Question

chapter ii.7
score: 1/5 answered: 4/5
question 5
rotate the point (6, 8) around (18, 17) by the angle (\frac{2}{8},\frac{\sqrt{60}}{8}).

Explanation:

1. First, translate the points so that the center of rotation is at the origin:

• Let the center of rotation \(C=(18,17)\) and the point to be rotated \(P=(6,8)\).
• Translate the point \(P\) relative to \(C\): \(P'=(6 - 18,8 - 17)=(- 12,-9)\).

1. Recall the rotation - matrix formula for a 2D rotation. If \(\cos\theta=\frac{2}{8}=\frac{1}{4}\) and \(\sin\theta=\frac{\sqrt{60}}{8}=\frac{\sqrt{15}}{4}\), the rotation matrix \(R\) is \(

$$\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$$

=

$$\begin{pmatrix}\frac{1}{4}&-\frac{\sqrt{15}}{4}\\\frac{\sqrt{15}}{4}&\frac{1}{4}\end{pmatrix}$$

\).

1. Multiply the translation - adjusted point \(P'\) by the rotation matrix:

• \(P'' = R\times P'=

$$\begin{pmatrix}\frac{1}{4}&-\frac{\sqrt{15}}{4}\\\frac{\sqrt{15}}{4}&\frac{1}{4}\end{pmatrix}$$

\times

$$\begin{pmatrix}-12\\-9\end{pmatrix}$$

\).

• Calculate the first component of \(P''\):
• \(x_{P''}=\frac{1}{4}\times(-12)-\frac{\sqrt{15}}{4}\times(-9)=-3+\frac{9\sqrt{15}}{4}\).
• Calculate the second component of \(P''\):
• \(y_{P''}=\frac{\sqrt{15}}{4}\times(-12)+\frac{1}{4}\times(-9)=-3\sqrt{15}-\frac{9}{4}\).

1. Translate the rotated point back to the original coordinate system:

• The final point \(P_{final}=(x_{P''}+18,y_{P''}+17)\).
• \(x = - 3+\frac{9\sqrt{15}}{4}+18=15+\frac{9\sqrt{15}}{4}\).
• \(y=-3\sqrt{15}-\frac{9}{4}+17 = 17-\frac{9}{4}-3\sqrt{15}= \frac{68 - 9}{4}-3\sqrt{15}=\frac{59}{4}-3\sqrt{15}\).

Answer:

\((15 + \frac{9\sqrt{15}}{4},\frac{59}{4}-3\sqrt{15})\)