In the diagram, square $P Q R S$ has side length $25, Q$ is located at $(0,7)$, and $R$ is on the $x$-axis. The square is rotated clockwise about $R$ until $S$ lies above the $x$-axis on the line with equation $x=39$. What are the new coordinates of $P$ after this rotation?

![](https://cdn.mathpix.com/cropped/2024_04_17_ab5f9723a8aff65a55ebg-2.jpg?height=464&width=439&top_left_y=1801&top_left_x=1190)
Since the side length of the square is 25 , then $Q R=R S=S P=P Q=25$.

Since $Q$ has coordinates $(0,7)$, then $Q O=7$.

Since $\triangle Q O R$ is right-angled at $O$, then by the Pythagorean Theorem,

$$
O R=\sqrt{Q R^{2}-O Q^{2}}=\sqrt{25^{2}-7^{2}}=\sqrt{576}=24
$$

since $O R>0$. (We could also recognize the special 7-24-25 right-angled triangle.)

Let $T$ be the point where the line $x=39$ crosses the $x$-axis. $T$ has coordinates $(39,0)$.

Since $O R=24$, then $R T=O T-O R=39-24=15$.

Now rotate the square about $R$ until $S$ lies on the line $x=39$.

Let $U$ be the point on the line $x=39$ so that $P U$ is perpendicular to $U T$. Note that $U$ is above $S$ on the line $x=39$ since $P$ is above $S$.

![](https://cdn.mathpix.com/cropped/2024_04_17_6308d40ff1469d7b994ag-04.jpg?height=472&width=458&top_left_y=157&top_left_x=888)

Since $\triangle R T S$ is right-angled at $T$, then by the Pythagorean Theorem,

$$
S T=\sqrt{R S^{2}-R T^{2}}=\sqrt{25^{2}-15^{2}}=\sqrt{400}=20
$$

since $S T>0$. (We could also recognize the special 15-20-25 right-angled triangle.)

Since $S T$ is vertical, then $S$ has coordinates $(39,20)$.

Finally, $\triangle S U P$ is congruent to $\triangle R T S$, because each is right-angled, they have equal hypotenuses, and

$$
\angle S P U=90^{\circ}-\angle P S U=90^{\circ}-\left(180^{\circ}-90^{\circ}-\angle R S T\right)=\angle R S T
$$

Therefore, $P U=S T=20$ and $S U=R T=15$.

Since $S$ has coordinates $(39,20)$, then $U$ has coordinates $(39,20+15)=(39,35)$ and so $P$ has coordinates $(39-20,35)=(19,35)$.

ANSWER: $(19,35)$