Problem: Given points P(−1,−2) and Q(4,2) in the xy-plane; point R(1,m) is taken so that PR+RQ is a minimum. Then m equals:
Answer Choices:
A. −53
B. −52
C. −51
D. 51
E. either −51 or 51.
Solution:
Method I. Since PR+RQ is a minimum, points P1,R1 and Q are collinear. Therefore, −1−1−2−m=−1−4−2−2 so that m=−52.
Method II. Since PR+RQ is a minimum, points P1,R1 and Q are collinear. Therefore, PR+RQ=PQ so that (1−(−1))2+(m+2)2+(4−1)2+(2−m)2=(4−(−1))2+(2+2)2