MCQ
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${30^o}$
- B${45^o}$
- C${60^o}$
- ✓${75^o}$
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 $(1 + r\cos \theta ,{\rm{ }}2 + r\sin \theta )$. If this point is at a distance $\frac{{\sqrt 6 }}{3}$ form (1, 2), then $r = \frac{{\sqrt 6 }}{3}.$
Therefore, the point is $\left( {1 + \frac{{\sqrt 6 }}{3}\cos \theta ,{\rm{ }}2 + \frac{{\sqrt 6 }}{3}\sin \theta } \right)$.
But this point lies on the line $x + y = 4$.
==> $\frac{{\sqrt 6 }}{3}(\cos \theta + \sin \theta ) = 1$ or $\sin \theta + \cos \theta = \frac{3}{{\sqrt 6 }}$
==> $\frac{1}{{\sqrt 2 }}\sin \theta + \frac{1}{{\sqrt 2 }}\cos \theta = \frac{{\sqrt 3 }}{2}$,
{Dividing both sides by $\sqrt 2 $}
==> $\sin (\theta + {45^o}) = \sin {60^o}$or sin ${120^o}$
==> $\theta = {15^o}$or ${75^o}$.
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$(A)$ $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ $(B)$ $\left(-\frac{9}{2 \sqrt{2}},-\frac{1}{\sqrt{2}}\right)$
$(C)$ $(3 \sqrt{3},-2 \sqrt{2})$ $(D)$ $(-3 \sqrt{3}, 2 \sqrt{2})$