2019 AMC 10B Problems/Problem 5
Problem
Triangle lies in the first quadrant. Points , , and are reflected across the line to points , , and , respectively. Assume that none of the vertices of the triangle lie on the line . Which of the following statements is not always true?
Triangle lies in the first quadrant.
Triangles and have the same area.
The slope of line is .
The slopes of lines and are the same.
Lines and are perpendicular to each other.
Solution
Let's analyze all of the options separately.
: Clearly is true, because a point in the first quadrant will have non-negative - and -coordinates, and so its reflection, with the coordinates swapped, will also have non-negative - and -coordinates.
: The triangles have the same area, since and are the same triangle (congruent). More formally, we can say that area is invariant under reflection.
: If point has coordinates , then will have coordinates . The gradient is thus , so this is true. (We know since the question states that none of the points , , or lies on the line , so there is no risk of division by zero).
: Repeating the argument for , we see that both lines have slope , so this is also true.
: By process of elimination, this must now be the answer. Indeed, if point has coordinates and point has coordinates , then and will, respectively, have coordinates and . The product of the gradients of and is , so in fact these lines are never perpendicular to each other (using the "negative reciprocal" condition for perpendicularity).
Thus the answer is .
Counterexamples
If and , then the slope of , , is , while the slope of , , is . is the reciprocal of , but it is not the negative reciprocal of . To generalize, let denote the coordinates of point , let denote the coordinates of point , let denote the slope of segment , and let denote the slope of segment . Then, the coordinates of are , and of are . Then, , and , so slopes arent negative reciprocals of each other.
Video Solution
~savannahsolver
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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