The angles made with the X-axis by the two li | vedclass

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(C) Let the slope of the line be $m$. The equation of the line passing through $P(1, 2)$ is $y-2 = m(x-1)$,which simplifies to $mx - y + (2-m) = 0$. T

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$\frac{\pi}{12}$ and $\frac{5 \pi}{12}$

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(C) Let the slope of the line be $m$. The equation of the line passing through $P(1, 2)$ is $y-2 = m(x-1)$,which simplifies to $mx - y + (2-m) = 0$. The intersection point of this line with $x+y=4$ is found by solving the system: $x + (mx + 2 - m) = 4$ $\Rightarrow x(1+m) = m+2$ $\Rightarrow x = \frac{m+2}{m+1}$. Then $y = 4 - x = 4 - \frac{m+2}{m+1} = \frac{4m+4-m-2}{m+1} = \frac{3m+2}{m+1}$. The distance between $P(1, 2)$ and the intersection point is given as $\frac{\sqrt{6}}{3}$. Using the distance formula: $\sqrt{(\frac{m+2}{m+1} - 1)^2 + (\frac{3m+2}{m+1} - 2)^2} = \frac{\sqrt{6}}{3}$. $\sqrt{(\frac{1}{m+1})^2 + (\frac{m}{m+1})^2} = \frac{\sqrt{6}}{3}$ $\Rightarrow \frac{\sqrt{1+m^2}}{|m+1|} = \frac{\sqrt{6}}{3}$. Squaring both sides: $\frac{1+m^2}{(m+1)^2} = \frac{6}{9} = \frac{2}{3}$. $3(1+m^2) = 2(m^2+2m+1)$ $\Rightarrow 3+3m^2 = 2m^2+4m+2$ $\Rightarrow m^2-4m+1 = 0$. Solving for $m$: $m = \frac{4 \pm \sqrt{16-4}}{2} = 2 \pm \sqrt{3}$. For $m = 2+\sqrt{3} = 	an(75^\circ) = 	an(\frac{5\pi}{12})$,the angle is $\frac{5\pi}{12}$. For $m = 2-\sqrt{3} = 	an(15^\circ) = 	an(\frac{\pi}{12})$,the angle is $\frac{\pi}{12}$.

The angles made with the $X$-axis by the two lines passing through the point $P(1, 2)$ and cutting the line $x+y=4$ at a distance $\frac{\sqrt{6}}{3}$ units from the point $P$ are

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