#### emmaiskool242

##### Junior Member

- Joined
- Mar 29, 2006

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- Thread starter emmaiskool242
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- Joined
- Mar 29, 2006

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I think I figured out what you're saying . . .

ExactlyI'm using the distance formula: \(\displaystyle \,D\:=\:\sqrt{(x_2-x_1)^2\,+\,(y_2-y_1)^2}\)

I am supposed to plug in the points \(\displaystyle \left(3,\,\frac{3}{7}\right)\) and \(\displaystyle \left(4,-\frac{2}{7}\right)\)

and I'm supposed to figure out what \(\displaystyle D\) equals.

\(\displaystyle \;\;\)You don't understand where the numbers go?

\(\displaystyle \;\;\)You can't handle fractions?

\(\displaystyle \;\;\)A square root gives you brain-freeze?

Try to "read" the Distance Formula.

Under the square root, it says:

\(\displaystyle \;\;\)Subtract the two x-values ... and square.

\(\displaystyle \;\;\)Subtract the two y-values ... and square.

\(\displaystyle \;\;\)Add the two quantities

\(\displaystyle \;\;\)Take the square root.

We have: \(\displaystyle \\:=\:\sqrt{(4 - 3)^2\,+\,\left(-\frac{2}{7}\,-\,\frac{3}{7}\right)^2} \:= \:\sqrt{1^2\,+\,\left(-\frac{5}{7}\right)^2}\:=\:\sqrt{1\,+\,\frac{25}{49}} \:=\:\sqrt{\frac{74}{49}}\)

Therefore: \(\displaystyle \,D\:=\:\frac{\sqrt{74}}{\sqrt{49}}\:=\:\frac{\sqrt{74}}{7}\)

- Joined
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