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Straight Line — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsStraight Line2 Marks
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
    $30^{\circ}$
  • B
    $45^{\circ}$
  • C
    $60^{\circ}$
  • ✓
    $75^{\circ}$
✓

Answer

Correct option: D.
$75^{\circ}$
(D)
Let the required line through the point $(1,2)$ be inclined at an angle 0 to the axis of X . Then its equation is $\frac{x-1}{\cos \theta}=\frac{y-2}{\sin \theta}=r$  ...(i)
The co-ordinates of a point on the line (i) are $(1+r \cos \theta, 2+r \sin \theta)$
If this point is at a distance $\frac{\sqrt{6}}{3}$ from $(1,2)$,then $r =\frac{\sqrt{6}}{3}$
Therefore, the point is $\left(1+\frac{\sqrt{6}}{3} \cos \theta, 2+\frac{\sqrt{6}}{3} \sin \theta\right)$
But this point lies on the line $x+y=4$
$\sin \theta+\cos \theta=\frac{3}{\sqrt{6}}$
$\Rightarrow \frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta=\frac{\sqrt{3}}{2}$ ....(Dividing both sides by $\sqrt{2}$ )
$\Rightarrow \sin \left(\theta+45^{\circ}\right)=\sin 60^{\circ}$ or $\sin 120^{\circ}$
$\Rightarrow \theta=15^{\circ}$ or $75^{\circ}$

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