What is the distance between $(5, –1)$ and $(3,7)$ ?

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Answer 1 · 2016-02-10T21:36:05

Use the distance formula: $d = sqrt((y_2-y_1)^2+(x_2-x_1)^2)$

This yields a distance of $sqrt 68$ units.

Use $d = sqrt((y_2-y_1)^2+(x_2-x_1)^2)$

$= sqrt((7-(-1))^2 +(3-5)^2) = sqrt(64+4) = sqrt 68$

Answer 2 · 2016-02-10T21:39:52

$"distance exactly"=2sqrt(17)" "$
$"distance approximately"~= 8.25" to 2 decimal places"$

Tony B

Now consider them as forming a triangle:
Tony B

From this you can see that Pythagoras will give us the answer for the distance between the points.

Let distance be $d$ then

$d^2 = (y_2-y_1)^2+(x_2-x_1)^2$

so $d = sqrt((-8)^2+(2)^2$

so $" " d= sqrt(68) = sqrt(2^2xx17)$

$=2sqrt(17)$

What is the distance between (5, –1)(5, –1) and (3,7)(3,7)?