Math Problem

emmaiskool242

Junior Member
Joined
Mar 29, 2006
Messages
66
I'm using the distance formula, its d=the square root of (x of 2-x of 1)2{squared} + ( y of 2 - y of 10 2{ squared} and I am supposed to plug in the points (3{=x of 1},3/7{=y of 1})(4{=x of 2},-2/7{=y of2}) and I'm supposed to figure out what D equals. I don't know if you can figure that out but please try and help me. ~THANKS
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, emmaiskool242!

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

I'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.
Exactly where is your difficulty?
\(\displaystyle \;\;\)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: \(\displaystyle \:D\:=\:\sqrt{(4 - 3)^2\,+\,\left(-\frac{2}{7}\,-\,\frac{3}{7}\right)^2} \:= \:\sqrt{1^2\,+\,\left(-\frac{5}{7}\right)^2}\:=\:\sqrt{1\,+\,\frac{25}{49}} \:=\:\sqrt{\frac{74}{49}}\)

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

emmaiskool242

Junior Member
Joined
Mar 29, 2006
Messages
66
Yes thats what I was saying and fractions just aren't my thing.....Thankyou for your help,and I will keep trying =)
 
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