrt2):(sqrt2,6sqrt2))=color(red)(sqrt(106)

#### Explanation:

The easiest way to do this is to observe that the given points are equivalent to $\left(- 1 , - 1\right)$ and $\left(1 , 6\right)$ scaled by a factor of $\sqrt{2}$

The distance between the given points will be the same as the distance between $\left(- 1 , - 1\right)$ and $\left(1 , 6\right)$ multiplied by $\sqrt{2}$

The horizontal distance between $\left(- 1 , - 1\right)$ and $\left(1 , 6\right)$ is $1 - \left(- 1\right) = 2$

The vertical distance between $\left(- 1 , - 1\right)$ and $\left(1 , 6\right)$ is $6 - \left(- 1\right) = 7$

The diagonal distance between $\left(- 1 , - 1\right)$ and $\left(1 , 6\right)$ is given by the Pythagorean Theorem as
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{2}^{2} + {7}^{2}} = \sqrt{4 + 49} = \sqrt{53}$

and the corresponding diagonal distance between $\left(- \sqrt{2} , - \sqrt{2}\right)$ and $\left(\sqrt{2} , 6 \sqrt{2}\right)$ is
$\textcolor{w h i t e}{\text{XXX}} \sqrt{2} d = \sqrt{2} \cdot \sqrt{53} = \sqrt{106}$