# How do you find a possible value for a if the points (a,0), (3,1) has a distance of d=sqrt2?

May 14, 2017

$a = 2$ or $a = 4$

#### Explanation:

The distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by $\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$.

Hence distance between $\left(a , 0\right)$ and $3 , 1$ is

$\sqrt{{\left(a - 3\right)}^{2} + {\left(0 - 1\right)}^{2}} = \sqrt{{a}^{2} - 6 a + 9 + 1} = \sqrt{{a}^{2} - 6 a + 10}$

as this is equal to $\sqrt{2}$, we have

${a}^{2} - 6 a + 10 = 2$ or ${a}^{2} - 6 a + 8 = 0$

or ${a}^{2} - 4 a - 2 a + 8 = 0$

i.e. $a \left(a - 4\right) - 2 \left(a - 4\right) = 0$

or $\left(a - 2\right) \left(a - 4\right) = 0$

i.e. $a = 2$ or $a = 4$

graph{((x-2)^2+y^2-0.01)((x-4)^2+y^2-0.01)((x-3)^2+(y-1)^2-0.01)=0 [-3.11, 6.89, -1.92, 3.08]}