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Math Help - distance

  1. #1
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    distance

    these are word for word out of the book. (d(A,B) means the distance from A to B) 1. if A, B, and C are three points in a plane, d(A,B) = 3 and d(B,C) = 2, what can you say about d(A,C)? for this one I just used the distance fomula using the lines lengths... so i did: d(A,C) = square root of(3^2 + 2^2) and i got the square root of 13... is this correct. the second one i have really no clue how to do. 2. let P1 = (-1,2) and P2 = (0,0). find the point P3 in quadrant 1 such that {P1,P2,P3} are vertices of an equilateral triangle.
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    Quote Originally Posted by pcgamer03 View Post
    1. if A, B, and C are three points in a plane, d(A,B) = 3 and d(B,C) = 2, what can you say about d(A,C)? for this one I just used the distance fomula using the lines lengths... so i did: d(A,C) = square root of(3^2 + 2^2) and i got the square root of 13... is this correct. the second one i have really no clue how to do.
    We aren't told anything about the orientations of ABC, so there are three cases to consider:
    1) The points line up ABC: Then d(AC) = d(AB) + d(BC) = 3 + 2 = 5.
    2) The points line up ACB: Then d(AC) = d(AB) - d(BC) = 3 - 2 = 1.
    3) Any other combination where the points are not co-linear will have distances between 1 and 5.

    So 1 < d(AB) < 5 is the best that we can do.

    -Dan
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    Quote Originally Posted by pcgamer03 View Post
    2. let P1 = (-1,2) and P2 = (0,0). find the point P3 in quadrant 1 such that {P1,P2,P3} are vertices of an equilateral triangle.
    Here's a start. Let P3 = (x, y). The distance from P3 to P1 is the same as the distance from P1 to P2 (call this distance d. You can find it.) So
    d = \sqrt{(x + 1)^2 + (y - 2)^2}

    This is also the distance from P3 to P2:
    d = \sqrt{x^2 + y^2}

    You now have two simultaneous equations in two unknowns. You can solve the system in any way you like. (Hint: square both equations before you start.)

    -Dan
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    not really sure how I should go about solving that. I used substitution and ended up with -5/2 = x+2y. i set the equation up like: x^2 + y ^2 = (x+1)^2 + (y-2)^2
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    Quote Originally Posted by pcgamer03 View Post
    not really sure how I should go about solving that. I used substitution and ended up with -5/2 = x+2y. i set the equation up like: x^2 + y ^2 = (x+1)^2 + (y-2)^2
    The substitution method works only when you solve one equation for one of the unknowns.

    For example, solve
    d = \sqrt{x^2 + y^2}
    for y:

    d^2 = x^2 + y^2

    y^2 = d^2 - x^2

    y = \pm \sqrt{d^2 - x^2}

    Now plug these values of y into the other distance equation (do the + and - cases separately.)

    -Dan
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    ok... i have absolutely no clue how to solve this. I substitued that in for y and for positive i got: x = (squarert(5) + 4 *d) / 2 * squarert(3) which makes no sense...
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pcgamer03 View Post
    ok... i have absolutely no clue how to solve this. I substitued that in for y and for positive i got: x = (squarert(5) + 4 *d) / 2 * squarert(3) which makes no sense...
    Just a quick thought, did you ever find the value for d? That simplifies the whole process enormously.

    -Dan
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  8. #8
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    d is the square root of 5
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    also... where did you get all of those nice looking math things from, it is much easier for people to understand problems when you have them looking like they would from the book.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by pcgamer03 View Post
    also... where did you get all of those nice looking math things from, it is much easier for people to understand problems when you have them looking like they would from the book.
    See here.

    -Dan
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