The right triangle in the coordinate plane is rotated 270° clockwise about the point (2, 1) and then reflected across the y-axis to form triangle A'B'C'. The pre-image of triangle ABC has coordinates A(2,1), B(2,4), and C(6,1). What are the new coordinates?
To rotate $270^\circ$ clockwise about $(2,1)$, we can translate the triangle so that $(2,1)$ is at the origin, rotate about the origin, and then translate the triangle back.
[asy]
pair A, B, C, O, Ap, Bp, Cp;
A = (2,1);
B = (2,4);
C = (6,1);
O = (2,1);
Ap = rotate(270)*(A-O)+O;
Bp = rotate(270)*(B-O)+O;
Cp = rotate(270)*(C-O)+O;
size(6cm);
draw((-1,0)--(7,0),Arrow);
draw((0,-1)--(0,5),Arrow);
draw((A)--(B)--(C)--cycle,dashed);
draw((Ap)--(Bp)--(Cp)--cycle);
draw((A - (2,1))--(A + (2,1)),dashed,Arrows);
draw((B - (2,1))--(B + (2,1)),dashed,Arrows);
draw((C - (2,1))--(C + (2,1)),dashed,Arrows);
label("$x$",(7,0),(2,0));
label("$y$",(0,5),(0,2));
dot("$(2,1)$",(O),SE,linewidth(4.5));
[/asy]
Since $(2,1)$ is already the origin, the effect of the pre-image under the translation is to move $A$, $B$, and $C$ so their new locations are $A(0,0)$, $B(0,3)$, and $C(4,0)$. We note that a $270^\circ$ clockwise rotation about the origin is the same as a $90^\circ$ counterclockwise rotation about the origin. Thus, we have the following image under this rotation:
[asy]
pair A, B, C, O, Ap, Bp, Cp;
A = (0,0);
B = (0,3);
C = (4,0);
O = (2,1);
Ap = rotate(270)*(A-O)+O;
Bp = rotate(270)*(B-O)+O;
Cp = rotate(270)*(C-O)+O;
size(6cm);
draw((-1,0)--(7,0),Arrow);
draw((0,-1)--(0,5),Arrow);
draw((A)--(B)--(C)--cycle,dashed);
draw((Ap)--(Bp)--(Cp)--cycle);
draw((O)--(A),dashed);
draw((O)--(B),dashed);
draw((O)--(C),dashed);
label("$x$",(7,0),(2,0));
label("$y$",(0,5),(0,2));
dot("$(2,1)$",(O),SE,linewidth(4.5));
[/asy]
To find the coordinates of $A'$, $B'$, and $C'$, we trace where they move after this reflection across the $y$-axis:
[asy]
pair A, B, C, O, Ap, Bp, Cp;
A = (2,1);
B = (2,4);
C = (6,1);
O = (0,0);
Ap = reflect((-1,0),(1,0))*(A);
Bp = reflect((-1,0),(1,0))*(B);
Cp = reflect((-1,0),(1,0))*(C);
size(6cm);
draw((-7,0)--(3,0),Arrow);
draw((0,-1)--(0,5),Arrow);
draw((Ap)--(Bp)--(Cp)--cycle,dashed);
draw((O)--(Ap),dashed);
draw((O)--(Bp),dashed);
draw((O)--(Cp),dashed);
label("$x$",(3,0),(2,0));
label("$y$",(0,5),(0,2));
dot("$(0,0)$",(0,0),NW,linewidth(4.5));
[/asy]
This reflection changes the sign of the $x$-coordinate, so triangle $ABC$ is reflected into triangle $A'B'C'$, where $A'\left(0,1\right)$, $B'\left(0,-4\right)$, and $C'\left(-4,1\right)$. Hence the answer is $\boxed{\left(0,1\right), \left(0,-4\right), \text{ and } \left(-4,1\right)}$.
To determine the new coordinates of the triangle after rotation and reflection, follow these steps:
Step 1: Rotate the original triangle 270° clockwise about the point (2, 1).
To rotate a point (x, y) 270° clockwise, we can use the following formulas:
x' = cx + (x - cx) * cos(θ) + (y - cy) * sin(θ)
y' = cy - (x - cx) * sin(θ) + (y - cy) * cos(θ)
Where (cx, cy) is the center of rotation (in this case, (2, 1)), and θ is the angle of rotation (270°).
For point A(2, 1), applying the rotation formulas:
x' = 2 + (2 - 2) * cos(270°) + (1 - 1) * sin(270°)
y' = 1 - (2 - 2) * sin(270°) + (1 - 1) * cos(270°)
Cos(270°) = 0, Sin(270°) = -1
x' = 2 + 0 + 0 = 2
y' = 1 - 0 + 0 = 1
So, the new coordinates for A' are (2, 1) after the rotation.
For point B(2, 4), applying the rotation formulas:
x' = 2 + (2 - 2) * cos(270°) + (4 - 1) * sin(270°)
y' = 1 - (2 - 2) * sin(270°) + (4 - 1) * cos(270°)
x' = 2 + 0 + 3 = 5
y' = 1 - 0 + 0 = 1
So, the new coordinates for B' are (5, 1) after the rotation.
For point C(6, 1), applying the rotation formulas:
x' = 2 + (6 - 2) * cos(270°) + (1 - 1) * sin(270°)
y' = 1 - (6 - 2) * sin(270°) + (1 - 1) * cos(270°)
x' = 2 + 4 * 0 + 0 = 2
y' = 1 - 4 * (-1) + 0 = 5
So, the new coordinates for C' are (2, 5) after the rotation.
Step 2: Reflect the triangle across the y-axis.
To reflect a point across the y-axis, we simply change the sign of the x-coordinate. In this case, we can reflect A', B', and C' to get A", B", and C" respectively.
For A'(2, 1), reflecting across the y-axis:
A" = (-2, 1)
For B'(5, 1), reflecting across the y-axis:
B" = (-5, 1)
For C'(2, 5), reflecting across the y-axis:
C" = (-2, 5)
So, the new coordinates after rotation and reflection are:
A"(-2, 1), B"(-5, 1), and C"(-2, 5).
To find the new coordinates of triangle A'B'C', we need to follow two transformations:
1. Rotating 270° clockwise about the point (2, 1)
2. Reflecting across the y-axis
Let's first perform the rotation:
1. Rotating 270° clockwise about the point (2, 1):
To rotate a point (x, y) about a point (h, k) by an angle θ clockwise, we can use the following formulas:
x' = (x - h) * cos(θ) - (y - k) * sin(θ) + h
y' = (x - h) * sin(θ) + (y - k) * cos(θ) + k
In this case, the rotation center (h, k) is (2, 1), and the angle θ is 270°.
For point A(2, 1):
x' = (2 - 2) * cos(270°) - (1 - 1) * sin(270°) + 2
y' = (2 - 2) * sin(270°) + (1 - 1) * cos(270°) + 1
Simplifying the equations:
x' = 2
y' = 0
So, the new coordinates of point A after the rotation are A'(2, 0).
Similarly, applying the rotation to points B(2, 4) and C(6, 1):
For point B(2, 4):
x' = (2 - 2) * cos(270°) - (4 - 1) * sin(270°) + 2
y' = (2 - 2) * sin(270°) + (4 - 1) * cos(270°) + 1
Simplifying the equations:
x' = -3
y' = 0
So, the new coordinates of point B after the rotation are B'(-3, 0).
For point C(6, 1):
x' = (6 - 2) * cos(270°) - (1 - 1) * sin(270°) + 2
y' = (6 - 2) * sin(270°) + (1 - 1) * cos(270°) + 1
Simplifying the equations:
x' = -3
y' = 3
So, the new coordinates of point C after the rotation are C'(-3, 3).
Now, let's perform the reflection across the y-axis:
To reflect a point (x, y) across the y-axis, we simply change the sign of the x-coordinate.
For point A':
A''(x', y') = (-2, 0)
For point B':
B''(x', y') = (-(-3), 0)
B''(x', y') = (3, 0)
For point C':
C''(x', y') = (-(-3), 3)
C''(x', y') = (3, 3)
So, the final coordinates of triangle A'B'C' after the reflection are A''(-2, 0), B''(3, 0), and C''(3, 3).