JEE Mains · Maths · STD 11 - 9. straight line
Let \(\mathrm{A}(4,-2), \mathrm{B}(1,1)\) and \(\mathrm{C}(9,-3)\) be the vertices of a triangle \(A B C\). Then the maximum area of the parallelogram AFDE , formed with vertices \(\mathrm{D}, \mathrm{E}\) and F on the sides \(\mathrm{BC}, \mathrm{CA}\) and AB of the triangle ABC respectively, is ______ .
- A 3
- B 5
- C 7
- D 9
Answer & Solution
Correct Answer
(A) 3
Step-by-step Solution
Detailed explanation
Area of \(\triangle \mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1\end{array}\right|\) \(=6\) square units Maximum area of \(\operatorname{AFDE}=\frac{1}{2} \times 6=3\) sq. units
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