# 1969 AHSME Problems/Problem 11

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## 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:

$\text{(A) } -\tfrac{3}{5}\quad \text{(B) } -\tfrac{2}{5}\quad \text{(C) } -\tfrac{1}{5}\quad \text{(D) } \tfrac{1}{5}\quad \text{(E) either }-\tfrac{1}{5}\text{ or} \tfrac{1}{5}.$

## Solution

$[asy] import graph; size(7.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-2.2,xmax=6.2,ymin=-4.2,ymax=4.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); dot((-1,-2)); label("P",(-1,-2),NW); dot((4,2)); label("Q",(4,2),NW); draw((-1,-2)--(4,2),dotted); dot((1,-0.4)); label("R",(1,-0.4),SE); [/asy]$ By the Triangle Inequality, $PR + QR \ge PR$, and equality holds if $R$ is on $PQ$. The equation of the line with $P$ and $Q$ is $y = \frac{4}{5}x - \frac{6}{5}$, so point $R$ is $(1,-\frac{2}{5})$. Thus, $m = \boxed{\textbf{(B) } -\frac{2}{5}}$.