Difference between revisions of "2019 AMC 10B Problems/Problem 5"
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<math>\textbf{(D)}</math>: Repeating the argument for <math>\textbf{(C)}</math>, we see that both lines have slope <math>-1</math>, so this is also true. | <math>\textbf{(D)}</math>: Repeating the argument for <math>\textbf{(C)}</math>, we see that both lines have slope <math>-1</math>, so this is also true. | ||
− | <math>\textbf{(E)}</math>: | + | <math>\textbf{(E)}</math>: This is the only one left, presumably the answer. To prove: if point <math>A</math> has coordinates <math>(p,q)</math> and point <math>B</math> has coordinates <math>(r,s)</math>, then <math>A'</math> and <math>B'</math> will, respectively, have coordinates <math>(q,p)</math> and <math>(s,r)</math>. The product of the gradients of <math>AB</math> and <math>A'B'</math> is <math>\frac{s-q}{r-p} \cdot \frac{r-p}{s-q} = 1 \neq -1</math>, so in fact these lines are '''never''' perpendicular to each other (using the "negative reciprocal" condition for perpendicularity). |
Thus the answer is <math>\boxed{\textbf{(E)}}</math>. | Thus the answer is <math>\boxed{\textbf{(E)}}</math>. |
Latest revision as of 21:24, 5 October 2023
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.
: This is the only one left, presumably the answer. To prove: 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
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Video Solution
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See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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