Q84 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics
If the lines $L_1 \equiv x-2 y+3=0, L_2 \equiv 2 x+y+1=0$ and $L_3 \equiv 3 x+y+c=0$ are concurrent and $\theta$ is the acute angle between the lines $L_1=0$ and $L_3=0$, then $\tan \theta=$
- A$c+2$
- B$c-5$
- Correct answerC$c+5$
- D$c-2$
Solution
Here $\mathrm{L}_1: \mathrm{x}-2 \mathrm{y}+3=0, \mathrm{~L}_2: 2 \mathrm{x}+\mathrm{y}+1=0$,
$L_3: 3 x+y+c=0$
Now from lines $\mathrm{L}_1 \& \mathrm{~L}_2$
$$
\begin{aligned}
2 x-4 y= & -6 \\
2 x+y & =-1 \\
-- & + \\
\hline-5 y & =-5 \Rightarrow y=1 \\
\Rightarrow x-2+3 & =0 \Rightarrow x=-1
\end{aligned}
$$
Now $-3+1+\mathrm{L}=0 \Rightarrow \mathrm{C}=2$
Now, $\tan \theta=\left|\frac{\frac{1}{2}+3}{1-\frac{3}{2}}\right|=\left|\frac{\frac{7}{2}}{-\frac{1}{2}}\right|=|-7|=7$
$$
\Rightarrow \tan \theta=7
$$
Hence, the correct answer is option: 'C'.