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Math Help - Need help on Easy Grade 10 Geometry question

  1. #16
    Senior Member topher0805's Avatar
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    Quote Originally Posted by LllamaGlama View Post
    Huh... i thought i was gonna find the distance from the offshre connecting to the onshore then add the distance from the refinery to the harbour?
    that may of been what u just said cuz idk wat a pythagoras is (is that (x2-x2)2+ (y2-y1)2 square rooted?

    lol my math typing skills are pretty bad
    I'm not trying to be rude at all when I ask this question:

    Do you honestly not know what Pythagoras is?
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  2. #17
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    Quote Originally Posted by topher0805 View Post
    I'm not trying to be rude at all when I ask this question:

    Do you honestly not know what Pythagoras is?
    Nah i dont take that rude at all. As u can see math isnt ny strongest... could it be a2+b2=c2?
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  3. #18
    Senior Member topher0805's Avatar
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    Quote Originally Posted by LllamaGlama View Post
    Nah i dont take that rude at all. As u can see math isnt ny strongest... could it be a2+b2=c2?
    Yes.

    a^2 = b^2 + c^2

    So, in the diagram that I drew, find the equation of the line connecting A and B in the form y = mx + b

    Then, take the negative reciprocal of the slope of AB. (flip the fraction and multiply it by -1).

    Next, use the slope of the line that connects C to AB and the point C to find the equation of that line also in the form y = mx + b

    Set the two equation equal to each other (mx + b = mx +b) and solve for x. The, plug that x value back into either equation to find y. This is the point at which the two pipes meet.

    Now, use the distance formula that topsquark provided to find the distance between C and the new point(lets call it D).

    Now use Pythagoras to find the distance from D to the refinery.

    Add these two distances:

    From C to D + From D to Refinery.

    And you have your answer!
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  4. #19
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    Quote Originally Posted by topher0805 View Post
    Yes.

    a^2 = b^2 + c^2

    So, in the diagram that I drew, find the equation of the line connecting A and B in the form y = mx + b

    Then, take the negative reciprocal of the slope of AB. (flip the fraction and multiply it by -1).

    Next, use the slope of the line that connects C to AB and the point C to find the equation of that line also in the form y = mx + b

    Set the two equation equal to each other (mx + b = mx +b) and solve for x. The, plug that x value back into either equation to find y. This is the point at which the two pipes meet.

    Now, use the distance formula that topsquark provided to find the distance between C and the new point(lets call it D).

    Now use Pythagoras to find the distance from D to the refinery.

    Add these two distances:

    From C to D + From D to Refinery.

    And you have your answer!
    Im pretty terrible with y=mx+b ... to find slope can i not just use y2-y/x2-x1?
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  5. #20
    Senior Member topher0805's Avatar
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    I have to go to class now but I'm sure someone will help you if you have any more questions.
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  6. #21
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    Quote Originally Posted by topher0805 View Post
    I have to go to class now but I'm sure someone will help you if you have any more questions.
    ya thanks for the help
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