Question

In what direction a line be drawn through the point $(1,2)$ , 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 $\theta +45^{\circ }=120^{\circ }$