Question

Angle made with x-axis by a straight line drawn through $(1,2)$ so that it intersects $x+y=4$ at a distance $\frac{\sqrt{6}}{3}$ from $(1,2)$ are :

• A. ${105}^{\circ }$
• B. ${75}^{\circ }$
• C. ${60}^{\circ }$
• D. ${15}^{\circ }$

Answer 1

Answer: (B), (D)

${15}^{\circ }$
${75}^{\circ }$

Let the angle be $\theta$. Then, equation of the given line is
$\frac{x-1}{\cos \theta }=\frac{y-2}{\sin \theta }$ ..... (i)
The coordinates of a point on (i) at a distance $\frac{\sqrt{6}}{3}$ from $(1,2)$ are $(1+\frac{\sqrt{6}}{3}\cos \theta ,2+\frac{\sqrt{6}}{3}\sin \theta )$. This point lies on $x+y=4$. Therefore,
$1+\frac{\sqrt{6}}{3}\cos \theta +2+\frac{\sqrt{6}}{3}\sin \theta =4$
$\Rightarrow \cos \theta +\sin \theta =\sqrt{\frac{3}{2}}$
$\Rightarrow \frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta =\frac{\sqrt{3}}{2}$
$\Rightarrow \sin (\theta +\frac{\pi }{4})=\sin \left(\frac{\pi }{3}\right)$ or $\sin \left(\frac{2\pi }{3}\right)$
$\Rightarrow \theta +\frac{\pi }{4}=\frac{\pi }{3},\frac{2\pi }{3}$
$\Rightarrow \theta ={75}^{\circ }$ or $\theta ={15}^{\circ }$