Q83 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics

Two lines $\mathrm{L}_1$ and $\mathrm{L}_2$ passing through the point $\mathrm{P}(1,2)$ cut the line $x+y=4$ at a distance of $\frac{\sqrt{6}}{3}$ units from $P$. Then the angles made by $\mathrm{L}_1, \mathrm{~L}_2$ with positive $\mathrm{X}$-axis are

  • Correct answer
    A
    $\frac{\pi}{3}, \frac{\pi}{6}$
  • B
    $\frac{\pi}{8}, \frac{3 \pi}{8}$
  • C
    $\frac{\pi}{12}, \frac{5 \pi}{12}$
  • D
    $\frac{\pi}{4}, \frac{\pi}{8}$

Solution

$$ P(1,2), r=\frac{\sqrt{6}}{3} $$ Let $L_1$ and $L_2$ cut the line $x+y=4$ at $\left(x_1, y_1\right)$ By parametric form $$ x_1=1+\sqrt{\frac{6}{3}} \cos \theta, y_1=2+\sqrt{\frac{6}{3}} \sin \theta $$ $$ \begin{aligned} & \because\left(x_1, y_1\right) \text { lies on } x+y=4 \\ & 1+\sqrt{\frac{6}{3}} \cos \theta+2+\sqrt{\frac{6}{3}} \sin \theta=4 \\ & \Rightarrow(\sin \theta+\cos \theta)=\frac{3}{\sqrt{6}} \end{aligned} $$ Squaring both sides, $$ \begin{aligned} & \Rightarrow \quad 1+\sin 2 \theta=\frac{3}{2} \Rightarrow \sin 2 \theta=\frac{1}{2} \\ & \therefore \quad \theta=\frac{\pi}{12}, \frac{5 \pi}{12} . \end{aligned} $$

Hence, the correct answer is option: 'A'.

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