# How do you find a possible value for a if the points (4,a), (8,4) has a distance of d=2sqrt5?

Apr 27, 2017

a = 6 or 2

#### Explanation:

We know the distance between two points = $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$ where $\left({x}_{1} , {y}_{1}\right) = \left(4 , a\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right) = \left(8 , 4\right)$

Hence $\sqrt{{\left(8 - 4\right)}^{2} + \left(4 - a\right)} = 2 \sqrt{5}$ [as per question]

or, ${\left(8 - 4\right)}^{2} + {\left(4 - a\right)}^{2} = {\left[2 \sqrt{5}\right]}^{2}$ [ squiring both sides]

$\Rightarrow {4}^{2} + 16 - 8 a + {a}^{2} = 20$

$\Rightarrow {a}^{2} - 8 a + 16 + 16 - 20 = 0$

$\Rightarrow {a}^{2} - 8 a + 12 = 0$

$\Rightarrow {a}^{2} - 2 a - 6 a + 12 = 0$

$\Rightarrow a \left(a - 2\right) - 6 \left(a - 2\right) = 0$

$\Rightarrow \left(a - 6\right) \left(a - 2\right) = 0$

$\Rightarrow \left(a - 6\right) = 0 , \left(a - 2\right) = 0$

$\Rightarrow a = 6 \mathmr{and} 2$