Coordinate Geometry- distance between two points

[zebra1707]

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

 

[Mentallic]

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 \sqrt{2b^2+2a^2-4ba}=\sqrt{2}(a-b)

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

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

Hence, \sqrt{b^2+a^2-2ba}=a-b

Can you see how this is possible? What must b^2+a^2-2ba be equivalent to such that when you take the square root of it, it is equal to a-b?

 

[eumyang]

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
\sqrt{2b^2+2a^2-4ba} = \sqrt{2(a^2 - 2ab + b^2)}
Can you take it from here?

 

[zebra1707]

Many thanks to all respondants, I appreciate all your help with this one.

Cheers

 

[Mentallic]

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)

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

d=\sqrt{(b-a)^2+(a-b)^2}

Since x^2=(-x)^2

d=\sqrt{(-(b-a))^2+(a-b)^2}=\sqrt{(a-b)^2+(a-b)^2}=\sqrt{2(a-b)^2}=\sqrt{2}(a-b)