Question

In what direction a line be drawn through the point $\left(\right.1,2\left.\right)$ , so that its point of intersection with the line $x+y=4$ is at a distance $\sqrt{6}/3$ from the given point?

• A. $30^\circ $
• B. $45^\circ $
• C. $60^\circ $
• D. $75^\circ $

Answer 1

Answer: (D)

Let the point be
$\left(1 + \frac{\sqrt{6}}{3} cos \theta , 2 + \frac{\sqrt{6}}{3} sin \theta \right)$
Given point lie on the line $x+y=4$
$\therefore 1+\frac{\sqrt{6}}{3}cos\theta +2+\frac{\sqrt{6}}{3}sin\theta =4$
$\frac{\sqrt{6}}{3}cos\theta +\frac{\sqrt{6}}{3}sin\theta =1$
$cos\theta +sin\theta =\frac{\sqrt{3}}{\sqrt{2}}$
$\frac{1}{\sqrt{2}}cos\theta +\frac{1}{\sqrt{2}}sin\theta =\frac{\sqrt{3}}{2}$
$sin45^\circ cos\theta +sin45^\circ sin\theta =\frac{\sqrt{3}}{2}$
$sin\left(\theta + 45 ^\circ \right)=\frac{\sqrt{3}}{2}$
$\left(\theta + 45 ^\circ \right)=60^\circ $ or $\theta +45^\circ =120^\circ $