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

Feb 10, 2016

Use the distance formula: $d = \sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}}$

This yields a distance of $\sqrt{68}$ units.

Explanation:

Use $d = \sqrt{{\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}}$

$= \sqrt{{\left(7 - \left(- 1\right)\right)}^{2} + {\left(3 - 5\right)}^{2}} = \sqrt{64 + 4} = \sqrt{68}$

Feb 10, 2016

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

Explanation:

Now consider them as forming a triangle:

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

Let distance be $d$ then

${d}^{2} = {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({x}_{2} - {x}_{1}\right)}^{2}$

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

so$\text{ } d = \sqrt{68} = \sqrt{{2}^{2} \times 17}$

$= 2 \sqrt{17}$