Q84 / 160 · TS EAMCET 2024 (09 May Shift 1) · Mathematics
$(a, b)$ is the point of concurrency of the lines $x-3 y+3=$ $0, k x+y+k=0$ and $2 x+y-8=0$. If the perpendicular distance from the origin to the line $L \equiv a x-b y+2 k=0$ is $p$, then the perpendicular distance from the point $(2,3)$ to $L=0$ is
- A$\frac{p}{2}$
- Correct answerB$p$
- C$2 p$
- D$3 p$
Solution
$x-3 y+3=0$
$2 x+y-8=0$
Solving we get $x=3, y=2 ; a=3, b=2$ $\begin{aligned} & (3,2) \text { lies on } k x+y+k=0 ; 3 k+2+k=0 \\ & \Rightarrow k=\frac{-1}{2} ; L \equiv 3 x-2 y-1=0 \\ & p=\frac{1}{\sqrt{3^2+2^2}}=\frac{1}{\sqrt{83}} \end{aligned}$ $\perp$ distance from $(2,3)$ to $L=0$ is $\left|\frac{6-6-1}{\sqrt{13}}\right|=\frac{1}{\sqrt{13}}=p$
Solving we get $x=3, y=2 ; a=3, b=2$ $\begin{aligned} & (3,2) \text { lies on } k x+y+k=0 ; 3 k+2+k=0 \\ & \Rightarrow k=\frac{-1}{2} ; L \equiv 3 x-2 y-1=0 \\ & p=\frac{1}{\sqrt{3^2+2^2}}=\frac{1}{\sqrt{83}} \end{aligned}$ $\perp$ distance from $(2,3)$ to $L=0$ is $\left|\frac{6-6-1}{\sqrt{13}}\right|=\frac{1}{\sqrt{13}}=p$
Hence, the correct answer is option: 'B'.