Question

A straight line drawn through $(1,2)$ intersects $x+y=4$ at a distance $\frac{\sqrt{6}}{3}$ from $(1,2)$. The angle made by the line with the positive direction of $x$-axis is $\alpha$. Find the greater of the two values of $\alpha .$

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

Answer 1

Answer: (B)

${75}^0$

$x=1+\frac{\sqrt{6}}{3}\cos \theta$

$y=2+\frac{\sqrt{6}}{3}\sin \theta$

It lies on $x+y=4$

$3+\frac{\sqrt{6}}{3}(\sin \theta +\cos \theta )=4$

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

Satisfies when $\theta =15°,75°$

Hence $\theta =75°$ is greater