# Maths

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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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

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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