Question 188716
Remember, the distance between any two points *[Tex \LARGE \left(x_{1},y_{1}\right)] and *[Tex \LARGE \left(x_{2},y_{2}\right)] is


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



In this case, *[Tex \LARGE \left(x_{1},y_{1}\right)] is the point (x,1) and *[Tex \LARGE \left(x_{2},y_{2}\right)] is the point (2,6)




Now matching corresponding coordinates, we get {{{x[1]=x}}}, {{{y[1]=1}}}, {{{x[2]=2}}}, and {{{y[2]=6}}}




{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} Start with the distance formula



{{{5=sqrt((x-2)^2+(1-6)^2)}}} Plug in {{{d=5}}} (the given distance) {{{x[1]=x}}}, {{{y[1]=1}}}, {{{x[2]=2}}}, and {{{y[2]=6}}} (the given coordinates)



The goal now is to solve for "x":



{{{5=sqrt((x-2)^2+(-5)^2)}}} Combine like terms.



{{{5=sqrt((x-2)^2+25)}}} Square -5 to get 25



{{{5^2=(x-2)^2+25}}} Square both sides (to eliminate the square root)



{{{25=(x-2)^2+25}}} Square 5 to get 25



{{{25-25=(x-2)^2}}} Subtract 25 from both sides.



{{{0=(x-2)^2}}} Combine like terms.



{{{(x-2)^2=0}}} Rearrange the equation



{{{x-2=0+-sqrt(0)}}} Take the square root of both sides.



{{{x-2=sqrt(0)}}} or {{{x-2=-sqrt(0)}}} Break up the "plus/minus" to form two equations.



{{{x-2=0}}} or {{{x-2=0}}}  Take the square root of {{{0}}} to get {{{0}}}.



{{{x=2+0}}} or {{{x=2-0}}} Add {{{2}}} to both sides.



{{{x=2}}} or {{{x=2}}} Combine like terms.



{{{x=2}}} Remove the duplicate solution



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Answer:



So the only solution is {{{x=2}}}.




So the point (2,1) is 5 units from (2,6)



Note: if you plot the points on a coordinate plane and count the number of units one point is from another, you'll find that the two points are 5 units apart.