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