Question

How does the distance formula ensure that the distance between two different points is positive?

Answer 1

Answer:

$(x_{2}-x_{1} ) ^{2} \geq 0 , always\\(y_{2}-y_{1} ) ^{2} \geq 0 , always\\so, \\\sqrt{(x_{2}-x_{1} ) ^{2} +(y_{2}-y\\_{1} ) ^{2} } \geq 0, always$

Step-by-step explanation:

Hello, I can help you with this. I like this kind of questions.

Step one

Let's remember the formula

Let

$P1(x_{1}, y_{1})\\P2(x_{2}, y_{2})$

the distance between P1 and P2 is given by:

$Distance=\sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1})^{2} }$

as you can see, the terms are powered to 2, so it does not matter the sign, it always will be  positive

$(x_{2}-x_{1} ) ^{2} \geq 0 , always\\(y_{2}-y_{1} ) ^{2} \geq 0 , always\\so, \\\sqrt{(x_{2}-x_{1} ) ^{2} +(y_{2}-y\\_{1} ) ^{2} } \geq 0, always$

I really hope it helps, have a nice day

Answer 2

Since the distance formula is square rooted, one of the number one things that will never ever happen in math... the square root a negative number. Try it in your calculator it will say ERROR.