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Geometry problem

  1. Jan 14, 2006 #1
    "Find all possible values of a and b given that y = ax + 14 is the perpendicular bisector of the line joining (1,2) to (b,6)" I'm totally stuck :confused: I've tried all the methods I could think of but they all lead to dead ends. I'm hoping someone can point me in the right direction.
    Many thanks.
     
  2. jcsd
  3. Jan 14, 2006 #2
    Ok guys I managed to solve it :D I was wondering if you could me whether my method was the best to use in the situation though and whether or not there was a simpler way. Here goes:

    A(1,2) B(b,6)

    I worked out the gradient of AB to be 4/(b-1)

    I knew that the product of the gradient of AB and the line y = ax + 14 had to be -1 as they are perpendicular, so 4/(b-1) had to be -1/a

    I rearranged that to get b = 1 - 4a

    I worked out the coordinates of the mid-point of AB to be ((1+b)/2 , 4) so I then substitued that into y = ax + 14 and rearranged to get b = -1-(20/a)

    So...

    #1 b = 1 - 4a
    #2 b = -1 - (20/a)

    I then solved them simultaneously to get the values of a and then substitued those into equation #1 to get the values of b.
     
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