JEE Mains · Maths · STD 11 - 9. straight line
Let the points of intersections of the lines \(x-y+1=0\), \(x-2 y+3=0\) and \(2 x-5 y+11=0\) are the mid points of the sides of a triangle \(A B C\). Then the area of the triangle \(\mathrm{ABC}\) is .... .
- A \(4\)
- B \(3\)
- C \(2\)
- D \(6\)
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
Detailed explanation
intersection point of give lines are \((1,2),(7,5)\), \((2,3)\) \(\Delta=\frac{1}{2}\left|\begin{array}{lll}1 & 2 & 1 \\ 7 & 5 & 1 \\ 2 & 3 & 1\end{array}\right|\) \(=\frac{1}{2}[1(5-3)-2(7-2)+1(21-10)]\) \(=\frac{1}{2}[2-10+11]\)…
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