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

If $(p, q)$ is a point on line $L$ , then by symmetry $(q, p)$ must be a point on $K$ . Therefore, the points on $K$ satisfy $x=ay+b$ .Solving for $y$ yields $y = \dfrac xa-\dfrac ba$ . $\Longrightarrow \boxed{E}$

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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