Question

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

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

Answer 1

The correct option is D ${75}^{\circ }$

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 $\frac{\sqrt{6}}{3}$ form (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.
$\Rightarrow \frac{\sqrt{6}}{3}(cos\theta +sin\theta )=1orsin\theta +cos\theta =\frac{3}{\sqrt{6}}$
$\Rightarrow \frac{1}{\sqrt{2}}sin\theta +\frac{1}{\sqrt{2}}cos\theta =\frac{\sqrt{3}}{2}$,
$\left\{Dividingbothsidesby\sqrt{2}\right\}$
$\Rightarrow sin(\theta +{45}^{\circ })=sin{60}^{\circ }orsin{120}^{\circ }$
$\Rightarrow \theta ={15}^{\circ }or{75}^{\circ }$.