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Difference between revisions of "1992 AHSME Problems/Problem 3"

(→‎Solution)
(Reworded solution 1 and added a second solution)
 
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We are give the points <math>(m,3)</math> and <math>(1,m)</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>  
 
Substituting into the slope formula, we get <math>\frac{m-3}{1-m}</math>  
−
After taking the cross-products and solving, we get <math>\text{(C) } \sqrt{3}\quad</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 22:42, 4 October 2016

Problem

If $m>0$ and the points $(m,3)$ and $(1,m)$ lie on a line with slope $m$, then $m=$

$\text{(A) } 1\quad \text{(B) } \sqrt{2}\quad \text{(C) } \sqrt{3}\quad \text{(D) } 2\quad \text{(E) } \sqrt{5}$

Solution

We know that the formula for slope is $m = \frac{y_2-y_1}{x_2-x_1}$ We are give the points $(m,3)$ and $(1,m)$. Substituting into the slope formula, we get $\frac{m-3}{1-m}$ After taking the cross-products and solving, we get $\sqrt{3}$, which is answer choice $\fbox{C}$.

Solution 2

Using the formula for slope as above, we know that $m=\frac{m-3}{1-m}$. Multiplying both sides of this equation by $1-m$, we get that $m(1-m)=m-3$ which expands to $m-m^2=m-3$. This forms the quadratic equation $-m^2+3=0$. This means that $m^2=3$, so $m=\sqrt{3}$, which corresponds to answer choice $\fbox{C}$.

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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