Question 742639
For convenience we work with d^2 instead of d



{{{
d^2=(x[2]-x[1])^2+(y[2]-y[1])^2
}}}


becomes:

{{{
5^2=(5-(a+2))^2+(-1-3)^2
}}}



then


{{{
25=(3-a)^2+(-4)^2
}}}

Notice that after we square a real quantity the result is positive so
(3-a)^2=(a-3)^2

and so we rewrite it as this:



{{{(a-3)^2=25-16}}}



{{{(a-3)^2=9}}}



{{{abs(a-3)=3}}}


this means  a-3=3  or a-3=-3


and so a=6 or a=0


checking...

(5-(0+2))^2+(-1-3)^2=3^2+4^2

and so 0 works


(5-(6+2))^2+(-1-3)^2=(-3)^2+(-4)^2

and so does 6



:)