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 $\boxed{\left(0,1\right), \left(0,-4\right), \text{ and } \left(-4,1\right)}$.
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?
1 answer
1 answer