Difference between revisions of "2021 Fall AMC 10B Problems/Problem 17"
(→Solution 2) |
m (→Solution 1) |
||
Line 10: | Line 10: | ||
Now, we note that <math>(4,1)</math> is a 90 degree rotation clockwise of <math>(-1,4)</math> about the origin, which is also where <math>l</math> and <math>m</math> intersect. So <math>m</math> is a 45 degree rotation of <math>l</math> about the origin clockwise. | Now, we note that <math>(4,1)</math> is a 90 degree rotation clockwise of <math>(-1,4)</math> about the origin, which is also where <math>l</math> and <math>m</math> intersect. So <math>m</math> is a 45 degree rotation of <math>l</math> about the origin clockwise. | ||
− | To rotate <math>l</math> 90 degrees clockwise, we build a square with adjacent vertices <math>(0,0)</math> and <math>(1,5)</math>. The other two vertices are at <math>(5,-1)</math> and <math>(6,4)</math>. The center of the square is at <math>(3,2)</math>, which is the midpoint of <math>(1,5)</math> and <math>(5,-1)</math>. The line <math>m</math> passes through the origin and the center of the square we built, namely at <math>(0,0)</math> and <math>(3,2)</math>. Thus the line is <math>y = \frac{2}{3} x</math>. The answer is '''(D)''' <math>\boxed{3y | + | To rotate <math>l</math> 90 degrees clockwise, we build a square with adjacent vertices <math>(0,0)</math> and <math>(1,5)</math>. The other two vertices are at <math>(5,-1)</math> and <math>(6,4)</math>. The center of the square is at <math>(3,2)</math>, which is the midpoint of <math>(1,5)</math> and <math>(5,-1)</math>. The line <math>m</math> passes through the origin and the center of the square we built, namely at <math>(0,0)</math> and <math>(3,2)</math>. Thus the line is <math>y = \frac{2}{3} x</math>. The answer is '''(D)''' <math>\boxed{2x - 3y = 0}</math>. |
− | ~hurdler | + | ~hurdler, minor edits by nightshade2526 |
==Solution 2== | ==Solution 2== |
Revision as of 21:56, 25 January 2022
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
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 (D)
.
~hurdler, minor edits by nightshade2526
Solution 2
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
Video Solution
~hurdler
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.