Problem:
Let R=(8,6). The lines whose equations are 8y=15x and 10y=3x contain points P and Q, respectively. such that R is the midpoint of PQ​. The length PQ equals m/n. where m and n are relatively prime positive integers. Find m+n.
Solution:
Let O=(0,0). The line through R that is parallel to OQ​ has equation 10y=3x+36. This line meets OP at A=(716​,730​). Because R is the midpoint of PQ​, it follows that A is the midpoint of OP. Then P=(732​,760​), and PQ=2PR=2(724​)2+(718​)2​=2⋅76​⋅42+32​=760​. Thus m+n=67​.
OR
Let P=(8t,15t) and Q=(10u,3u). Because R is the midpoint of PQ​, it follows that
8t+10u15t+3u​=16 and =12​
The solution to this system is t=74​ and u=78​, so P=(732​,760​),Q=(780​,724​), and PQ=71​482+362​=712​42+32​=760​. Thus m+n=67​.
