TS EAMCET · Maths · Straight Lines
The area of the quadrilateral formed by the lines \(x+2 y+3=0,2 x+4 y+9=0\), \(x-2 y+3=0\) and \(3 x-6 y+11=0\) is
- A \(\frac{5}{12}\)
- B \(\frac{1}{4}\)
- C \(\frac{3}{4}\)
- D \(\frac{7}{12}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{4}\)
Step-by-step Solution
Detailed explanation
\(L_1: x+2y+3=0 \implies 2x+4y+6=0\) \(L_2: 2x+4y+9=0\) \(L_3: x-2y+3=0 \implies 3x-6y+9=0\) \(L_4: 3x-6y+11=0\) \(C_1=6, C_2=9\) \(C_3=9, C_4=11\) \(A_1=2, B_1=4\) \(A_2=3, B_2=-6\) \(Area = \frac{|(C_1-C_2)(C_3-C_4)|}{|A_1B_2-A_2B_1|}\)…
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