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=71482+362=71242+32=760. Thus m+n=67.
