Question 91263
"Find the distance between (7,-2) and (-5,3)"

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(7,-2\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(-5,3\right)]

{{{d=sqrt((7--5)^2+(-2-3)^2)}}} Plug in {{{x[1]=7}}}, {{{x[2]=-5}}}, {{{y[1]=-2}}}, {{{y[2]=3}}}

{{{d=sqrt((12)^2+(-5)^2)}}} Evaluate {{{7--5}}} to get 12. Evaluate {{{-2-3}}} to get -5.

{{{d=sqrt(144+25)}}} Square each value

{{{d=sqrt(169)}}} Add

{{{d=13}}} Simplify the square root

So the distance between (7,-2) and (-5,3) is 13 units

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"Find the distance between (-3,-2) and (1,4)"

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(-3,-2\right)] and *[Tex \Large \left(x_{2}, y_{2}\right)] is the second point *[Tex \Large \left(1,4\right)]

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

{{{d=sqrt((-4)^2+(-6)^2)}}} Evaluate {{{-3-1}}} to get -4. Evaluate {{{-2-4}}} to get -6.

{{{d=sqrt(16+36)}}} Square each value

{{{d=sqrt(52)}}} Add

{{{d=2*sqrt(13)}}} 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 approximates to

{{{d=7.21110255092798}}}

which rounds to

7.21

So the distance between (-3,-2) and (1,4) is approximately 7.21 units