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)
---------------------------------------
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
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.
3 answers
Correction
at
75m+12n=80×6
3(25m+4n)=80×2
25m+4n=160
typing this wasn't easy
at
75m+12n=80×6
3(25m+4n)=80×2
25m+4n=160
typing this wasn't easy
a+b=11463+175=11638