Let P be the point of intersection of the lin | vedclass

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(D) The point of intersection $P$ of lines $L_1: x-y-7=0$ and $L_2: x+y-5=0$ is $(6, -1)$.For point $A(x_1, y_1)$ on $L_1$ with $PA=3\sqrt{2}$,the sym

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$\cos^{-1}\left(\frac{9}{\sqrt{85}}\right)$

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(D) The point of intersection $P$ of lines $L_1: x-y-7=0$ and $L_2: x+y-5=0$ is $(6, -1)$.For point $A(x_1, y_1)$ on $L_1$ with $PA=3\sqrt{2}$,the symmetric form is $\frac{x-6}{\cos 	heta_1} = \frac{y+1}{\sin 	heta_1} = \pm 3\sqrt{2}$,where $	an 	heta_1 = 1$ (slope of $L_1$ is $1$).Thus,$x_1 = 6 \pm 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 6 \pm 3$ and $y_1 = -1 \pm 3\sqrt{2} \cdot \frac{1}{\sqrt{2}} = -1 \pm 3$.Since $x_1, y_1 \geq 0$,we choose the positive sign: $A = (9, 2)$.For point $B(x_2, y_2)$ on $L_2$ with $PB=\sqrt{2}$,the symmetric form is $\frac{x-6}{\cos 	heta_2} = \frac{y+1}{\sin 	heta_2} = \pm \sqrt{2}$,where $	an 	heta_2 = -1$ (slope of $L_2$ is $-1$).Thus,$x_2 = 6 \pm \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 6 \pm 1$ and $y_2 = -1 \pm \sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}}ight) = -1 \mp 1$.Since $x_2, y_2 \geq 0$,we choose $x_2 = 6-1=5$ and $y_2 = -1+1=0$,so $B = (5, 0)$.Let $O$ be the origin $(0, 0)$. The vectors are $\vec{OA} = (9, 2)$ and $\vec{OB} = (5, 0)$.The angle $	heta$ at the origin is given by $\cos 	heta = \frac{\vec{OA} \cdot \vec{OB}}{|\vec{OA}| |\vec{OB}|} = \frac{9 \cdot 5 + 2 \cdot 0}{\sqrt{9^2+2^2} \sqrt{5^2+0^2}} = \frac{45}{\sqrt{85} \cdot 5} = \frac{9}{\sqrt{85}}$.Therefore,$	heta = \cos^{-1}\left(\frac{9}{\sqrt{85}}ight)$.

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