Question

The larger of the two angles made with the x-axis of a straight line drawn through $(1,2)$ so that it intersects $x+y=4$ at a point distant $\frac{\sqrt{6}}{3}$ from $(1,2)$ is

• A. ${105}^0$
• B. ${75}^0$
• C. ${60}^0$
• D. ${15}^0$

Answer 1

Answer: (B)

${75}^0$

Let $AP=r=\frac{\sqrt{6}}{3}$

Equation of $AP:\frac{x-1}{\cos \theta }=\frac{y-2}{\sin \theta }=r$

$\therefore P\equiv (1+r\cos \theta ,2+r\sin \theta )$

$P$ will staisfy $x+y=4$

$\therefore 1+\frac{\sqrt{6}}{3}\cos \theta +2+\frac{\sqrt{6}}{3}\sin \theta =4$

$\Rightarrow \cos (\theta -\frac{\pi }{4})=\frac{\sqrt{3}}{2}=\cos {30}^0$

$\therefore \theta ={75}^0$ or ${15}^0$