_{2 }- X

_{1})

^{2}+ (Y

_{2}- Y

_{1})

^{2}]

_{1}= 8, Y

_{1}= 4, X

_{2}= 5, Y

_{2}= -2

^{2}+ (-2 - 4)

^{2}]

_{ }

^{2}+ (-6)

^{2}]

on a standard coordinate plane, what is the distance between the two points(8,4) (5,-2)

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The distance of two points is: d = √[(X_{2 }- X_{1})^{2} + (Y_{2} - Y_{1})^{2}]

So, for your problem, X_{1} = 8, Y_{1} = 4, X_{2} = 5, Y_{2} = -2

Just plug those numbers into the formula:

d = √[(5 - 8)^{2} + (-2 - 4)^{2}]_{
}

d = √[(-3)^{2} + (-6)^{2}]

d = √[9 + 36]

d = √[45]

d = 3√[5]

OR

d = 6.708

Use the distance formula

x_{2} = 8

x_{1} = 5

y_{2} = 4

y_{1} = -2

Sqrt( (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} )

Sqrt( (8 - 5 )^{2} + (4 - (-2))^{2} )

Sqrt( 3^{2} + 6^{2 })

sqrt (9 + 36)

sqrt(45)

sqrt(9) * sqrt(5)

3 sqrt(5) or **3√5**

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