Rotations: Practice

Apply $(x,y)\mapsto(y,-x)$ to every vertex.

$A'=(-1,4)$

$B'=(-1,1)$

$C'=(2,2)$

Keep these correspondences when joining the image.

Remove one full turn when locating the final image.

$450-360=90$

The remaining counterclockwise quarter-turn uses $(-y,x)$.

$(5,-2)\mapsto(2,5)$

A half-turn negates both starting coordinates. The equations are $-x=-6$ and $-y=-8$.

$x=6$

$y=8$

Negating these coordinates checks the given image.

Relative to O, T is three units right and two up. A clockwise quarter-turn changes three right to three down and two up to two right.

$x'=-2+2=0$

$y'=3-3=0$

The origin rule applied directly to T would use the wrong center.

V is four left and six above C. A half-turn places its image four right and six below C.

$x'=3+4=7$

$y'=-2-6=-8$

Distance preservation makes corresponding lengths equal.

$6r+5=53$

Subtract 5, then divide by 6.

$6r=48$

$r=8$

A rotation preserves angle measure, so

$2t+17=95$

Subtract 17 and divide by 2.

$2t=78$

$t=39$

The horizontal connector AA-prime has midpoint (1,0). Its perpendicular bisector is the vertical line $x=1$.

The vertical connector BB-prime has midpoint (2,2). Its perpendicular bisector is the horizontal line $y=2$.

These distinct lines meet at candidate $O(1,2)$, which is not either connector's midpoint. Check the directed turn from this candidate.

Relative to O, A is two left and two down, while A-prime is two right and two down. A counterclockwise quarter-turn sends left to down and down to right:

$(-2,-2)\mapsto(2,-2).$

Relative to O, B is one right and one down, while B-prime is one right and one up. The same counterclockwise quarter-turn gives

$(1,-1)\mapsto(1,1).$

The taught quarter-turn rule keeps each radius unchanged. It matches both pairs with the same directed angle, so a 90-degree counterclockwise rotation about (1,2) makes both matches.

Each point keeps its own radius, so their paths are different circles. Rotation preservation compares their distance before and after the turn; it does not require equal travel distances for points at different radii.

For a nonidentity turn in this range, only the center remains fixed. An off-center point would move around its positive-radius circle, so the fixed point F must be the center.

A 270-degree counterclockwise turn uses $(x,y)\mapsto(y,-x)$. Thus the image is $(-4,-a)$.

$-a=-9$

$a=9$

The first image coordinate already agrees.

The radius is fixed. Turning 80 degrees clockwise gives one ray, and turning 80 degrees counterclockwise gives a different ray.

$80+80=160$

The two directions have not completed a full turn between them, so the image points are distinct. Each ray has one point at radius OP.