Parametric Representation of Lines
Let the two straight lines passing through the point $P(2, 3)$ make angles $\theta_1$ and $\theta_2$ with the positive $x$-axis. The parametric equations of these lines are:
$x = 2 + r \cos \theta$
$y = 3 + r \sin \theta$
where $r$ is the distance from point $P$ to a point $(x, y)$ on the line.
Intersection with Given Line
The points $(x, y)$ lie on the line $x + y = 6$. Substituting the parametric equations into the line equation:
$(2 + r \cos \theta) + (3 + r \sin \theta) = 6$
$5 + r(\cos \theta + \sin \theta) = 6$
$r(\cos \theta + \sin \theta) = 1$
Using the Given Distance
We are given that the distance $r = \sqrt{\frac{2}{3}}$. Substituting this value:
$\sqrt{\frac{2}{3}} (\cos \theta + \sin \theta) = 1$
$\cos \theta + \sin \theta = \sqrt{\frac{3}{2}}$
Solving the Trigonometric Equation
Squaring both sides of the equation:
$(\cos \theta + \sin \theta)^2 = \left(\sqrt{\frac{3}{2}}\right)^2$
$\cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta = \frac{3}{2}$
Using the identities $\cos^2 \theta + \sin^2 \theta = 1$ and $2 \sin \theta \cos \theta = \sin(2\theta)$:
$1 + \sin(2\theta) = \frac{3}{2}$
$\sin(2\theta) = \frac{3}{2} - 1 = \frac{1}{2}$
The general solutions for $2\theta$ are $2\theta = n\pi + (-1)^n \frac{\pi}{6}$, where $n$ is an integer.
- For $n=0$: $2\theta_1 = \frac{\pi}{6} \implies \theta_1 = \frac{\pi}{12}$.
- For $n=1$: $2\theta_2 = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \implies \theta_2 = \frac{5\pi}{12}$.
These two values of $\theta$ satisfy the original condition $\cos \theta + \sin \theta = \sqrt{\frac{3}{2}}$. Other solutions arising from squaring do not satisfy this condition.
Calculating the Sum of Angles
The two angles are $\theta_1 = \frac{\pi}{12}$ and $\theta_2 = \frac{5\pi}{12}$.
The sum is:
$(\theta_1 + \theta_2) = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$
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Important Questions from Coordinate Geometry
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