Let the angles made with the positive x-axis by two straight lines drawn from the point P(…

Question

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 $\sqrt{\dfrac23}$ from the point $P$ be $	heta_1$ and $	heta_2$. Then the value of $(	heta_1+	heta_2)$ is:

Accepted Answer

Let $Q(t,6-t)$ be a point on $x+y=6$ with $PQ^2 = \dfrac23$:
$$(2-t)^2+(t-3)^2 = \frac23 \implies 6t^2-30t+37=0$$
$$t_1+t_2 = 5, \qquad t_1t_2 = \frac{37}{6}$$

With $\theta_i = \tan^{-1}\left(\dfrac{t_i-3}{2-t_i}\right)$ (slope of $PQ_i$),
$$\theta_1+\theta_2 = \tan^{-1}\left(\frac{5(t_1+t_2)-2t_1t_2-12}{(t_1+t_2)-5}\right)$$
The denominator $(t_1+t_2)-5 = 0$, so the argument tends to $\infty$, giving
$$\theta_1+\theta_2 = \tan^{-1}(\infty) = \frac{\pi}{2}$$

Solutions · 1