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# Math Help - Distance Between Two Points

1. ## Distance Between Two Points

Use the distance formula to find the distance between the two points
(a, b) (b, a).

2. Originally Posted by sologuitar
Use the distance formula to find the distance between the two points
(a, b) (b, a).
$d = \sqrt{(b-a)^2 + (a-b)^2}$

all set up for you ... finish it.

3. $A(a,b)$ and $B(b,a)$
The distance between the two points A and B:
$|\vec{AB}|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}=\sqrt{(b-a)^2+(a-b)^2}=\sqrt{2}.|a-b|$
I don't know is it true or false ? So if it's false, please tell me !

4. Originally Posted by lovemath
$A(a,b)$ and $B(b,a)$
The distance between the two points A and B:
$|\vec{AB}|=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}=\sqrt{(b-a)^2+(a-b)^2}=\sqrt{2}.|a-b|$
I don't know is it true or false ? So if it's false, please tell me !
That is true. Notice $(a-b)^2=(b-a)^2$ and $\sqrt{(a-b)^2}=|a-b|=|b-a|$

5. ## lovemath

Username lovemath got the right answer. I tried getting the right answer over and over again but ended up with something totally different. How does the answer become sqrt{2}*|a - b|? In the radicand we have (a - b)^2 + (b - a)^2, which equals 2a^2 - 4ab +2b^2

6. Originally Posted by sologuitar
Username lovemath got the right answer. I tried getting the right answer over and over again but ended up with something totally different. How does the answer become sqrt{2}*|a - b|? In the radicand we have (a - b)^2 + (b - a)^2, which equals 2a^2 - 4ab +2b^2

Yes, and that is $2(a^2- 2ab+ b^2)= 2(a-b)^2$. $\sqrt{2(a-b)^2}= \sqrt{2}\sqrt{(a-b)^2}$ and $\sqrt{(a-b)^2}= |a-b|$

7. ## Yes but...

Originally Posted by HallsofIvy
Yes, and that is $2(a^2- 2ab+ b^2)= 2(a-b)^2$. $\sqrt{2(a-b)^2}= \sqrt{2}\sqrt{(a-b)^2}$ and $\sqrt{(a-b)^2}= |a-b|$
I understand everything you said here. However, I don't understand how the absolute value comes into play.

8. Originally Posted by sologuitar
I understand everything you said here. However, I don't understand how the absolute value comes into play.
graph $y = \sqrt{x^2}$ in your calculator ... what do you see?

9. Originally Posted by sologuitar
I understand everything you said here. However, I don't understand how the absolute value comes into play.
$\sqrt{2^2}= \sqrt{4}= 2$ $\sqrt{(-2)^2}= \sqrt{4}= 2$. Since $\sqrt{x}$ is defined as "the non-negative number a such that $a^2= x$, if x is itself non-negative, then $\sqrt{x^2}= x$ but if x is negative, $\sqrt{x^2}= -x$. In either case $\sqrt{x^2}= |x|$.

10. ## graphing

Originally Posted by skeeter
graph $y = \sqrt{x^2}$ in your calculator ... what do you see?
After graphing y = sqrt{x^2}, I got the line y = x, which crosses through the origin. What's your point?

11. Originally Posted by sologuitar
After graphing y = sqrt{x^2}, I got the line y = x, which crosses through the origin. What's your point?