# How do you find a possible value for a if the points (-2,a), (6,1)has a distance of d=4sqrt5?

Dec 31, 2017

$a = - 3 \text{ or } a = 5$

#### Explanation:

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

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

$\text{let "(x_1,y_1)=(6,1)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 2 , a\right)$

$d = \sqrt{{\left(- 2 - 6\right)}^{2} + {\left(a - 1\right)}^{2}} = 4 \sqrt{5}$

$\Rightarrow \sqrt{64 + {\left(a - 1\right)}^{2}} = 4 \sqrt{5}$

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

$\Rightarrow 64 + {\left(a - 1\right)}^{2} = 80$

$\text{subtract 64 from both sides}$

${\left(a - 1\right)}^{2} = 16$

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

$\Rightarrow a - 1 = \pm \sqrt{16} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow a = 1 \pm 4$

$\Rightarrow a = 1 - 4 = - 3 \text{ or } a = 1 + 4 = 5$

Dec 31, 2017

Possible values of $a$ are $- 3 \mathmr{and} 5$.

#### Explanation:

Distance between two points $\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$ is

D= sqrt((x_1-x_2)^2+(y_1-y_2)^2)) Therefore,

Distance between two points $\left(- 2 , a\right) \mathmr{and} \left(6 , 1\right)$ is

$D = \sqrt{{\left(- 2 - 6\right)}^{2} + {\left(a - 1\right)}^{2}} = 4 \sqrt{5}$ , Squaring both

sides we get , $64 + {a}^{2} - 2 a + 1 = 80$ or

${a}^{2} - 2 a + 1 + 64 - 80 = 0$ or

${a}^{2} - 2 a - 15 = 0 \mathmr{and} {a}^{2} - 5 a + 3 a - 15 = 0$ or

$a \left(a - 5\right) + 3 \left(a - 5\right) = 0 \therefore \left(a + 3\right) \left(a - 5\right) = 0$

$\therefore a = - 3 , a = 5$

Possible values of $a$ are $- 3 \mathmr{and} 5$ [Ans]