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Math Help - Triangle problem in coordinate geometry

  1. #1
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    Triangle problem in coordinate geometry

    I have a triangle with two vertices known (1,3) and (-4,1).

    I know a formula relating the acute angle formed by two lines and their slopes. It is

    tanA=\lvert\frac{m1-m2}{1+m1m2}\rvert

    I used this formula for the hypotenuse and one of the sides(m1) and then again used this formula for the hypotenuse and the other side(m2). I used the result m1m2=-1 in the second result and got two values of m1. But when I equate those values I get tanA=0.
    What is wrong?
    Attached Thumbnails Attached Thumbnails Triangle problem in coordinate geometry-untitled.jpg  
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  2. #2
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    Re: Triangle problem in coordinate geometry

    What do you want to find out, for drawing a triangle we need to know three elements?
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  3. #3
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    Re: Triangle problem in coordinate geometry

    I don't want to find anything. I am just not able to understand why I am getting tanA=0 when it could be anything.
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  4. #4
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    Re: Triangle problem in coordinate geometry

    Please show your work that will make this clear. In fact where are the slopes of tow other sides of the triangle? I Sri think some information is missing. With only this much information we gat infinitely many triangles and hence the problem?
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    Re: Triangle problem in coordinate geometry

    there is no other information. I don't know what I am doing wrong with the information that is provided.
    slope of hypotenuse=2/5

    putting this in the formula for tan A.We get

     m1=\frac{2-5tanA}{2tanA+5}

    Similarly, I found m2 in terms of tan A and replaced m2 with -1/m1 to get
    m1=\frac{2+5tanA}{5-2tanA}

    On equating these two values of tanA I get tanA=0.
    Last edited by AaPa; November 16th 2013 at 09:01 PM.
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  6. #6
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    Re: Triangle problem in coordinate geometry

    m_1 appears to be the slope of a vertical line. But, if it is not, then you can calculate it using (x,y). m_1 = \dfrac{3-y}{1-x} (note: this slope is not defined if x=1).

    Then m_2 appears to be the slope of a horizontal line. Still, we can calculate it: m_2 = \dfrac{1-y}{-4-x} (note: this slope is not defined if x = -4).

    Then, you would have \tan A = \left| \dfrac{\tfrac{3-y}{1-x} - \tfrac{1-y}{-4-x}}{1+\tfrac{3-y}{1-x}\tfrac{1-y}{-4-x}} \right| = \left| \dfrac{(4+x)(3-y)+(1-y)(1-x)}{(4+x)(x-1)+(1-y)(3-y)} \right|

    Simplifying, you get \tan A = \left|\dfrac{13 + 2x -5y}{x^2+y^2+3x-4y-1} \right|

    Is that what you were looking for?
    Last edited by SlipEternal; November 16th 2013 at 09:52 PM.
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  7. #7
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    Re: Triangle problem in coordinate geometry

    I was not looking for anything. I am just getting a wrong result that is tanA=0 but I am not able to find mistake in my work.
    Last edited by AaPa; November 16th 2013 at 10:02 PM.
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  8. #8
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    Re: Triangle problem in coordinate geometry

    Oh, I misread the problem. This may not be the solution. The calculation I gave was for the angle between the lines with slopes m_1,m_2, not angle A. Also, if you are suggesting that the two lines are perpendicular, then \dfrac{3-y}{1-x} = -\dfrac{-4-x}{1-y}

    So, possibly this would be helpful \tan A = \left|\dfrac{\tfrac{3-y}{1-x} - \tfrac{2}{5}}{1+\tfrac{3-y}{1-x}\tfrac{2}{5}} \right| = \dfrac{\text{opp}}{\text{adj}} = \sqrt{\dfrac{(x+4)^2 + (y-1)^2}{(1-x)^2 + (3-y)^2}} The second equation just came from finding the length of each segment.

    Then, simplifying the LHS and squaring both sides, you have:

    \left(\dfrac{13-5y+2x}{11-5x-2y}\right)^2 = \dfrac{(x+4)^2 + (y-1)^2}{(x-1)^2 + (3-y)^2}

    Apparently, this has solutions at x=-4, y=3 over the integers, and over the reals, it has solutions x \approx -4.19 or x \approx 1.19 and y \approx 2.
    Last edited by SlipEternal; November 16th 2013 at 10:26 PM.
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  9. #9
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    Re: Triangle problem in coordinate geometry

    How did you get the solutions. It has infinite solutions.
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  10. #10
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    Re: Triangle problem in coordinate geometry

    Quote Originally Posted by AaPa View Post
    How did you get the solutions. It has infinite solutions.
    A system of two equations in two variables may be solvable. This one is.

    If m_1 and m_2 are the slopes of perpendicular lines, then m_1 = -\dfrac{1}{m_2}. So you have \dfrac{3-y}{1-x} = -\dfrac{-4-x}{1-y}

    From calculating \tan A in two different ways: \left(\dfrac{13-5y+2x}{11-5x-2y}\right)^2 = \dfrac{(x+4)^2 + (y-1)^2}{(x-1)^2 + (3-y)^2}

    Then, I plugged those two into Wolframalpha, and it gave me the solutions.
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  11. #11
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    Re: Triangle problem in coordinate geometry

    But all that was given was a line segment with an arbitrary point(x,y). It could have lied at any point on the circle with the line segment as the diameter.
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  12. #12
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    Re: Triangle problem in coordinate geometry

    That's a good point. I dunno. Weird that Wolframalpha gave solutions...
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  13. #13
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    Re: Triangle problem in coordinate geometry

    please tell me my mistake in post 5.
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  14. #14
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    Re: Triangle problem in coordinate geometry

    The line with slope m_2 is not adjacent to angle A. Your formula requires the slopes of two lines that intersect at angle A.
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    Re: Triangle problem in coordinate geometry

    thank you!!!!!!!
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