1981 AHSME Problems/Problem 10

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 $y = \boxed{(\textbf{E})\ \frac{1}{a}x-\frac{b}{a}}$ .

1981 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1981 AHSME Problems/Problem 10

Problem 10

Solution

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