# 2720.11 – Minimizing the Sum of Distances

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

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

Bonus: draw a clear diagram illustrating your solution.

Solution

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

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

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

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

Therefore, the line $PQ$ has the equation $y=0.8x-1.2$.

To find $R$, we simply plug in $1$ to the equation, and our y-value will be $n$:

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

Point $R$ is therefore $(1,-0.4)$ and the correct answer is b.