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Math Help - Unknown Coordinates

  1. #1
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    Unknown Coordinates

    Well, I have a problem, We're given 3 points(sorry if i've defined/called it wrong) which is A B and C, which also has each a coordinates, but the problem is, only A and B has, C doesn't have any, and we're tasked to find it, but I don't know the formula in finding it, so I can't proceed in measuring the distance between AC and BC, only AB.

    Here:

    A(-1, 2)
    B(4,2)
    C(?,?)

    Could someone please tell me how to find the C coordinates?

    NOTE: My professor said it is an equilateral triangle when we're going to measure it.
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  2. #2
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    Quote Originally Posted by dissidia View Post
    Well, I have a problem, We're given 3 points(sorry if i've defined/called it wrong) which is A B and C, which also has each a coordinates, but the problem is, only A and B has, C doesn't have any, and we're tasked to find it, but I don't know the formula in finding it, so I can't proceed in measuring the distance between AC and BC, only AB.

    Here:

    A(-1, 2)
    B(4,2)
    C(?,?)

    Could someone please tell me how to find the C coordinates?

    NOTE: My professor said it is an equilateral triangle when we're going to measure it.
    The information that the triangle is equilateral is very important here.
    We know that the distance from point A to point B (which is a side of a triangle here) can be calculated using the following formula:

    AB = sqrt( (x2-x1) + (y2-y1) )

    By applying the formula we can calculate the distance between points A(-1,2) and B(4,2)

    AB = sqrt ( (-1-4)) + (2-2) ) = sqrt(25+0) = 5

    Since the triangle ABC is equilateral, we know that AB = BC = AC = 5

    If we mark the unknown point as C(a,b), we'll get two equations that we can use to determine the coordinates a and b.

    Here are a few tips to get you started. I'll be posting the full solution a bit later.
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  3. #3
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    Isn't a,b somehow related to points AB? couldn't really get the equations.
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  4. #4
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    Okay, to continue from where I left:
    The points were A(-1,2), B(4,2) and C(a,b)
    We'll get two equations

    The length of AC
    5 = sqrt( (-1 -a) + (2-b) )
    The length of BC
    5= sqrt( (4-a) + (2-b) )

    From this, we'll get the equation AC = BC

    sqrt( (-1 -a) + (2-b) ) = sqrt( (4-a) + (2-b) )
    (-1 -a) + (2-b) = (4-a) + (2-b)

    We're a bit lucky here that the y-coordinate for both A and B is the same,
    so we'll notice that the (2-b) on both sides will cancel each other out, and
    we'll be left with a simple equation to calculate a

    (-1 -a) = (4-a) <=> 1+2a+a = 16 -8a + a <=> 10a = 15 <=> a= 3/2

    After we've solved a, you can use either one of the equations AC = 5 or BC = 5 to solve b by replacing a with 3/2.
    Let's use BC for example:
    5 = sqrt( (4-3/2) + (2-b) ) <=> 25 = (5/2) + (2-b) <=> 25 = 25/4 + 4 -4b + b <=> b - 4b - 59/4 = 0
    Solve the equation and you'll get b= 1/2 * (4 5 sqrt(3))

    This means that there are actually two possible y-coordinates for point C, which makes sense if you think about it geometrically.

    I hope this helped!
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  5. #5
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    Quote Originally Posted by dissidia View Post
    Isn't a,b somehow related to points AB? couldn't really get the equations.
    You could really give the x and y coordinates in C(a,b) any names. It doesn't have to be C(a,b), it could be C(x0,y0), C(u,v) or whatever.

    The equation to calculate the distance of two points is really just the Pythagoran theorem c = a + b.
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  6. #6
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    A very brilliant mind of yours, though it took me some minutes to understand what you've wrote. Thanks a lot, this really helps.
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  7. #7
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    Just a question bro, how did you manage to get off that b here(b - 4b - 59/4 = 0 ) so that you can get the value of b?
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  8. #8
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    Quote Originally Posted by dissidia View Post
    Just a question bro, how did you manage to get off that b here(b - 4b - 59/4 = 0 ) so that you can get the value of b?
    Since we have a quadratic equation, the simplest way to solve it is to use the quadratic formula.
    The solutions the polynom ax + bx + c = 0 are

    x= ( -bsqrt(b - 4ac) ) / 2a

    With the equation b - 4b - 59/4 = 0
    a=1 , b=4 and c = 59/4

    For more info, see this link:
    The Quadratic Formula Explained
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  9. #9
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    You're cooler than I thought. Thanks again bro.
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  10. #10
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    unknown coordinates

    posted by dissidia

    Given two coordinates find the third assuming you meant to say AB is one side of an equilateral triangle.The answers thus far only define the lenght of a side of the triangle .The solution to either question can be derived more simply.If the midpoint of an equalateral is 2.5 the side is 5 and the altitude is 2.5 times radial 3 so the latter dimension is used to get the two new values of y and x is 2.5 in both cases


    bjh
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  11. #11
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    unknown coordinates

    Hi isosky
    I do not understand your message. I have supplied a method to find the two unknown coordinates Tell me how i can help you to understand this.Remember that the question is to supply the missing coordinates.


    bjh
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