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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)

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  1. anonymous
    • 4 years ago
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    thats it roughly find lengths of the 3 sides and Check that AC^2 = AB^2 + BC^2 so proving ABC is rt angled

  2. anonymous
    • 4 years ago
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    No negative x values here, but not that important

  3. anonymous
    • 4 years ago
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    oh - right - i messed up there1

  4. anonymous
    • 4 years ago
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    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.

  5. anonymous
    • 4 years ago
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    for example BC^2 = 0^2 + (4-2)^2 = 4

  6. anonymous
    • 4 years ago
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    Thanks guys :)

  7. anonymous
    • 4 years ago
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    |dw:1328130277370:dw|

  8. anonymous
    • 4 years ago
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    yw

  9. amistre64
    • 4 years ago
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    we could also create 3 vectors and dot them to see if any 2 vectors = 0

  10. anonymous
    • 4 years ago
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    This like a different language to me when it comes to math. lol What do you mean?

  11. anonymous
    • 4 years ago
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    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.

  12. anonymous
    • 4 years ago
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    I see...that is a little confusing to me right now.

  13. anonymous
    • 4 years ago
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    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

  14. anonymous
    • 4 years ago
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    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.

  15. anonymous
    • 4 years ago
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    Thank you for all the help :) I finally figured it out!!

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