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Difference between revisions of "1981 AHSME Problems/Problem 10"

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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>
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==Problem 10==
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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>
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<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>
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==Solution==
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Recall that the reflection of any point <math>(p, q)</math> across the line <math>y = x</math> is <math>(q, p)</math>.  Therefore, any point <math>(p, q)</math> lies on <math>L</math> if and only if <math>(q, p)</math> lies on <math>K</math>. Therefore, any point <math>(r, s)</math> lies on <math>K</math> if and only if <math>(s, r)</math> lies on <math>L</math>, that is, <math>r=as+b</math>. Therefore, line <math>K</math> has equation <math>x = ay + b</math>, and isolating <math>y</math> yields <math>y = \boxed{(\textbf{E})\ \frac{1}{a}x-\frac{b}{a}}</math>.
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==See also==
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{{AHSME box|year=1981|num-b=9|num-a=11}}
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{{MAA Notice}}

Latest revision as of 16:29, 18 August 2025

Problem 10

The lines $L$ and $K$ are symmetric to each other with respect to the line $y=x$. If the equation of the line $L$ is $y=ax+b$ with $a\neq 0$ and $b\neq 0$, then the equation of $K$ is $y=$

$\textbf{(A)}\ \dfrac{1}{a}x+b\qquad\textbf{(B)}\ -\dfrac{1}{a}x+b\qquad\textbf{(C)}\ \dfrac{1}{a}x-\dfrac{b}{a}\qquad\textbf{(D)}\ \dfrac{1}{a}x+\dfrac{b}{a}\qquad\textbf{(E)}\ \dfrac{1}{a}x-\dfrac{b}{a}$

Solution

Recall that the reflection of any point $(p, q)$ across the line $y = x$ is $(q, p)$. Therefore, any point $(p, q)$ lies on $L$ if and only if $(q, p)$ lies on $K$. Therefore, any point $(r, s)$ lies on $K$ if and only if $(s, r)$ lies on $L$, that is, $r=as+b$. Therefore, line $K$ has equation $x = ay + b$, and isolating $y$ yields $y = \boxed{(\textbf{E})\ \frac{1}{a}x-\frac{b}{a}}$.

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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