# What is the distance between (1,-4) and (7,5)?

May 18, 2018

$3 \sqrt{13}$ or 10.81665383

#### Explanation:

make a right angle triangle with the two points being the end points of the hypotenuse.

The distance between the $x$ values is 7-1=6

The distance between the $y$ values is 5- -4=5+4=9

So our triangle has two shorter sides 6 and 9 and we need to find the length of the hypotenuse, use Pythagoras.

${6}^{2} + {9}^{2} = {h}^{2}$

$36 + 81 + 117$

$h = \sqrt{117} = 3 \sqrt{13}$

May 18, 2018

$\sqrt{117} \approx 10.82 \text{ to 2 dec. places}$

#### Explanation:

$\text{calculate the distance d using the "color(blue)"distance formula}$

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(1,-4)" and } \left({x}_{2} , {y}_{2}\right) = \left(7 , 5\right)$

$d = \sqrt{{\left(7 - 1\right)}^{2} + {\left(5 - \left(- 4\right)\right)}^{2}}$

$\textcolor{w h i t e}{d} = \sqrt{{6}^{2} + {9}^{2}} = \sqrt{36 + 81} = \sqrt{117} \approx 10.82$

May 18, 2018

$\sqrt[]{117}$

#### Explanation:

If you were to draw a right triangle so that the hypotenuse is the line between $\left(1 , - 4\right)$ and $\left(7 , 5\right)$, you would observe that the two legs of the triangle would be of length $6$ (i.e. the distance between $x = 7$ and $x = 1$) and $9$ (i.e. the distance between $y = 5$ and $y = - 4$). By applying the pythagorean theorem,

${a}^{2} + {b}^{2} = {c}^{2}$,

where $a$ and $b$ are the lengths of the legs of a right triangle and $c$ is the length of the hypotenuse, we obtain:

${6}^{2} + {9}^{2} = {c}^{2}$.

Solving for the length of the hypotenuse (i.e. the distance between the points $\left(1 , - 4\right)$ and $\left(7 , 5\right)$), we get:

$c = \sqrt[]{117}$.

The process of finding the distance between two points by use of a right triangle can be formulated thusly:

Distance = root()((x_2−x_1)^2+(y_2−y_1)^2).

This is called the distance formula, and can be used to expedite the solving of this sort of problem.