# Coordinate Geometry- distance between two points

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
Homework Helper
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
Homework Helper
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?

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

Cheers

Mentallic
Homework Helper
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 $$x^2=(-x)^2$$

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