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:

$(\theta_1 + \theta_2) = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$