Show that (-1, 1), (3, 3), and (1, -3) form right triangle

[SCSmith]

Show that the points are vertices of a right triangle.

Where Points A(-1,1) B (3,3) and A(-1,1) C (1,-3) are the sides of a Right Triangle and Points B (3,3) C(1,-3) is the Base.

Given, d = sqrt (x[2] - x[1])^2 + y[2] -y[1]^2 then

sqrt [3 - (-1)]^2 + (3 -1)^2
sqrt (3 + 1)^2 + 2^2
sqrt 4^2 + 2^2
sqrt 16 + 4
sqrt 20 and

sqrt [1 - (-1)]^2 + (-3 - 1)^2
sqrt 2^2 + -4^2
sqrt 4 + 16
sqrt 20

sqrt (1 - 3)^2 + (-3 - 3)^2
sqrt -2^2 + -6^2
sqrt 4 +36
sqrt 40

so,

sqrt20^2 + sqrt20^2 = 20 +20 = 40
sqrt40^2 = 40

so these vertices are angles of a right triangle.

Please confirm if correct

 

[soroban]

Re: Geometry Problem.

Hello, SCSmith!

Your work and your reasoning is absolutely correct.
There is, however, a faster proof.

Show that the points are vertices of a right triangle:
. . A(-1,1), B(3,3), C(1,-3)

The slope of \(\displaystyle AB\) is: \(\displaystyle \L\:m_{_{AB}} \:=\:\frac{3\,-\,1}{3\,-\,(-1)} \:=\:\frac{2}{4}\:=\:\frac{1}{2}\)

The slope of \(\displaystyle AC\) is: \(\displaystyle \L\:m_{_{AC}} \:=\:\frac{-3\,-\,1}{1\,-\,(1)}\:=\:\frac{-4}{2}\:=\:-2\)

Since \(\displaystyle \,\left(m_{_{AB}}\right)\left(m_{_{AC}}\right) \:=\:\left(\frac{1}{2}\right)(-2)\:=\:-1\), then: \(\displaystyle \,AB\,\perp\,AC\)

Therefore, \(\displaystyle \Delta ABC\) is a right triangle.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I am not laughing.
I did the very same thing on a quiz ... many years ago.

After the class, I was quite proud of my use of Pythagorus.
Then a classmate said, "Didn't you check the slopes?"
. . . . . ack! . . . (slap head, kick self)