Respuesta :

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.

Answer:

[tex](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[/tex]

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

[tex]P1(x_{1}, y_{1})\\P2(x_{2}, y_{2})\\[/tex]

the distance between P1 and P2 is given by:

[tex]Distance=\sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1})^{2} }[/tex]

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

[tex](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[/tex]

I really hope it helps, have a nice day

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