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Math,Geometry

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What are the equations of the lines through (-5,-3) and passing at distance 2sqrt5 from (5,7)

Please I really need your help. I do not know how to do this problem.

  • Math,Geometry -

    Make a sketch, let the point of contact be P(x,y)
    (there will be two of them, let the math take care of it)
    label A(-5,-3) and B(5,7)

    point P must be on a circle with centre B(7,5) and radius of BP = 2√5
    the equation of that circle is
    (x-7)^2 + (y-5)^2 = 20
    which when expanded gets us
    x^2 + y^2 = 10x+14y-54

    also the slope of AP must be the negative reciprocal of the slope of BP
    slope AP = (y+3)/(x+5)
    slope BP = (y-7)/(x-5)

    then (x-5)/-(y-7) = (x+5)/(y+3)
    x^2 - 25 = -y^2 + 7y + 21 - 3y
    x^2 + y^2 = 4y + 46

    So 10x + 14y - 54 = 4y+46
    10x = -10y + 100
    x = 10- y

    Then in x^2 + y^2 = 4y + 46
    (10-y)^2 + y^2 = 4y + 46
    100 - 20y + y^2 + y^2 = 4y + 46
    2y^2 - 24y + 54 = 0
    y^2 - 12y + 27 = 0
    (y-3)(y-9)= 0
    y = 3 or y = 9

    if y = 3, then x = 10-3 = 7 ...... P is (7,3)
    if y = 9, then x = 10-9 = 1 ...... P is (1,9)

    almost done ......

    for P as (7,3)
    slope of AP = (3+3)/(7+5) = 6/12 = 1/2
    equation of AP:
    y+3 = (1/2)(x+5)
    2y + 6 = x+5
    x - 2y = 1 is one of the line equations

    for P as (1,9)
    .......
    I will let you have the pleasure of finishing it.

    (check my work, but it came out so nice to have an error in it)

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