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Popular >

distance (3,-2),(13,10)

Solution

distance

(3, - 2), (13, 10)

Solution

261

Solution steps

The distance between

(x_{1}, y_{1}), (x_{2}, y_{2})

is:

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

The distance between

(3, - 2)

and

(13, 10)

is

= (13 - 3)^{2} + (10 - (- 2))^{2}

Simplify

= 261

Popular Examples

perpendicular y=3x-2,(-1,3)perpendicular

y = 3 x - 2, (- 1, 3)

midpoint (-8,2),(-8+4sqrt(3),6)midpoint

(- 8, 2), (- 8 + 43, 6)

range of (x(2x^2-3x+1))/(x^3+1)range

\frac{x ( 2 x ^{2} - 3 x + 1 )}{x ^{3} + 1}

line y=x+1 line

y = x + 1

inverse of h(x)=\sqrt[3]{x-2}+3 inverse

Frequently Asked Questions (FAQ)

What is the distance (3,-2),(13,10) ?
The distance (3,-2),(13,10) is 2sqrt(61)