Difference between revisions of "1957 AHSME Problems/Problem 23"

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Plugging this into either of the original equations, we get <math>(0,10)</math> and <math>(1,9)</math>. The distance between those two points is <math>\boxed{\textbf{(C) }\sqrt{2}}</math>
 
Plugging this into either of the original equations, we get <math>(0,10)</math> and <math>(1,9)</math>. The distance between those two points is <math>\boxed{\textbf{(C) }\sqrt{2}}</math>
 
==See Also==
 
==See Also==
{{AHSME box|year=1957|num-b=22|num-a=24}}
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{{AHSME 50p box|year=1957|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 
[[Category:AHSME]][[Category:AHSME Problems]]
 
[[Category:AHSME]][[Category:AHSME Problems]]

Revision as of 10:09, 25 July 2024

The graph of $x^2 + y = 10$ and the graph of $x + y = 10$ meet in two points. The distance between these two points is:

$\textbf{(A)}\ \text{less than 1} \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \sqrt{2}\qquad \textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{more than 2}$

Solution

We can merge the two equations to create $x^2+y=x+y$. Using either the quadratic equation or factoring, we get two solutions with $x$-coordinates $0$ and $1$.

Plugging this into either of the original equations, we get $(0,10)$ and $(1,9)$. The distance between those two points is $\boxed{\textbf{(C) }\sqrt{2}}$

See Also

1957 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AHSME Problems and Solutions

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