Question 112073
#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(1,1\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(3,2\right)]


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


{{{d=sqrt((-2)^2+(-1)^2)}}} Evaluate {{{1-3}}} to get -2. Evaluate {{{1-2}}} to get -1. 


{{{d=sqrt(4+1)}}} Square each value


{{{d=sqrt(5)}}} Add


So the distance approximates to


{{{d=2.23606797749979}}}


which rounds to

2.2


So the distance between (1,1) and (3,2) is approximately 2.2 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(4,0\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(6,3\right)]


{{{d=sqrt((4-6)^2+(0-3)^2)}}} Plug in {{{x[1]=4}}}, {{{x[2]=6}}}, {{{y[1]=0}}}, {{{y[2]=3}}}


{{{d=sqrt((-2)^2+(-3)^2)}}} Evaluate {{{4-6}}} to get -2. Evaluate {{{0-3}}} to get -3. 


{{{d=sqrt(4+9)}}} Square each value


{{{d=sqrt(13)}}} Add


So the distance approximates to


{{{d=3.60555127546399}}}


which rounds to

3.6


So the distance between (4,0) and (6,3) is approximately 3.6 units



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#3

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(5,2\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(5,-2\right)]


{{{d=sqrt((5-5)^2+(2--2)^2)}}} Plug in {{{x[1]=5}}}, {{{x[2]=5}}}, {{{y[1]=2}}}, {{{y[2]=-2}}}


{{{d=sqrt((0)^2+(4)^2)}}} Evaluate {{{5-5}}} to get 0. Evaluate {{{2--2}}} to get 4. 


{{{d=sqrt(0+16)}}} Square each value


{{{d=sqrt(16)}}} Add


{{{d=4}}} 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 (5,2) and (5,-2) is 4 units


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#4


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(1,3\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(1,5\right)]


{{{d=sqrt((1-1)^2+(3-5)^2)}}} Plug in {{{x[1]=1}}}, {{{x[2]=1}}}, {{{y[1]=3}}}, {{{y[2]=5}}}


{{{d=sqrt((0)^2+(-2)^2)}}} Evaluate {{{1-1}}} to get 0. Evaluate {{{3-5}}} to get -2. 


{{{d=sqrt(0+4)}}} Square each value


{{{d=sqrt(4)}}} Add


{{{d=2}}} 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 (1,3) and (1,5) is 2 units