# How do you find the possible values for a if the points (-5,a), (3,1) has a distance of sqrt89?

Jun 24, 2018

$a = 6$ or $a = - 4$

#### Explanation:

The formula for distance of two Points is given by

$d \left({P}_{1} , {P}_{2}\right) = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
so we get

$\sqrt{{\left(- 5 - 3\right)}^{2} + {\left(a - 1\right)}^{2}} = \sqrt{89}$

squaring we get

$64 + {a}^{2} - 2 a + 1 = 89$
combining like Terms

${a}^{2} - 2 a - 24 = 0$

using the quadratic formula we get

${a}_{1 , 2} = 1 + \pm \sqrt{25}$

so
${a}_{1} = 6$
${a}_{2} = - 4$

Jun 24, 2018

$a = - 4 \text{ or } a = 6$

#### Explanation:

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

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

$\text{let "(x_1,y_1)=(3,1)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 5 , a\right)$

$d = \sqrt{{\left(- 5 - 3\right)}^{2} + {\left(a - 1\right)}^{2}} = \sqrt{89}$

$\sqrt{64 + {\left(a - 1\right)}^{2}} = \sqrt{89}$

$\textcolor{b l u e}{\text{square both sides}}$

$64 + {\left(a - 1\right)}^{2} = 89$

$\text{subtract 64 from both sides}$

${\left(a - 1\right)}^{2} = 25$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(a - 1\right)}^{2}} = \pm \sqrt{25} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$a - 1 = \pm 5$

$\text{add 1 to both sides}$

$a = 1 \pm 5$

$a = 1 - 5 = - 4 \text{ or } a = 1 + 5 = 6$