# How can the distance formula be derived from the pythagorean theorem?

##### 1 Answer

#### Answer:

Let's see.

#### Explanation:

I have drawn a graph in which there are two points

#color(red)(p_1(x_1,y_1))" and "color(red)(p_2(x_2,y_2)# .

- We can easily say that

#" "bar(OD)=x_1" ; "bar(OE)=x_2" ; "bar(AD)=y_1" ; "bar(EB)=y_2# We also have a rectangle

#square OCED# . So,#color(red)(bar(AC)=bar(DE)) " and "color(red)(bar(AD)=bar(CE)# Now,

#bar(AC)=bar(DE)=bar(OE)-bar(OD)=(x_2 -x_1)#

#bar(BC)=bar(BE)-bar(CE)=bar(BE)-bar(AD)=(y_2-y_1)# With the help of

Pythagorean theorem,

#bar(AB)^2=bar(BC)^2+bar(AC)^2#

#bar(AB)^2=(x_2-x_1)^2+(y_2-y_1)^2#

#bar(AB)=sqrt((x_2-x_1)^2+(y_2-y_1)^2#

N.B:- As it is a square value , you may take#(x_1-x_2)# or,#(x_2-x_1)# . I mean you have to take difference.That's#(x_1~x_2)#

So, the required formula is proved that

If the distance between two points,#color(green)(p_1(x_1,y_1)# and#color(green)(p_2(x_2,y_2)# is#color(red)(r# then,

#color(red)(ul(bar(|color(green)(r=sqrt((x_1-x_2)^2+(y_1-y_2)^2))|# Hope it helps...

Thank you...