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

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== See also ==
{{AHSME box|year=1969|num-b=10|num-a=12}}   
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[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
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{{MAA Notice}}

Revision as of 17:02, 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 AHSC (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 31 32 33 34 35
All AHSME Problems and Solutions

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