2021 Fall AMC 10B Problems/Problem 17
Contents
Problem
Distinct lines and lie in the -plane. They intersect at the origin. Point is reflected about line to point , and then is reflected about line to point . The equation of line is , and the coordinates of are . What is the equation of line
Solution 1
Denote as the origin.
Even though the problem is phrased as a coordinate bash, that looks disgusting. Instead, let's try to phrase this problem in terms of Euclidean geometry, using the observation that , and that both and must pass through in order to preserve the distance from to the origin. ( and are just defined as points on lines and .) Because of how reflections work, we have that and ; adding these two equations together and using angle addition, we have that . Since the sum of both sides combined must be by angle addition, This is helpful! We can now return to using coordinates, with this piece of information in mind: The angle is a little bit unwieldy in the coordinate plane, so we should try to make a triangle. Let be a point on ; to make fit nicely in the diagram, let it be . Now, let's draw a perpendicular to through point , intersecting at point . is a triangle, so is a degree counterclockwise rotation from about . Therefore, the coordinates of are So, is a point on line , which we already know passes through the origin; therefore, 's equation is
~ihatemath123
(We never actually had to use the information of the exact coordinates of ; as long as , when we move around, this will not affect 's equation.)
Solution 2
It is well known that the composition of 2 reflections , one after another, about two lines and , respectively, that meet at an angle is a rotation by around the intersection of and .
Now, we note that is a 90 degree rotation clockwise of about the origin, which is also where and intersect. So is a 45 degree rotation of about the origin clockwise.
To rotate 90 degrees clockwise, we build a square with adjacent vertices and . The other two vertices are at and . The center of the square is at , which is the midpoint of and . The line passes through the origin and the center of the square we built, namely at and . Thus the line is . The answer is .
~hurdler
~minor edits by nightshade2526
Solution 3
We know that the equation of line is . This means that is reflected over the line . This means that the line with and is perpendicular to , so it has slope . Then the equation of this perpendicular line is , and plugging in for and yields .
The midpoint of and lies at the intersection of and . Solving, we get the x-value of the intersection is and the y-value is . Let the x-value of be - then by the midpoint formula, . We can find the y-value of the same way, so .
Now we have to reflect over to get to . The midpoint of and will lie on , and this midpoint is, by the midpoint formula, . must satisfy this point, so .
Now the equation of line is
~KingRavi
Solution 4
First, use Solution 1's method to get and that the angle between lines and is . From here, note that the slope of line is less than that of line as otherwise wouldn't even be close to . Thus, line is a clockwise rotation of line . Line makes an angle of with the positive x axis. Thus, line makes an angle of with the positive x axis. Thus, the slope of line is by the tangent addition formula. Since the slope of line is , its equation is , which is choice .
Solution 5(cheese)
When we graph all the lines and points with a ruler, you can see that a slope of is too big while is too small. We also see that the slope cannot be negative, therefore the answer is ~ agentdabber
Video Solution
~hurdler
Video Solution 2 (by Interstigation)
https://www.youtube.com/watch?v=KdrYlPmqqv0
~Interstigation
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=PgFX55o6h1g
~IceMatrix
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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All AMC 10 Problems and Solutions |
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