TS EAMCET · Maths · Straight Lines
\((a, b)\) 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 \(\frac{p}{2}\)
- B \(p\)
- C \(2 p\)
- D \(3 p\)
Answer & Solution
Correct Answer
(B) \(p\)
Step-by-step Solution
Detailed explanation
\(x-3 y+3=0\) \(2 x+y-8=0\) 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…
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