MHT CET · Maths · Pair of Lines
The auxiliary equation of the lines passing through the origin and having slopes
\(\sqrt{3}+1\) and \(\sqrt{3}-1\) is
- A \(m^{2}-2 \sqrt{3} m+2=0\)
- B \(m^{2}-2 \sqrt{3} m-2=0\)
- C \(m^{2}+2 \sqrt{3} m-2=0\)
- D \(m^{2}+2 \sqrt{3} m+2=0\)
Answer & Solution
Correct Answer
(A) \(m^{2}-2 \sqrt{3} m+2=0\)
Step-by-step Solution
Detailed explanation
Equations of required lines are
\(y=(\sqrt{3}+1) x \text { and } y=(\sqrt{3}-1) x\)
\(\therefore\) their joint equation is \([(\sqrt{3}+1) x-y][(\sqrt{3}-1) x-y]=0\)
\(\therefore 2 x^{2}-2 \sqrt{3} x y+y^{2}=0\)
Dividing both sides by \(x^{2}\), we get
\(\begin{aligned}
\left(\frac{y}{x}\right)^{2}-2 \sqrt{3} \frac{y}{x}+2 &=0 \\
\therefore m^{2}-2 \sqrt{3} m+2 &=0
\end{aligned}\)
\(y=(\sqrt{3}+1) x \text { and } y=(\sqrt{3}-1) x\)
\(\therefore\) their joint equation is \([(\sqrt{3}+1) x-y][(\sqrt{3}-1) x-y]=0\)
\(\therefore 2 x^{2}-2 \sqrt{3} x y+y^{2}=0\)
Dividing both sides by \(x^{2}\), we get
\(\begin{aligned}
\left(\frac{y}{x}\right)^{2}-2 \sqrt{3} \frac{y}{x}+2 &=0 \\
\therefore m^{2}-2 \sqrt{3} m+2 &=0
\end{aligned}\)
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