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Math Help - Co-ordinate Geometry Q

  1. #1
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    Red face Co-ordinate Geometry Q

    A question I'm stuck on, I'm never quite sure how to start off questions that require you to 'prove' things ... I don't have a very mathematical brain I think! :/ so here it is:

    if p is the length if the perpendicular from the origin to the line x/a + y/b = 1, prove that 1/p = 1/a + 1/b.

    I assume I'll have to use the perpendicular distance formula somehow, but I don't really know what to do ...
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  2. #2
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    Re: Co-ordinate Geometry Q

    We know that the length of the perpendicular from the point (a', b') to the line ax+by+c=0 is (aa'+bb'+c)/[sqrt(a^2+b^2)]
    In our case a'=0, b'=0
    and the line is x/a+y/b=1
    thus substituting in the formula we get
    (0+0-1)/[sqrt[(1/a^2)+(1/b^2)]= p
    -1/sqrt[1/a^2+1/b^2] =p
    resiprocating both the sides
    sqrt[1/a^2+1/b^2]=1/p
    square both the sides we get
    1/a^2+1/b^2=1/p^2
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  3. #3
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    Re: Co-ordinate Geometry Q

    The following may further help in understanding
    From the step
    (-1)/(√(1/a^2 + 1/b^2 ) )=p By squaring both the sides we get

    1/((a^2+ b^2)/(a^2 b^2 ) )=p^2
    1/p^2 =(a^2+ b^2)/(a^2 b^2 )
    1/p^2 = a^2/(a^2 b^2 )+ ( b^2)/(a^2 b^2 )

    1/p^2 = 1/b^2 + ( 1)/a^2
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  4. #4
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    Re: Co-ordinate Geometry Q

    Ok I understand everything up to here but how to do you get this when you square it?:

    1/((a^2+ b^2)/(a^2 b^2 ) )=p^2 ([/QUOTE]
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  5. #5
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    Re: Co-ordinate Geometry Q

    Quote Originally Posted by HelenMc9 View Post
    Ok I understand everything up to here but how to do you get this when you square it?:
    1/((a^2+ b^2)/(a^2 b^2 ) )=p^2
    Using the distance formula for a point to a line we get:
    \frac{1}{{\sqrt {a^{ - 2}  + b^{ - 2} } }} = p
    which can be written as
    \sqrt {a^{ - 2}  + b^{ - 2} }  = \frac{1}{p}.

    Now square both sides.
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  6. #6
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    Re: Co-ordinate Geometry Q

    The denominator was under the square root. It was Sqr root √(1/a^2 + 1/b^2 ). I hope it is clear now
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  7. #7
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    Re: Co-ordinate Geometry Q

    Please refer to the sketch:

    Co-ordinate Geometry Q-perpdistancep.jpg

    The sum of the areas of the two smaller triangles is equal to the area of the largest triangle, giving us (after multiplying by 2):

    cp+dp=ab

    p(c+d)=ab

    \frac{1}{p}=\frac{c+d}{ab}

    \frac{1}{p^2}=\frac{(c+d)^2}{a^2b^2}

    By Pythagoras, we have:

    (c+d)^2=b^2+a^2 and so:

    \frac{1}{p^2}=\frac{b^2+a^2}{a^2b^2}=\frac{1}{a^2}  +\frac{1}{b^2}
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