Is triangle ABC a right triangle???

L

lectromagnet94

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Oct 22, 2008
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27
IDK how to answer this... i didnt pay attention in class...

Is ABC with vertices A(–2, –5), B(–3, 2), and C(1, –2) a right triangle?

Below are my choices...
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ABC is a right triangle because .
ABC is a right triangle because .
ABC is a right triangle because .
ABC is not a right triangle.
 
lectromagnet94 said:
IDK how to answer this... i didnt pay attention in class...

Is ABC with vertices A(–2, –5), B(–3, 2), and C(1, –2) a right triangle?

Below are my choices...
|
|
|
|
V

ABC is a right triangle because .
ABC is a right triangle because .
ABC is a right triangle because .
ABC is not a right triangle.
Click to expand...
Hint:

Calculate the lengths of AB, BC and CA and use Pythagoras's theorem .

Please us show your work, indicating exactly where you are stuck - so that we know where to begin to help you.
 
Can you just give me the answer?!
p.s.
havnt learned the P. theorem. went over it in algebra 1... but didnt really learn anything.
 
lectromagnet94 said:
Can you just give me the answer?!-- No I don't work that way
p.s.
havnt learned the P. theorem. went over it in algebra 1... but didnt really learn anything.
Click to expand...

Do a google search - or dust off your old algebra book.

Do you know how to find length of a line between two points? That equation is derived using Pythagoras's theorem.
 
You can believe anything you want :shock:

If you keep up not paying attention in class, you'll end up flipping burgers at Burger King.
 
"...A(–2, –5), B(–3, 2), and C(1, –2)"

AB = sqrt((-3 - (-2))^2 + (2 - (-5))^2) = sqrt (5^2 + 7^2) = sqrt (25 + 49) = sqrt 74

BC = sqrt ((1 - (-3))^2 + (-2 - 2)^2) = sqrt (4^2 + (-4)^2) = sqrt (16 + 16) = sqrt 32

AC = sqrt ((1 - (-2))^2 + (-2 - (-5)^2) = sqrt (3^2 + 3^2) = sqrt (9 + 9) = sqrt 18

I purposely did not simplify the square roots.

74 ? 32 + 18, so ABC is not a right triangle.
 
There.....Fastone is making sure you'll flip burgers :shock:

Now copy his solution and hand it in to your teacher. And all's well...
 
Every Indian and Chinese student should be thankful to the fast one - he would make sure they are needed here.
 
Hello, lectromagnet94!

Here's an easier way . . .

\(\displaystyle \text{Is }\Delta ABC\text{ with vertices }A(\text{-}2,\, \text{-}5),\;B(\text{-}3,\,2),\:C(1,\,\text{-}2)\text{ a right triangle?}\)
Click to expand...

\(\displaystyle \text{The lines are perpendicular if their slopes are }negative\;reciprocals.\)

.\(\displaystyle \text{That is: }\;m_1 \:=\:-\frac{1}{m_2}\)


But you went over that in Algebra 1 ... where you didn't really learn anything, right?
.
 
soroban said:
But you went over that in Algebra 1 ... where you didn't really learn anything, right?
.
Click to expand...
WRONG Soroban: he became a whizz on the pinball machine :idea:
 
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