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anonymous
 4 years ago
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)
anonymous
 4 years ago
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)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats it roughly find lengths of the 3 sides and Check that AC^2 = AB^2 + BC^2 so proving ABC is rt angled

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No negative x values here, but not that important

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh  right  i messed up there1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The distance formula is \[d=\sqrt{(y2y1)+(x2x1)}\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0for example BC^2 = 0^2 + (42)^2 = 4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1328130277370:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0we could also create 3 vectors and dot them to see if any 2 vectors = 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This like a different language to me when it comes to math. lol What do you mean?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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
 4 years ago
Best ResponseYou've already chosen the best response.0I see...that is a little confusing to me right now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is this correct or am I way off? D(A,B)=sqrt (44)^2+(31)^2=0+16= 16 D(B,C)=sgrt (42)^2(11)^2=4+0=4 D(A,C)=sqrt (42)^2+(31)^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
 4 years ago
Best ResponseYou've already chosen the best response.0A =(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. \[d^2=(\sqrt{(y2y1)^2+(x2x1)^2})^2=(y2y1)^2+(x2x1)^2\] So you can evaluate thisand see if it results in a true answer: \[[(1+3)^2+(24)^2]=[(1+3)^2+(44)^2]+[(11)^2+(42)^2]\] \[4^2+(2)^2=4^2+0^2+0^2+2^2\] \[20=16+4\] This expression is TRUE; therefore, it is a right triangle.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you for all the help :) I finally figured it out!!
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