Pythagorean Theorem graphing help?

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1 Answer
May 5, 2018

The distance #d# between two points #(x_1, y_1)# and #(x_2, y_2)# is equal to

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

How is this related to the Pythagorean Theorem?

If we square both sides, we get

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

Then let's make a couple quick variable changes. Let #a = x_2-x_1# and #b=y_2-y_1#. After we swap these out, we get:

#d^2 = a^2+b^2#

And hey, look! There's the Pythagorean Theorem!

Since #d# represents the distance between our two points, it makes sense that we can think of #d# as the length of the hypotenuse of a right triangle, whose sides are parallel to the #x# and #y# axes.

Thus, #a# (which is #x_2-x_1#) is the horizontal difference between our two points (aka the bottom leg of our triangle), and #b# (which is #y_2-y_1#) is the vertical difference between the two points (aka the side leg of our triangle).

Let's use question 1 as an example. We'll call the bottom-left point #(x_1, y_1)# and the top-right point #(x_2,y_2)#.

Thus, #(x_1, y_1)=(–2, –2)# and #(x_2,y_2)=(1, 3)#. Substituting these into the distance formula gives:

#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#
#color(white)d=sqrt([1-(–2)]^2+[3-(–2)]^2)#
#color(white)d=sqrt((1+2)^2+(3+2)^2)#
#color(white)d=sqrt(3^2+5^2)#
#color(white)d=sqrt(9+25)#
#color(white)d=sqrt(34)#

I'll leave the others as an exercise.

Area of a right triangle is easy: #A=1/2 b h#, where #b# is the base (which, for us, is the horizontal distance #abs(x_2-x_1)#) and #h# is the height (which would be #abs(y_2-y_1)#). Thus,

#A=1/2abs(x_2-x_1)abs(y_2-y_1)#

Perimeter is the sum of all three side lengths, which for us, would be

#P = b + h + d#

where

#b = abs(x_2-x_1)#,
#h = abs(y_2-y_1)#, and
#d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)#.