Minimize Sum of Distances

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:

  1. -3/5
  2. -2/5
  3. -1/5
  4. 1/5
  5. either -1/5 or 1/5


Show/Hide Solution

If PR + RQ is a minimum, then R is between P and Q, that is, it is on the segment PQ. So we tease out the equation of the line PQ:

We start with P (-1,-2), Q (4,2), and y=mx + b.

m = (2 – (-2))/(4 – (-1)) = 4/5.

So y = 4/5x + b.

Plug in Q: 2 = 4 · 4/5 + b = 16/5 + b = 10/5 → b = -6/5.

So line PQ is y = 4x/5 – 6/5.

R is (1,m), where m is the y-coordinate, not a slope m.

Then y = 4/5 · 1 – 6/5 = -2/5, so R is (1,-2/5) and our answer is (b).

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