# How do you find two values of x such that point (x, 5) is five units from the point (-1, 2)?

Mar 16, 2018

$x = 3$
$x = - 5$

#### Explanation:

The distance formula is
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
$5 = \sqrt{{\left(- 1 - x\right)}^{2} + {\left(2 - 5\right)}^{2}}$
$5 = \sqrt{{\left(x + 1\right)}^{2} + {\left(- 3\right)}^{2}}$
$5 = \sqrt{{\left(x + 1\right)}^{2} + 9}$
$25 = {\left(x + 1\right)}^{2} + 9$
$16 = {\left(x + 1\right)}^{2}$
$\pm 4 = x + 1$

$x = 3$
$x = - 5$

A faster way to do this is to plot the point $\left(- 1 , 2\right)$ and use a
3-4-5 right triangle to find x where y=5.