Question 1137813


given: The point ({{{x}}},{{{-1}}})  is {{{13}}} units from the point ({{{3}}},{{{11}}}). 

find: What are the possible values of {{{x}}}? 



use formula for  distance between two points

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

=> given distance {{{d=13}}}, points ({{{x}}},{{{-1}}}) and ({{{3}}},{{{11}}})


{{{13=sqrt((x-3)^2+(-1-11)^2)}}}

 {{{13=sqrt((x-3)^2+(-12)^2)}}}

{{{13=sqrt((x-3)^2+144)}}}........square both sides

{{{13^2=(sqrt((x-3)^2+144))^2}}}

{{{169=(x-3)^2+144}}}

{{{169-144=(x^2-6x+9)}}}

{{{25=x^2-6x+9}}}

{{{0=x^2-6x+9-25}}}

{{{x^2-6x-16=0}}}...factor

{{{x^2+2x-8x-16=0}}}

{{{(x^2+2x)-(8x+16)=0}}}

{{{x(x+2)-8(x+2)=0}}}

{{{(x-8)(x+2)=0}}}

solutions:

{{{x=8}}} or {{{x=-2}}}-> the possible values of {{{x}}}


so, your point are: ({{{8}}},{{{-1}}}) and ({{{-2}}},{{{-1}}})


check the distance between:

({{{8}}},{{{-1}}}) and ({{{3}}},{{{11}}})

*[invoke formula_distance 8, -1, 3, 11]

({{{-2}}},{{{-1}}}) and ({{{3}}},{{{11}}})

*[invoke formula_distance -2, -1, 3, 11]