Difference between revisions of "1992 AHSME Problems/Problem 3"
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(Reworded solution 1 and added a second solution) |
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== Solution == | == Solution == | ||
− | <math>\fbox{C}</math> | + | We know that the formula for slope is <math>m = \frac{y_2-y_1}{x_2-x_1}</math> |
+ | We are give the points <math>(m,3)</math> and <math>(1,m)</math>. | ||
+ | Substituting into the slope formula, we get <math>\frac{m-3}{1-m}</math> | ||
+ | After taking the cross-products and solving, we get <math>\sqrt{3}</math>, which is answer choice <math>\fbox{C}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | Using the formula for slope as above, we know that <math>m=\frac{m-3}{1-m}</math>. Multiplying both sides of this equation by <math>1-m</math>, we get that <math>m(1-m)=m-3</math> which expands to <math>m-m^2=m-3</math>. This forms the quadratic equation <math>-m^2+3=0</math>. This means that <math>m^2=3</math>, so <math>m=\sqrt{3}</math>, which corresponds to answer choice <math>\fbox{C}</math>. | ||
== See also == | == See also == |
Latest revision as of 21:42, 4 October 2016
Contents
Problem
If and the points and lie on a line with slope , then
Solution
We know that the formula for slope is We are give the points and . Substituting into the slope formula, we get After taking the cross-products and solving, we get , which is answer choice .
Solution 2
Using the formula for slope as above, we know that . Multiplying both sides of this equation by , we get that which expands to . This forms the quadratic equation . This means that , so , which corresponds to answer choice .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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 |
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