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

Apr 6, 2018

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

$\text{ "bar(OD)=x_1" ; "bar(OE)=x_2" ; "bar(AD)=y_1" ; } \overline{E B} = {y}_{2}$

We also have a rectangle $\square O C E D$. So, color(red)(bar(AC)=bar(DE)) " and "color(red)(bar(AD)=bar(CE)

Now,

• $\overline{A C} = \overline{D E} = \overline{O E} - \overline{O D} = \left({x}_{2} - {x}_{1}\right)$

• $\overline{B C} = \overline{B E} - \overline{C E} = \overline{B E} - \overline{A D} = \left({y}_{2} - {y}_{1}\right)$

With the help of Pythagorean theorem,

${\overline{A B}}^{2} = {\overline{B C}}^{2} + {\overline{A C}}^{2}$

${\overline{A B}}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}$

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

N.B:- As it is a square value , you may take $\left({x}_{1} - {x}_{2}\right)$ or, $\left({x}_{2} - {x}_{1}\right)$. 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...