Q111 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
Find the direction in which a straight line must be drawn through the point $(1,2)$ so that its point of intersection with the line $x+y=4$ may be at a distance of $\sqrt{\dfrac{2}{3}}$ from this point.
- A$30^{\circ} \text { or } 150^{\circ}$
- Correct answerB$15^{\circ} \text { or } 75^{\circ}$
- C$60^{\circ} \text { or } 120^{\circ}$
- D$50^{\circ} \text { or } 100^{\circ}$
Solution
Let the line pass through the point $P(1, 2)$ at an angle $\theta$ with the positive $x$-axis. The equation of the line in parametric form is $\dfrac{x-1}{\cos \theta} = \dfrac{y-2}{\sin \theta} = r$, where $r$ is the distance from $P$.
Any point on this line is given by $(1 + r \cos \theta, 2 + r \sin \theta)$.
The intersection point lies on the line $x + y = 4$. Substituting the coordinates of the point into this equation:
$(1 + r \cos \theta) + (2 + r \sin \theta) = 4$
$3 + r(\cos \theta + \sin \theta) = 4$
$r(\cos \theta + \sin \theta) = 1$
Given $r = \sqrt{\dfrac{2}{3}}$, we have $\sqrt{\dfrac{2}{3}}(\cos \theta + \sin \theta) = 1$.
$\cos \theta + \sin \theta = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{2}$.
Multiplying by $\dfrac{1}{\sqrt{2}}$ on both sides:
$\dfrac{1}{\sqrt{2}}\cos \theta + \dfrac{1}{\sqrt{2}}\sin \theta = \dfrac{\sqrt{6}}{2\sqrt{2}} = \dfrac{\sqrt{3}}{2}$.
$\sin(\theta + 45^{\circ}) = \sin 60^{\circ}$ or $\sin(\theta + 45^{\circ}) = \sin 120^{\circ}$.
Case 1: $\theta + 45^{\circ} = 60^{\circ} \Rightarrow \theta = 15^{\circ}$.
Case 2: $\theta + 45^{\circ} = 120^{\circ} \Rightarrow \theta = 75^{\circ}$.
Thus, the required angles are $15^{\circ}$ and $75^{\circ}$.
Answer: $15^{\circ} \text { or } 75^{\circ}$
Any point on this line is given by $(1 + r \cos \theta, 2 + r \sin \theta)$.
The intersection point lies on the line $x + y = 4$. Substituting the coordinates of the point into this equation:
$(1 + r \cos \theta) + (2 + r \sin \theta) = 4$
$3 + r(\cos \theta + \sin \theta) = 4$
$r(\cos \theta + \sin \theta) = 1$
Given $r = \sqrt{\dfrac{2}{3}}$, we have $\sqrt{\dfrac{2}{3}}(\cos \theta + \sin \theta) = 1$.
$\cos \theta + \sin \theta = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{2}$.
Multiplying by $\dfrac{1}{\sqrt{2}}$ on both sides:
$\dfrac{1}{\sqrt{2}}\cos \theta + \dfrac{1}{\sqrt{2}}\sin \theta = \dfrac{\sqrt{6}}{2\sqrt{2}} = \dfrac{\sqrt{3}}{2}$.
$\sin(\theta + 45^{\circ}) = \sin 60^{\circ}$ or $\sin(\theta + 45^{\circ}) = \sin 120^{\circ}$.
Case 1: $\theta + 45^{\circ} = 60^{\circ} \Rightarrow \theta = 15^{\circ}$.
Case 2: $\theta + 45^{\circ} = 120^{\circ} \Rightarrow \theta = 75^{\circ}$.
Thus, the required angles are $15^{\circ}$ and $75^{\circ}$.
Answer: $15^{\circ} \text { or } 75^{\circ}$
Hence, the correct answer is option: 'B'.