Can any one help with this? I am just stumped on this problem!
plot each point and form the triangle ABC. Verify that the triangle is a right triangle. Find its area.
A =(4, −3); B =(4, 1); C = (2, 1)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

thats it roughly
find lengths of the 3 sides
and Check that AC^2 = AB^2 + BC^2 so proving ABC is rt angled

- anonymous

No negative x values here, but not that important

- anonymous

oh - right - i messed up there1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

The distance formula is \[d=\sqrt{(y2-y1)+(x2-x1)}\]
You don't need to find the square root though because you're just going to be squaring them again when using the Pythagorean theorem.

- anonymous

for example BC^2 = 0^2 + (4-2)^2 = 4

- anonymous

Thanks guys :)

- anonymous

|dw:1328130277370:dw|

- anonymous

yw

- amistre64

we could also create 3 vectors and dot them to see if any 2 vectors = 0

- anonymous

This like a different language to me when it comes to math. lol What do you mean?

- anonymous

you need to know something about vectors to understand what amis said
it would take too long to explain here - briefly
when two vectors are multiplied (or 'dotted'), if the answer is zero then they are at right angles to each other.

- anonymous

I see...that is a little confusing to me right now.

- anonymous

Is this correct or am I way off?
D(A,B)=sqrt (4-4)^2+(-3-1)^2=0+16= 16
D(B,C)=sgrt (4-2)^2(1-1)^2=4+0=4
D(A,C)=sqrt (4-2)^2+(-3-1)^2=4+16=20
D(A,B)^2+(B,C)^2=16^2+4^2=256+16=272
D(A,C)=20^2= 400
D(A,B)^2+D(B,C)^2 does not equal D(A,C)^2
It is not a right triangle

- anonymous

A =(4, −3); B =(4, 1); C = (2, 1)
If ABC is a right triangle,
\[d(AC)^2=d(AB)^2+d(BC)^2\]
You can go without using the square root of the distance formula here because you will be later squaring that distance when plugging it into the Pythagorean theorem.
\[d^2=(\sqrt{(y2-y1)^2+(x2-x1)^2})^2=(y2-y1)^2+(x2-x1)^2\]
So you can evaluate thisand see if it results in a true answer:
\[[(1+3)^2+(2-4)^2]=[(1+3)^2+(4-4)^2]+[(1-1)^2+(4-2)^2]\]
\[4^2+(-2)^2=4^2+0^2+0^2+2^2\]
\[20=16+4\]
This expression is TRUE; therefore, it is a right triangle.

- anonymous

Thank you for all the help :) I finally figured it out!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.