MHT CET · Maths · Straight Lines
If the slope of one of the lines given by \(\mathrm{K} x^2+6 x y+y^2=0\) is three times the order, then the value of K is
- A \(\frac{9}{4}\)
- B \(\frac{4}{9}\)
- C \(\frac{27}{4}\)
- D \(\frac{4}{27}\)
Answer & Solution
Correct Answer
(C) \(\frac{27}{4}\)
Step-by-step Solution
Detailed explanation
Given equation of pair of lines is
\(K x^2+6 x y+y^2=0 \)
\( \therefore A=K, H=3, B=1\)
Let the slopes of the lines be \(m_1\) and \(m_2\)
\(m_1+m_2=\frac{-2 H}{B} \text { and } m_1 m_2=\frac{A}{B}\)
Given that \(\mathrm{m}_2=3 \mathrm{~m}_1\)
\(\therefore m_1+3 m_1=\frac{-2 H}{B}=-6 \Rightarrow m_1=\frac{-3}{2}\)
and \(m_1 \times 3 m_1=\frac{A}{B}=K \Rightarrow 3 m_1^2=K \Rightarrow K=\frac{27}{4}\)
\(K x^2+6 x y+y^2=0 \)
\( \therefore A=K, H=3, B=1\)
Let the slopes of the lines be \(m_1\) and \(m_2\)
\(m_1+m_2=\frac{-2 H}{B} \text { and } m_1 m_2=\frac{A}{B}\)
Given that \(\mathrm{m}_2=3 \mathrm{~m}_1\)
\(\therefore m_1+3 m_1=\frac{-2 H}{B}=-6 \Rightarrow m_1=\frac{-3}{2}\)
and \(m_1 \times 3 m_1=\frac{A}{B}=K \Rightarrow 3 m_1^2=K \Rightarrow K=\frac{27}{4}\)
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