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

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Answer 1 · 2017-05-14T08:11:46

$a=2$ or $a=4$

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $sqrt((x_2-x_1)^2+(y_2-y_1)^2)$ .

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

$sqrt((a-3)^2+(0-1)^2)=sqrt(a^2-6a+9+1)=sqrt(a^2-6a+10)$

as this is equal to $sqrt2$ , we have

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

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

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

or $(a-2)(a-4)=0$

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]}

How do you find a possible value for a if the points (a,0), (3,1) has a distance of d=sqrt2d=sqrt2?