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TS EAMCET · Maths · Pair of Lines

For \(a, b, c \in \mathbb{R}\), if \(6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0\) and \(|a|+|b| \neq 0\), then all the lines given by \(a x+b y+c=0\) are

  1. A concurrent at \((3,1)\) or \((1,3)\)
  2. B parallel to each other \(\forall a, b, c \in \mathbb{R}\)
  3. C concurrent at \((-2,-3)\) or \((3,-1)\)
  4. D concurrent at \((2,3)\) or \((-3,1)\)
Verified Solution

Answer & Solution

Correct Answer

(C) concurrent at \((-2,-3)\) or \((3,-1)\)

Step-by-step Solution

Detailed explanation

We have \(6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0\) where a, b, c, \(\in \mathbf{R}\) \( \Rightarrow 6 {a}^2+{a}(7 {~b}-{c})-3 {~b}^2-{c}^2+4 {bc}=0 \) On solving we get \( \Rightarrow a=\frac{(c-7 b) \pm(11 b-5 c)}{12} \) Solving equations (i) and (ii) with \({ax}+{by}+{c}=0\) one by…