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Math Help - Locus question.

  1. #1
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    Locus question.

    how abot this question:
    The locus of points the sum of whose distances from the points (1,0) and (-1,0) equals 2.

    What is the locus of points and how is the distance?
    Last edited by mr fantastic; May 15th 2010 at 04:05 PM. Reason: Moved from another thread.
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  2. #2
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    Quote Originally Posted by sandy View Post
    how abot this question:
    The locus of points the sum of whose distances from the points (1,0) and (-1,0) equals 2.

    What is the locus of points and how is the distance?
    Do the obvious thing and start by plotting the points (1, 0) and (-1, 0). Now think some more about the given condition.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Do the obvious thing and start by plotting the points (1, 0) and (-1, 0). Now think some more about the given condition.
    i did put them on the plot
    and we get a line on the x axis from -1 to 1 so the distance is 2 .
    but what do we do next?
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    Quote Originally Posted by sandy View Post
    i did put them on the plot
    and we get a line on the x axis from -1 to 1 so the distance is 2 .
    but what do we do next?
    No. No. Not that way.

    Let P (x, y) be the point. Find the distance between (x, y) and (1, 0), and (x, y) and (-1, 0). Then add the distances and equate it to the given value. Then simplify the equation to get the locus.
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  5. #5
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    Quote Originally Posted by sandy View Post
    i did put them on the plot
    and we get a line on the x axis from -1 to 1 so the distance is 2 .
    but what do we do next?
    You're menat to note that any point on the line segment joining (-1, 0) and (1, 0) satisfies the given condition. Therefore ....
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  6. #6
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    Hi sandy,

    you can also ask if it is possible for the point "d" located anywhere off the line segment from (-1,0) to (1,0) to be a total distance of 2 away from both points.
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  7. #7
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    Quote Originally Posted by sa-ri-ga-ma View Post
    No. No. Not that way.

    Let P (x, y) be the point. Find the distance between (x, y) and (1, 0), and (x, y) and (-1, 0). Then add the distances and equate it to the given value. Then simplify the equation to get the locus.
    i done that it will be
    sqrt[ (x-1)^2 + y^2 ] + sqrt[ (x+1)^2 + y^2 ] =2

    bt how do we solve this?
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  8. #8
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    Quote Originally Posted by sandy View Post
    i done that it will be
    sqrt[ (x-1)^2 + y^2 ] + sqrt[ (x+1)^2 + y^2 ] =2

    bt how do we solve this?
    Hi sandy,

    \sqrt{(x-1)^2+y^2}+\sqrt{(x+1)^2+y^2}=2

    \sqrt{(1+x)^2+y^2}=2-\sqrt{(1-x)^2+y^2}

    Square both sides

    \sqrt{(1+x)^2+y^2}\sqrt{(1+x)^2+y^2}=\left(2-\sqrt{(1-x)^2+y^2}\right)\left(2-\sqrt{(1-x)^2+y^2}\right)

    (1+x)^2+y^2=4-4\sqrt{(1-x)^2+y^2}+(1-x)^2+y^2

    y^2 is common

    1+2x+x^2=4-4\sqrt{(1-x)^2+y^2}+1-2x+x^2

    1+x^2 is common

    2x=4-4\sqrt{(1-x)^2+y^2}-2x

    4x=4\left(1-\sqrt{(1-x)^2+y^2}\right)

    x=1-\sqrt{(1-x)^2+y^2}

    1-x=\sqrt{(1-x)^2+y^2}

    Square both sides

    (1-x)^2=(1-x)^2+y^2

    The only way these can be equal is if y=0.

    You should note that you can see the solution geometrically,
    by just following mr fantastic's hint.

    The distance from (-1,0) to (1,0) is 1-(-1)=2.

    The shortest distance between 2 points is a straight line.
    Hence any point not between (-1,0) and (1,0) must be a combined distance >2 from those two points.

    Any point between (-1,0) and (1,0) is a combined distance of 2 from the two points.

    Visually, placing any point above the line segment, you can see that the
    combined hypotenuses of the two right-angled triangles formed
    have sum of lengths > the combined lengths of the bases.

    Hence, any point not on the line segment is more than 2 combined units from those points.
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