# What is the distance between (15,-4) and (7,5)?

Jun 27, 2017

See a solution process below:

#### Explanation:

The formula for calculating the distance between two points is:

$d = \sqrt{{\left(\textcolor{red}{{x}_{2}} - \textcolor{b l u e}{{x}_{1}}\right)}^{2} + {\left(\textcolor{red}{{y}_{2}} - \textcolor{b l u e}{{y}_{1}}\right)}^{2}}$

Substituting the values from the points in the problem gives:

$d = \sqrt{{\left(\textcolor{red}{7} - \textcolor{b l u e}{15}\right)}^{2} + {\left(\textcolor{red}{5} - \textcolor{b l u e}{- 4}\right)}^{2}}$

$d = \sqrt{{\left(\textcolor{red}{7} - \textcolor{b l u e}{15}\right)}^{2} + {\left(\textcolor{red}{5} + \textcolor{b l u e}{4}\right)}^{2}}$

$d = \sqrt{{\left(- 8\right)}^{2} + {9}^{2}}$

$d = \sqrt{64 + 81}$

$d = \sqrt{145}$

Or

$d = 12.042$ rounded to the nearest thousandth.

Jun 27, 2017

It might not seem like it, but this question just invooves simple Pythagorus on a graph. Instead of getting the two lengths of the known sides, it has to be worked out by finding the length.

However, this is super easy, just fin the change in $x$ and the change in $y$.

To get from 15 $\to$ 7 we go back by 8, however, we are talking about length, so we take it as $\left\mid - 8 \right\mid = 8$, and not $- 8$. Pur horizontal side has a length of 8.

To get from -4 $\to$ 5 we go up by 9. This will give us a verticle length of 9.

Now we have a right-angled triangle of lengths 8, 9, and $h$, $h$ being the hypotenuse (longest side) of the triangle.

To find the length of $h$, we use ${a}^{2} = {b}^{2} + {c}^{2}$, where #a=sqrt(b^2+c^2)

We add our values in to get $h = \sqrt{{8}^{2} + {9}^{2}} = \sqrt{64 + 81} = \sqrt{145} = 12.0415946 \approx 12.0$