MHT CET · Maths · Pair of Lines
The joint equation of pair of lines through the origin and having slopes \((1+\sqrt{2})\) and \(\frac{1}{(1+\sqrt{2})}\) is
- A \(x^2+2 x y+y^2=0\)
- B \(x^2-2 \sqrt{2} x y-y^2=0\)
- C \(x^2-2 \sqrt{2} x y+y^2=0\)
- D \(x^2+2 x y-y^2=0\)
Answer & Solution
Correct Answer
(C) \(x^2-2 \sqrt{2} x y+y^2=0\)
Step-by-step Solution
Detailed explanation
\(
\frac{1}{1+\sqrt{2}}=\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1
\)
Hence equations of lines passing through origin and having slopes \((\sqrt{2}+1)\) and \((\sqrt{2}-1)\) are \(\mathrm{y}=(\sqrt{2}+1) \mathrm{x}\) and \(\mathrm{y}=(\sqrt{2}-1) \mathrm{x}\)
Their joint equation is \([(\sqrt{2}+1) \mathrm{x}-\mathrm{y}][(\sqrt{2}-1) \mathrm{x}-\mathrm{y}]=0\) \(x^2-2 \sqrt{2} x y+y^2=0\)
\frac{1}{1+\sqrt{2}}=\frac{(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1
\)
Hence equations of lines passing through origin and having slopes \((\sqrt{2}+1)\) and \((\sqrt{2}-1)\) are \(\mathrm{y}=(\sqrt{2}+1) \mathrm{x}\) and \(\mathrm{y}=(\sqrt{2}-1) \mathrm{x}\)
Their joint equation is \([(\sqrt{2}+1) \mathrm{x}-\mathrm{y}][(\sqrt{2}-1) \mathrm{x}-\mathrm{y}]=0\) \(x^2-2 \sqrt{2} x y+y^2=0\)
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