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Suppose that R is at (8, 6). Let P be a point on the line 8y = 15x and Q be a point on the line 10y = 3x, and suppose that R is the midpoint of PQ. Then the length of PQ can be written as a / b, where a and b have no common factors other than 1. What is a + b.

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Maurice

  1. If P is on 8y = 15x
    then we could call P (p,15p/8)
    similarly if Q is on 10y = 3x
    then let's refer to Q as (q,3q/10)
    given that R(8,6) is the midpoint of PQ
    (p+q)/2 = 8
    p + q =16 -----> q = 16-p
    and ((15p/8 + 3q/10)/2 = 6
    15p/8 + 3q/10 = 12
    times 40
    75p + 12q = 480 or
    25p + 4q = 160

    25p + 4(16-p) = 160
    21p = 96
    p= 32/7
    P is (32/7 , 60/7)

    q = 16-p = 16-32/7 = 80/7
    Q is (80/7 , 24/7)

    PQ = √(80/7 - 32/7)^2 + (24/7-60/7)^2)
    = .. you do the button pushing,
    I got a fraction of the form a/b, then do a+b

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    Reiny

  2. let P = (m,15/8 m)
    Q = (n,3/10 n)
    Then since PR=QR,
    (8-m)^2 + (6 - 15/8 m)^2 = (8-n)^2 + (6 - 3/10 n)^2
    and, since the slope of PR = slope of QR,
    (15/8 m - 6)/(m-8) = (3/10 n - 6)/(n-8)
    Solving for m and n, we have
    m = 32/7, n = 80/7
    Plugging that into the distance formula, we see that PQ = 60/7

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    oobleck

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