Q85 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics
If the line $4 x+3 y-k=0,2 x+y+3=0$ and $3 x+2 y+k=0$ are concurrent, then the perpendicular distance from the point of concurrency of these lines to the line $3 x+4 y+2=0$ is
- A$\frac{3}{5}$
- B1
- C$\frac{13}{5}$
- Correct answerD3
Solution
$$
\begin{aligned}
& \text { (d) }\left|\begin{array}{ccc}
4 & 3 & -k \\
2 & 1 & 3 \\
3 & 2 & k
\end{array}\right|=0 \\
& \Rightarrow \quad 4(k-6)-3(2 k-9)-k(4-3)=0 \\
& \Rightarrow \quad 4 k-24-6 k+27-4 k+3 k=0 \Rightarrow 3 k=3 \\
& \Rightarrow \quad k=1 \\
& \therefore \quad L_1: 4 x+3 y-1=0 \\
& \quad L_2: 2 x+y+3=0
\end{aligned}
$$
Eqn. (1) - (Eqn. (2) $\times 2$ ),
$$
\begin{aligned}
& y=7 \\
\therefore \quad x & =-5
\end{aligned}
$$
$\therefore \quad$ Point of concurrency is $(-5,7)$
Distance from $(-5,7)$ to $3 x+4 y+2=0$
$$
\begin{aligned}
& \quad d=\left|\frac{3(-5)+4(7)+2}{\sqrt{3^2+4^2}}\right| \\
& \therefore \quad d=3 .
\end{aligned}
$$
Hence, the correct answer is option: 'D'.