Question

Pythagorean Theorem graphing help?

Answer 1

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)`.