Question 981021
The first point is ({{{x[1]}}},{{{y[1]}}}). The second point is ({{{x[2]}}},{{{y[2]}}}).


Since the first point is ({{{x}}},{{{1}}}), we can say ({{{x[1]}}},{{{y[1]}}}) = ({{{x}}},{{{1}}})
So {{{x[1] =x}}}, {{{y[1]=1}}}


Since the second point is ({{{-2}}},{{{5}}}), we can also say ({{{x[2]}}},{{{y[2]}}}) = ({{{-2}}},{{{5}}})
So {{{x[2]=-2}}}, {{{y[2]=5}}}.

Now use the distance formula:


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

since given {{{d=2sqrt(7)}}}, {{{x[1] =x}}},{{{y[1]=1}}},{{{x[2]=-2}}}, and {{{y[2]=5}}}, substitute these values 


{{{2sqrt(7)=sqrt((x-(-2))^2+(1-5)^2)}}}

{{{2sqrt(7)=sqrt((x+2)^2+(-4)^2)}}}

{{{2sqrt(7)=sqrt((x+2)^2+16)}}}

{{{2sqrt(7)=sqrt(x^2+4x+4+16)}}}.......square both sides

{{{(2sqrt(7))^2=(sqrt(x^2+4x+4+16))^2}}}

{{{4*7=x^2+4x+20}}}

{{{28=x^2+4x+20}}}

{{{0=x^2+4x+20-28}}}

{{{0=x^2+4x-8}}}...........use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

{{{x = (-4 +- sqrt( 4^2-4*1*(-8) ))/(2*1) }}}

{{{x = (-4 +- sqrt( 16+32 ))/2 }}}

{{{x = (-4 +- sqrt( 48 ))/2 }}}

{{{x = (-4 +- sqrt( 4^2*3 ))/2 }}}

{{{x = (-cross(4)2 +- cross(4)2sqrt( 3 ))/cross(2) }}}

{{{x = (-2 +- 2sqrt( 3 )) }}}

solutions:

{{{x = -2 + 2sqrt( 3 ) }}}
{{{x = -2(1 -sqrt( 3 )) }}}

or

{{{x = -2 - 2sqrt( 3 ) }}}
{{{x = -2(1 +sqrt( 3 )) }}}


so, the first point is ({{{x}}},{{{1}}}) could be:  

({{{-2(1 -sqrt( 3 ))}}},{{{1}}}) 
or 
({{{-2(1+sqrt(3 ))}}},{{{1}}})