1981 AHSME Problems/Problem 10: Difference between revisions
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==Problem 10== | |||
The lines <math>L</math> and <math>K</math> are symmetric to each other with respect to the line <math>y=x</math>. If the equation of the line <math>L</math> is <math>y=ax+b</math> with <math>a\neq 0</math> and <math>b\neq 0</math>, then the equation of <math>K</math> is <math>y=</math> | |||
<math> \textbf{(A)}\ \dfrac{1}{a}x+b\qquad\textbf{(B)}\ -\dfrac{1}{a}x+b\qquad\textbf{(C)}\ \dfrac{1}{a}x-\dfrac{b}{a}\qquad\textbf{(D)}\ \dfrac{1}{a}x+\dfrac{b}{a}\qquad\textbf{(E)}\ \dfrac{1}{a}x-\dfrac{b}{a} </math> | |||
==Solution== | |||
If <math>(p, q)</math> is a point on line <math>L</math>, then by symmetry <math>(q, p)</math> must be a point on <math>K</math>. Therefore, the points on <math>K</math> satisfy <math>x=ay+b</math>.Solving for <math>y</math> yields <math>y = \dfrac xa-\dfrac ba</math>. <math>\Longrightarrow \boxed{E}</math> | If <math>(p, q)</math> is a point on line <math>L</math>, then by symmetry <math>(q, p)</math> must be a point on <math>K</math>. Therefore, the points on <math>K</math> satisfy <math>x=ay+b</math>.Solving for <math>y</math> yields <math>y = \dfrac xa-\dfrac ba</math>. <math>\Longrightarrow \boxed{E}</math> | ||
==See also== | |||
{{AHSME box|year=1981|num-b=9|num-a=11}} | |||
{{MAA Notice}} | |||
Revision as of 12:42, 28 June 2025
Problem 10
The lines 
and 
are symmetric to each other with respect to the line 
. If the equation of the line is 
with 
and 
, then the equation of is 
Solution
If 
is a point on line , then by symmetry 
must be a point on . Therefore, the points on satisfy 
.Solving for 
yields 
. 
See also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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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