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 will be at a Distance of $\frac{\sqrt{6}}{3}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Let the required line through the point (1, 2) be inclined at an angle $\theta$ to the axis of x. Then its equation is
$\frac{x-1}{cos\theta }=\frac{y-2}{sin\theta }=r$ ......(i)
where r is distance of any point (x, y) on the line from the point (1, 2)
The coordinates of any point on the line (i) are $(1+rcos\theta ,2+rsin\theta )$. If this point is at a distance $\frac{\sqrt{6}}{3}$ from (1, 2), then $r=\frac{\sqrt{6}}{3}$
Therefore, the point is $(1+\frac{\sqrt{6}}{3}cos\theta ,2+\frac{\sqrt{6}}{3}sin\theta )$.
But this point lies on the line x + y = 4.
$\Rightarrow \frac{\sqrt{6}}{3}(cos\theta +sin\theta )=1orsin\theta +cos\theta =\frac{3}{\sqrt{6}}\Rightarrow \frac{1}{\sqrt{2}}sin\theta +\frac{1}{\sqrt{2}}cos\theta =\frac{\sqrt{3}}{2},$ [Dividing both sides by $\sqrt{2}$]
$\Rightarrow sin(\theta +{45}^{\circ })=sin{60}^{\circ }orsin{120}^{\circ }\Rightarrow \theta ={15}^{\circ }or{75}^{\circ }$