Difference between revisions of "1969 AHSME Problems/Problem 11"

(Created page with "== Problem == Given points <math>P(-1,-2)</math> and <math>Q(4,2)</math> in the <math>xy</math>-plane; point <math>R(1,m)</math> is taken so that <math>PR+RQ</math> is a minimum...")
(No difference)

Revision as of 17:34, 30 September 2014

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

$\fbox{B}$

See also

1969 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png