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Math Help - Coordinate Geometry

  1. #1
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    Coordinate Geometry



    I'm sorry my semi circle isn't drawn very well, but it is meant to be a semi circle.

    this is what I did, could you please tell me where I went wrong?

    Because the distance from the point of (0,0) to both the bottom right and bottom left point is 10 units, so I assumed that the distance from that point to any point on the semi circle would be 10 units.

    From here, I used the distance formula (d=square root of (x2-x1)^2+(y2-y1)^2) putting in (0-s)^2+(0-8)^2 and putting in "10" as d.

    So:
    10=SQUAREROOT[(0-s)^2+(0-8)^2]

    From here I solved until I got d=-2.

    But the answer in the back of the book is "6".

    Could you please tell me where I went wrong in the process and how to fix it?

    If I made a mistake in solving for d and I did everything else right can you show me how to accurately solve for d?

    Thank you very much!
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  2. #2
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    Are you looking for the distance from the origin (0,0) to a point at (s,8) which lies on the semi-circle

    By, the looks of it this seem somewhat to what you are doing.

    Recall that the formula of a Circle is  x^2 + y^2 = r^2

    If the distance from the middle of the circle to the end is 10 then this is your Radius or R
    and you are already give your Y value

    plug in everything and solve
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by positiveion View Post


    I'm sorry my semi circle isn't drawn very well, but it is meant to be a semi circle.

    this is what I did, could you please tell me where I went wrong?

    Because the distance from the point of (0,0) to both the bottom right and bottom left point is 10 units, so I assumed that the distance from that point to any point on the semi circle would be 10 units.

    From here, I used the distance formula (d=square root of (x2-x1)^2+(y2-y1)^2) putting in (0-s)^2+(0-8)^2 and putting in "10" as d.

    So:
    10=SQUAREROOT[(0-s)^2+(0-8)^2]

    From here I solved until I got d=-2.

    But the answer in the back of the book is "6".

    Could you please tell me where I went wrong in the process and how to fix it?

    If I made a mistake in solving for d and I did everything else right can you show me how to accurately solve for d?

    Thank you very much!
    draw a vertical line from the point (s,8) to the x-axis. do you see that you have a right-triangle, with base length s (which is what you want to find), height 8 and hypotenuse 10. you can use Pythagoras' theorem to find s, which will be 6. or notice that this is a 3-4-5 triangle, all the sides are multiplied by 2. so the base has to be 2*3 = 6, since we have the hypotenuse is 2*5 and the height is 2*4
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  4. #4
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    I'm supposed to find what "s" is.
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  5. #5
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    Quote Originally Posted by positiveion View Post

    this is what I did, could you please tell me where I went wrong?

    Because the distance from the point of (0,0) to both the bottom right and bottom left point is 10 units, so I assumed that the distance from that point to any point on the semi circle would be 10 units.

    From here, I used the distance formula (d=square root of (x2-x1)^2+(y2-y1)^2) putting in (0-s)^2+(0-8)^2 and putting in "10" as d.

    So:
    10=SQUAREROOT[(0-s)^2+(0-8)^2]

    From here I solved until I got d=-2.

    ...
    1. Your equation is OK.

    2. There isn't any "d" in your equation, so where did you get this value from?

    3.
    10=\sqrt{(0-s)^2+(0-8)^2}~\implies~ 10 = \sqrt{s^2+64}<br />

    Now square both sides:

    100 = s^2 +64~\implies~36 = s^2~\implies~s = -6~\vee~\boxed{s=6}

    The negative solution isn't very plausible here.
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  6. #6
    Member arpitagarwal82's Avatar
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    positiveion your approach 10=SQUAREROOT[(0-s)^2+(0-8)^2] is perfect.
    You just did some calculation mistake.
    If you solve the above thing properly s will come out to be 6 not 2.

    Remember you have to find the value of s.
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