How do you find the distance between (10,19), (-13,9)?

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Answer 1 · 2017-08-17T12:08:01

Use $a^{2} + b^{2} = c^{2}$ $a^2 + b^2 = c^2$

$c = 25.1$ $c = 25.1$

Let the change in x be $a^{2}$ $a^2$
Let the change in y be $b^{2}$ $b^2$

The the square root of $c^{2}$ $c^2$ = the distance

$a^{2} = {(19 - 9)}^{2}$ $a^2 = (19-9)^2$

$a^{2} = 100$ $a^2 = 100$

$b^{2} = {(10 - - 13)}^{2}$ $b^2 = ( 10- -13)^2$

$b^{2} = 529$ $b^2 = 529$

$100 + 529 = 629$ $100 + 529 = 629$

$629 = c^{2}$ $629 = c^2$

square root of $629 = c$ $629 = c$

$c = \sqrt{629}$ $c= sqrt629$

$c = 25.1$ $c= 25.1$

Answer 2 · 2017-08-17T13:10:44

$= 25.08$ $=25.08$

The formula for the distance between two points is based on the Theorem of Pythagoras.

$| A B | = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}})$ $|AB| = sqrt((x_2-x_1)^2 + (y_2-y_1)^2))$

We have the points $(10, 19) and (- 13, 9)$ $(10,19) and (-13,9)$

Distance = $\sqrt{{(10 - (- 13))}^{2} + {(19 - 9)}^{2}}$ $sqrt((10-(-13))^2 +(19-9)^2)$

$= \sqrt{23^{2} + 10^{2}}$ $= sqrt(23^2 +10^2)$

$= \sqrt{629}$ $=sqrt629$

$= 25.08$ $=25.08$