Question

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. Compute a + b.

Answer 1

Point p a point on the line
8y=15x

Well assuming x=m

8y=15m

y=15m/8

Co-ordinate of p=(m,15m/8)

And Q a point on the line

10y=3x

Y=3n/10

Asumming x=n
Coordinate of Q (n,3n/10)

Mid point through p&q=(8,6)

((M+n)/2,((75m+12n)/80)=(8,6)


(m+n)/2=8

m+n=16.....(1)


75m+12n=80×6

3(25m+3n)=80×3

25m+4n=160......(2)
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m+n=16.....(1)
25m+4n=160......(2)

n=16-m.......(3)

Plug 3 into 2

25m+4(16-m)=160

25m+64-4m=160

21m=96

m=96/21=32/7

N=16-32/7=(112-32)/7=80/7

Point p set it as (32/7,480/7)

And Q set up as (80/7,24/7)

Distance from p to q=√[(x2-x1)²+(y2-y1)²)]

PQ=√[(80/7-32/7)²+(24/7-480/7)²]=√(48)²+(456)²]/7=√(2304+207936)/7=√(210240)/7=(45852)/100

a/b=45852/700

a:b=45652:700=11463:175

a=11463+175=11638

That what I could think off

Answer 2

Correction
at
75m+12n=80×6

3(25m+4n)=80×2

25m+4n=160

typing this wasn't easy

Answer 3

a+b=11463+175=11638