TS EAMCET · Maths · Straight Lines
The equations of the sides \(A B, A C\) and \(B C\) of a \(\triangle A B C\) are respectively \(x-3 y=0,3 x-y=0\), \(x+y+4=0\). If \(P\) and \(Q\) are the points on the line \(3 x-y+k=0\) passing through \(B\) such that \(P B: B Q=1: 1\), then \(k=\)
- A \(8\)
- B \(12\)
- C \(-8\)
- D \(-12\)
Answer & Solution
Correct Answer
(A) \(8\)
Step-by-step Solution
Detailed explanation
We have, equation of \(A B, A C\) and \(B C\) are \(x-3 y=0,3 x-y=0, x+y+4=0\) on solving equation of \(A B\) and \(B C\) we get \(B(-3,-1)\) Since \(B\) lies on \(3 x-y+k=0\) \(\therefore \quad 3(-3)-(-1)+k=0 \Rightarrow k=8\)
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