2021 AMC 10 B Fall

Complete problem set with solutions and individual problem pages

Problem 17 Hard

Distinct lines \ell and m lie in the x y-plane. They intersect at the origin. Point P(-1,4) is reflected about line \ell to point P^{\prime}, and then P^{\prime} is reflected about line m to point P^{\prime \prime}. The equation of line \ell is 5 x-y=0, and the coordinates of P^{\prime \prime} are (4,1). What is the equation of line m ?(2021 AMC Fall 10B, Question #17)

  • A.

    5 x+2 y=0

  • B.

    3 x+2 y=0

  • C.

    x-3 y=0

  • D.

    2 x-3 y=0

  • E.

    5 x-3 y=0

Answer:D

Solution 1:

It is well known that the composition of 2 reflections, one after another, about two lines l and m, respectively, that meet at an angle \theta is a rotation by 2 \theta around the intersection of l and m. Now, we note that (4,1) is a 90 degree rotation clockwise of (-1,4) about the origin, which is also where l and m intersect. So m is a 45 degree rotation of l about the origin clockwise. To rotate l 90 degrees clockwise, we build a square with adjacent vertices (0,0) and (1,5). The other two vertices are at (5,-1) and (6,4). The center of the square is at (3,2), which is the midpoint of (1,5) and (5,-1). The line m passes through the origin and the center of the square we built, namely at (0,0) and (3,2). Thus the line is y=\frac{2}{3} x. The answer is (D) 2 x-3 y=0.

Solution 2:

We know that the equation of line \ell is y=5 x. This means that P^{\prime} is (-1,4) reflected over the line y=5 x. This means that the line with P and P^{\prime} is perpendicular to \ell, so it has slope -\frac{1}{5}. Then the equation of this perpendicular line is y=-\frac{1}{5} x+c, and plugging in (-1,4) for x and y yields c=\frac{19}{5}.

The midpoint of P^{\prime} and P lies at the intersection of y=5 x and y=-\frac{1}{5} x+\frac{19}{5}. Solving, we get the x-value of the intersection is \frac{19}{26} and the y-value is \frac{95}{26}. Let the x-value of P^{\prime} be x^{\prime} then by the midpoint formula, \frac{x^{\prime}-1}{2}=\frac{19}{26} \Longrightarrow x^{\prime}=\frac{32}{13}. We can find the y-value of P^{\prime} the same way, so P^{\prime}=\left(\frac{32}{13}, \frac{43}{13}\right).

Now we have to reflect P^{\prime} over m to get to (4,1). The midpoint of P^{\prime} and P^{\prime \prime} will lie on m, and this midpoint is, by the midpoint formula, \left(\frac{42}{13}, \frac{28}{13}\right) . y=m x must satisfy this point, so m=\frac{\frac{28}{13}}{\frac{42}{13}}=\frac{28}{42}=\frac{2}{3} Now the equation of line m is y=\frac{2}{3} x \Longrightarrow 2 x-3 y=0=D

Related Problems
Problem 15 2021 AMC 10 B Fall
Problem 16 2021 AMC 10 B Fall
Problem 18 2021 AMC 10 B Fall
Problem 19 2021 AMC 10 B Fall
Think Academy Powered by ChatGPT