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Math Help - Coordinate of a point in a triangle

  1. #1
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    Coordinate of a point in a triangle

    Hi, I have the following exercise and I am kind of stuck in the last bit now... There exercise says

    Coordinate of a point in a triangle-mathhelpimage.jpg

    "Consider the simple right-angled triangle ABC in the diagram above. The points P, Q, R divide the sides of the triangle in the ratios indicated. Find the coordinates of the points P, Q, R in terms of s, t, u. (Q having 1 and t on each side is only the ratio, it doesn't mean that the length between Q and C is 1)

    Well...so far I have Q=(0, t/(t+1)) and R=(1/(1+u), 0)... How do I get P though? Thanks a lot!
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  2. #2
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    Quote Originally Posted by juanma101285 View Post
    Hi, I have the following exercise and I am kind of stuck in the last bit now... There exercise says

    Click image for larger version. 

Name:	mathhelpimage.jpg 
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ID:	20619

    "Consider the simple right-angled triangle ABC in the diagram above. The points P, Q, R divide the sides of the triangle in the ratios indicated. Find the coordinates of the points P, Q, R in terms of s, t, u. (Q having 1 and t on each side is only the ratio, it doesn't mean that the length between Q and C is 1)

    Well...so far I have Q=(0, t/(t+1)) and R=(1/(1+u), 0)... How do I get P though? Thanks a lot!
    Drop a perpendicular from P to AB, let the foot of the perpendicular be M.
    AM = the x - coordinate of P.

    Now use similar triangles:

    triangle ABC similar with triangle PMB:

    \frac{BP}{BC}=\frac{BM}{AB}
    \frac{1}{1+s}=\frac{1-x}{1}

    1-x=\frac{1}{1+s}

    and find x.

    In the same way, use similar triangles to find the y coordinate of P.
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  3. #3
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    Quote Originally Posted by juanma101285 View Post
    Hi, I have the following exercise and I am kind of stuck in the last bit now... There exercise says

    Click image for larger version. 

Name:	mathhelpimage.jpg 
Views:	36 
Size:	7.2 KB 
ID:	20619

    "Consider the simple right-angled triangle ABC in the diagram above. The points P, Q, R divide the sides of the triangle in the ratios indicated. Find the coordinates of the points P, Q, R in terms of s, t, u. (Q having 1 and t on each side is only the ratio, it doesn't mean that the length between Q and C is 1)

    Well...so far I have Q=(0, t/(t+1)) and R=(1/(1+u), 0)... How do I get P though? Thanks a lot!
    The horizontal co-ordinate of P is

    \displaystyle\frac{s}{1+s}

    since if you drop a perpendicular from P to [OB], you will get the horizontal co-ordinate.

    You get the vertical co-ordinate if you draw a horizontal from P to [OC].

    This is

    \displaystyle\frac{1}{1+s}
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  4. #4
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    Cool, thanks a lot! I have other questions in the same exercise but I think I can solve them by myself now
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  5. #5
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    OK... I am stuck again . How do you solve these parametric equations? (they need to be in terms of t and u). I'd just like to have a tip so that I can work on it myself and learn...

    So, I got for BQ:

    (x, y)=(1,0)+m(-1, t/(1+t))

    and for CR I got:

    (x, y)=(0,1)+n(1/(1+u), -1)

    I know I have to equate the x and y, but I still can't get rid of m and n...

    So I got

    1-m=n/(1+u) and mt/(t+1)=1-n

    So I substituted n=1+u-m-mu into

    mt/(t+1)=1-n, but I get mt=m+mu-u-tu+mut+mt

    How am I supposed to get rid of n and m then? :-/
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