#3**+5 **

I think what heureka is trying to show is that the distance forumula, which looks really horrible, is really just Pythagoras' Theorum!

If you draw it the right angled triangle here so that the interval is the hypotenuse then

the two short sides are 5units and 5 units.

The hypotenuse which is the desired length is $$\sqrt{5^2+5^2}$$ etc.

Melody
Jul 3, 2014

#1**+5 **

If you have graph paper the easiest way to see this is to plot the points and connect them to make a right triangle. One side will be 5 long because 2-(-3)=5 and the other will be 5 as well since 0-(-5)=5. Now we can use the pythagorean equation to find the hypotenuse which is equivalent to the distance between the points.

5^2+5^2=h^2

50=h^2

sqrt(50)=h

5sqrt(2)=h If you desire a decimal approximation of this it can easily be typed into any calculator

jboy314
Jul 2, 2014

#2**+5 **

distance between points (2,-5) and (-3,0) ?

$$\begin{array}{crcrccr}

\mbox{Point 1:} \; &(x_1&=&-3,& y_1&=&0)\\

\mbox{Point 2:} \; &(x_2&=&2,& y_2&=&-5)

\end{array}$$

$$\begin{array}{ccl}

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

distance&=&\sqrt{[2-(-3)]^2+[(-5)-0]^2}\\

&=&\sqrt{(2+3)^2+(-5)^2}\\

&=&\sqrt{(5)^2+(-5)^2}\\

&=&\sqrt{25+25}\\

&=&\sqrt{2*25}\\

&=&\sqrt{25*2}\\

&=&\sqrt{25}\sqrt{2}\\

distance&=&5\sqrt{2}\\

\end{array}$$

heureka
Jul 3, 2014

#3**+5 **

Best Answer

I think what heureka is trying to show is that the distance forumula, which looks really horrible, is really just Pythagoras' Theorum!

If you draw it the right angled triangle here so that the interval is the hypotenuse then

the two short sides are 5units and 5 units.

The hypotenuse which is the desired length is $$\sqrt{5^2+5^2}$$ etc.

Melody
Jul 3, 2014