Coordinate Geometry- distance between two points

  • Thread starter zebra1707
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  • #1
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Hi

Homework Statement



Two points X(a,b) and Y(b,a)

Prove that the distance = sqrt2 (a-b)


Homework Equations



d = sqrt(x2-x1)^2 + (y2-y1)^2)

The Attempt at a Solution



I have drilled it down to

(sqrt 2b^2+2a^2-4ba) but unable to drill it down to sqrt2 (a-b)


Help appreciated - many thanks
 

Answers and Replies

  • #2
Mentallic
Homework Helper
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Hi

Homework Statement



Two points X(a,b) and Y(b,a)

Prove that the distance = sqrt2 (a-b)


Homework Equations



d = sqrt(x2-x1)^2 + (y2-y1)^2)

The Attempt at a Solution



I have drilled it down to

(sqrt 2b^2+2a^2-4ba) but unable to drill it down to sqrt2 (a-b)


Help appreciated - many thanks
Well if [tex]\sqrt{2b^2+2a^2-4ba}=\sqrt{2}(a-b)[/tex]

Then dividing through by [tex]\sqrt{2}[/tex] gives [tex]\frac{\sqrt{2(b^2+a^2-2ba)}}{\sqrt{2}}=\frac{\sqrt{2}(a-b)}{\sqrt{2}}[/tex]

And remember that [tex]\sqrt{ab}=\sqrt{a}\sqrt{b}[/tex] so we have [tex]\frac{\sqrt{2}\sqrt{b^2+a^2-2ba}}{\sqrt{2}}=)=(a-b)[/tex]

Hence, [tex]\sqrt{b^2+a^2-2ba}=a-b[/tex]

Can you see how this is possible? What must [tex]b^2+a^2-2ba[/tex] be equivalent to such that when you take the square root of it, it is equal to a-b?
 
  • #3
eumyang
Homework Helper
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Hi

Homework Statement



Two points X(a,b) and Y(b,a)

Prove that the distance = sqrt2 (a-b)

Homework Equations



d = sqrt(x2-x1)^2 + (y2-y1)^2)
Another relevant equation would be
x2 - 2xy + y2 = (x - y)2
(This is one of those perfect square trinomial formulas from Algebra I.)

Note that
[tex]\sqrt{2b^2+2a^2-4ba} = \sqrt{2(a^2 - 2ab + b^2)}[/tex]
Can you take it from here?
 
  • #4
107
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Many thanks to all respondants, I appreciate all your help with this one.

Cheers
 
  • #5
Mentallic
Homework Helper
3,798
94
Well since you found the answer already, just in case you're curious this is how you should have worked on the answer:

Two points, X(a,b) and Y(b,a)

[tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

[tex]d=\sqrt{(b-a)^2+(a-b)^2}[/tex]

Since [tex]x^2=(-x)^2[/tex]

[tex]d=\sqrt{(-(b-a))^2+(a-b)^2}=\sqrt{(a-b)^2+(a-b)^2}=\sqrt{2(a-b)^2}=\sqrt{2}(a-b)[/tex]
 

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