2720.11 – Minimizing the Sum of Distances

Given points P(−1,−2)P(-1,-2) and Q(4,2)Q(4,2) in the cartesian plane, point R(1,n)R(1,n) is taken so that the sum of the distances PR+RQPR+RQ is the smallest it can be. What is the value of nn?

  1. n=−0.6,n = -0.6,
  2. n=−0.4,n = -0.4,
  3. n=−0.2,n = -0.2,
  4. n=0.2,n = 0.2,
  5. n=±0.2.n = \pm0.2.

Bonus: draw a clear diagram illustrating your solution.

Solution

If PR+RQPR+RQ is at its smallest, then RR must be some point on the line PQPQ, so let’s start by finding the equation y=mx+by = mx + b of that line. First the slope:

m=△y△x=2−(−2)4−(−1)=45.m = \frac{\triangle y}{\triangle x} = \frac{2-(-2)}{4-(-1)} = \frac{4}{5}.

So far the equation is y=0.8x+by = 0.8x+b. To find bb, we plug in a point. We’ll use QQ.

y=0.8x+b→2=0.8(4)+b→2=3.2+b→b=−1.2.y = 0.8x+b \rightarrow 2=0.8(4)+b \rightarrow 2=3.2+b \rightarrow b=-1.2.

Therefore, the line PQPQ has the equation y=0.8x−1.2y=0.8x-1.2.

To find RR, we simply plug in 11 to the equation, and our y-value will be nn:

y=0.8x−1.2→n=0.8(1)−1.2→n=0.8−1.2→n=−0.4.y = 0.8x-1.2 \rightarrow n = 0.8(1)-1.2 \rightarrow n = 0.8-1.2 \rightarrow n = -0.4.

Point RR is therefore (1,−0.4)(1,-0.4) and the correct answer is b.