Question

A straight line is drawn through the point $A(1,2)$ such that its point of intersection with the straight line $x+y=4$ is at a distance $\sqrt{6} / 3$ from the given point ' $A$ '. Find the angle which the lines makes with the positive direction of $x$-axis.

• A. $\theta=15^{\circ} \& 75^{\circ}$
• B. $\theta=75^{\circ} \& 45^{\circ}$
• C. $\theta=45^{\circ} \& 60^{\circ}$
• D. $\theta=60^{\circ} \& 30^{\circ}$

Answer 1

Answer: (A)

Let the angle of inclination of line $\theta$, and as passed through point $A(1,2)$, so equation of line is
$\frac{x-1}{\cos \theta}=\frac{y-2}{\sin \theta}=\pm \frac{\sqrt{6}}{3}$
$\therefore $ General point on the line is
$P\left(1 \pm \frac{\sqrt{6}}{3} \cos \theta, 2 \pm \frac{\sqrt{6}}{3} \sin \theta\right)$
Let the point $P$ on the straight line $x+y=4$, so
$3 \pm \frac{\sqrt{6}}{3}(\sin \theta+\cos \theta)=4$
$\Rightarrow \pm \frac{\sqrt{6}}{3}(\sin \theta+\cos \theta)=1$
On squaring both sides, we get
$2(1+\sin 2 \theta)=3$
$\Rightarrow \sin 2 \theta=\frac{1}{2}$
$\Rightarrow 2 \theta=30^{\circ}$ and $150^{\circ}$
$\Rightarrow \theta=15^{\circ} \text { and } 75^{\circ}$