Question 119510
#1



Start with the given distance formula

{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} where *[Tex \Large \left(x_{1}, y_{1}\right)] is the first point *[Tex \Large \left(0,8\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(0,-4\right)]


{{{d=sqrt((0-0)^2+(8--4)^2)}}} Plug in {{{x[1]=0}}}, {{{x[2]=0}}}, {{{y[1]=8}}}, {{{y[2]=-4}}}


{{{d=sqrt((0)^2+(12)^2)}}} Evaluate {{{0-0}}} to get 0. Evaluate {{{8--4}}} to get 12. 


{{{d=sqrt(0+144)}}} Square each value


{{{d=sqrt(144)}}} Add


{{{d=12}}} Simplify the square root  (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)


So the distance between (0,8) and (0,-4) is 12 units



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#2


Start with the given distance formula

{{{d=sqrt((x[1]-x[2])^2+(y[1]-y[2])^2)}}} where *[Tex \Large \left(x_{1}, y_{1}\right)] is the first point *[Tex \Large \left(2,6\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(-3,4\right)]


{{{d=sqrt((2--3)^2+(6-4)^2)}}} Plug in {{{x[1]=2}}}, {{{x[2]=-3}}}, {{{y[1]=6}}}, {{{y[2]=4}}}


{{{d=sqrt((5)^2+(2)^2)}}} Evaluate {{{2--3}}} to get 5. Evaluate {{{6-4}}} to get 2. 


{{{d=sqrt(25+4)}}} Square each value


{{{d=sqrt(29)}}} Add


So the distance approximates to


{{{d=5.3851648071345}}}


which rounds to

5.39


So the distance between (2,6) and (-3,4) is approximately 5.39 units