AP EAMCET · Maths · Straight Lines
If the lines \(2 x+y-3=0,3 x+2 y-2=0, k x-3 y-23=0\) are concurrent, then the roots of the equation \(6 x^2-7 x+k=0\) are
- A \(\frac{1}{2}, \frac{2}{3}\)
- B \(2,3\)
- C \(3,4\)
- D \(6,2\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}, \frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{vmatrix} 2 & 1 & -3 \\ 3 & 2 & -2 \\ k & -3 & -23 \end{vmatrix} = 0\) \(2(-46-6) - 1(-69+2k) - 3(-9-2k) = 0\) \(-104 + 69 - 2k + 27 + 6k = 0\) \(4k - 8 = 0 \implies k = 2\) \(6x^2 - 7x + 2 = 0\) \(x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(2)}}{2(6)}\)…
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