Difference between revisions of "2021 Fall AMC 10B Problems/Problem 17"

Revision as of 11:03, 14 December 2021

Problem

Distinct lines $\ell$ and $m$ lie in the $xy$ -plane. They intersect at the origin. Point $P(-1, 4)$ is reflected about line $\ell$ to point $P'$ , and then $P'$ is reflected about line $m$ to point $P''$ . The equation of line $\ell$ is $5x - y = 0$ , and the coordinates of $P''$ are $(4,1)$ . What is the equation of line $m?$

$(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0$

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) $\boxed{3y - 2x = 0}$ .

~hurdler

Solution 2

We know that the equation of line $\ell$ is $y = 5x$ . This means that $P'$ is $(-1,4)$ reflected over the line $y = 5x$ . This means that the line with $P$ and $P'$ 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 $P'$ and $P$ lies at the intersection of $y = 5x$ 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 $P'$ be $x'$ - then by the midpoint formula, $\frac{x' - 1}{2} = \frac{19}{26} \implies x' = \frac{32}{13}$ . We can find the y-value of $P'$ the same way, so $P' = (\frac{32}{13},\frac{43}{13})$ .

Now we have to reflect $P'$ over $m$ to get to $(4,1)$ . The midpoint of $P'$ and $P''$ will lie on $m$ , and this midpoint is, by the midpoint formula, $(\frac{42}{13},\frac{28}{13})$ . $y = mx$ 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 $y = \frac{2}{3}x \implies 2x-3y = 0 = \boxed{D}$

~KingRavi

Video Solution

Solution 2021 Fall 10B #17

~hurdler

@@ Line 26: / Line 26: @@
 ~KingRavi
+==Video Solution==
+[https://youtu.be/bqCacR8aXmc|Video Solution 2021 Fall 10B #17]
+~hurdler
 ==See Also==
 {{AMC10 box|year=2021 Fall|ab=B|num-a=18|num-b=16}}
 {{MAA Notice}}

2021 Fall AMC 10B (Problems • Answer Key • Resources)
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