Compositions and Congruence: Practice

The clockwise quarter-turn uses $(y,-x)$, so the intermediate point is $P'(4,3)$.

Apply the translation to that point.

$x''=4+2=6$

$y''=3-5=-2$

Thus $P''(6,-2)$.

Undo the slide first: subtract 3 from the horizontal coordinate and 1 from the height.

$x'=-5-3=-8$

$y'=4-1=3$

Reflect $N'(-8,3)$ across the same y-axis to recover $N(8,3)$.

The second mirror is five units left of the first, so the translation is ten left.

$2(5)=10$

The horizontal coordinate becomes $7-10=-3$, and height stays 2. A direct reflection check gives $J'(1,2)$, then $J''(-3,2)$.

Two parallel reflections translate through twice the gap. Divide the travel distance by 2.

$18\div2=9$

The mirrors are 9 cm apart.

Double the directed angle between the chosen rays.

$2(120^\circ)=240^\circ$

The composition turns counterclockwise about O. Choosing the opposite ray of a mirror would change the result by a full turn, without changing the final image.

Across the x-axis first, the point becomes $(6,2)$; reflecting next across the y-axis gives $(-6,2)$.

Across the y-axis first, it becomes $(-6,-2)$; the x-axis reflection again gives $(-6,2)$.

Both coordinates reverse signs.

$(x,y)\mapsto(-x,-y)$

The perpendicular mirrors commute as a half-turn about the origin.

The horizontal mirror changes height to twice 2 minus the old height.

$y'=4-(-3)=7$

The intermediate point is $R'(-1,7)$. Add 5 to its horizontal coordinate.

$-1+5=4$

Thus $R''(4,7)$.

The vertical mirror changes only the horizontal coordinate.

$x'=2(-1)-x=-2-x$

The parallel translation adds 4 to the height. Since one step acts across the mirror and the other along it, either order gives the displayed rule.

The endpoints J and K correspond to U and V, so distance preservation gives $UV=8$ cm.

The vertex K corresponds to V. Angle preservation gives

$m\angle UVW=49^\circ.$

Both properties survive every rigid step.

The given steps are actual rigid motions, which take each straight side to the segment between its image endpoints. The same fold and slide carry the enclosed interior with that boundary. Thus this sequence supplies the required whole-triangle match.

Each step preserves the segment length, so the initial and final lengths agree.

$3t+4=28$

Subtract 4 and divide by 3.

$3t=24$

$t=8$

Reverse the steps and invert each motion. First undo the parallel translation with its opposite displacement. Then reflect in the same mirror, which undoes the original reflection.