1992 AHSME Problems/Problem 3

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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