Let the angles made with the positive x-axis | vedclass

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(C) Let the line passing through $P(2,3)$ with angle $	heta$ be represented in parametric form as $(x, y) = (2 + r\cos	heta, 3 + r\sin	heta)$, wher

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(C) Let the line passing through $P(2,3)$ with angle $	heta$ be represented in parametric form as $(x, y) = (2 + r\cos	heta, 3 + r\sin	heta)$, where $r = \sqrt{\frac{2}{3}}$.Since this point lies on the line $x+y=6$, we substitute the coordinates:$(2 + r\cos	heta) + (3 + r\sin	heta) = 6$$r(\cos	heta + \sin	heta) + 5 = 6$$\sqrt{\frac{2}{3}}(\cos	heta + \sin	heta) = 1$$\cos	heta + \sin	heta = \sqrt{\frac{3}{2}}$Squaring both sides:$(\cos	heta + \sin	heta)^2 = \frac{3}{2}$$1 + \sin(2	heta) = \frac{3}{2}$$\sin(2	heta) = \frac{1}{2}$Thus, $2	heta = \frac{\pi}{6}$ or $2	heta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.Therefore, $	heta_1 = \frac{\pi}{12}$ and $	heta_2 = \frac{5\pi}{12}$.Summing the angles: $	heta_1 + 	heta_2 = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$.

Let the angles made with the positive x-axis by two straight lines drawn from the point $P(2,3)$ and meeting the line $x+y=6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_{1}$ and $\theta_{2}$. Then the value of $(\theta_{1}+\theta_{2})$ is:

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