TS EAMCET · Maths · Straight Lines
If the equation of the circumcircle of the triangle formed by the lines \(\mathrm{L}_1 \equiv x+y=0\), \(\mathrm{L}_2 \equiv 2 x+y-1=0, \mathrm{~L}_3 \equiv x-3 y+2=0\) is \(\lambda_1 \mathrm{~L}_1 \mathrm{~L}_2+\lambda_2 \mathrm{~L}_2 \mathrm{~L}_3+\lambda_3 \mathrm{~L}_3 \mathrm{~L}_1=0\), then \(\frac{7 \lambda_1}{\lambda_2}+\frac{\lambda_3}{\lambda_1}=\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
Coefficients for the circumcircle \(\lambda_1 L_1 L_2 + \lambda_2 L_2 L_3 + \lambda_3 L_3 L_1 = 0\): Coeff of \(x^2\): \(2\lambda_1 + 2\lambda_2 + \lambda_3\) Coeff of \(y^2\): \(\lambda_1 - 3\lambda_2 - 3\lambda_3\) Coeff of \(xy\): \(3\lambda_1 - 5\lambda_2 - 2\lambda_3\) For…
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