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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Straight Lines

BITSAT 2013 · Mathematics · Straight Lines

If the lines x+my+n=0, mx+ny+ =0 and nx+ y+m=0 are concurrent then

If the lines ℓx+my+n=0,mx+ny+ℓ=0\ell \mathrm{x}+\mathrm{my}+\mathrm{n}=0, \mathrm{mx}+\mathrm{ny}+\ell=0 and nx+ℓy+m=0\mathrm{nx}+\ell \mathrm{y}+\mathrm{m}=0 are concurrent then

Options

  1. Aℓ+m+n=0\ell+m+n=0
  2. Bℓ−m−n=0\ell-m-n=0
  3. Cℓ+m−n=0\ell+m-n=0
  4. Dm+n−ℓ=0m+n-\ell=0

Correct answer

A. ℓ+m+n=0\ell+m+n=0

Step-by-step solution

Since the lines are concurrent, so \Rightarrow \begin{array}{l}\left|\begin{array}{lll} \ell \mathrm{m} \mathrm{n}\\ \mathrm{m} \mathrm{n}& \ell \\ \mathrm{n}& \ell \mathrm{m} \end{array}\right|=0 \Rightarrow 3 \ell \mathrm{mn}-\ell^{3}-\mathrmm^{3}-\mathrmn^{3}=0 \\ (\ell+\mathrm{m}+\mathrm{n})\left(\ell^{2}+\mathrmm^{2}+\mathrmn^{2}-(\mathrm{m}-\mathrm{mn}-\mathrm{n} \ell)=0\right. \end{array} ⇒ℓ+\Rightarrow \quad \ell+\mathrm{m}+\mathrm{n}=0 \left[\because \ell^{2}+\mathrmm^{2}+\mathrmn^{2}$ >6 \mathrm{ m}+\mathrm{mn}+\mathrm{n}\right]
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