JEE Mains · Maths · STD 11 - 9. straight line
Let \(B\) and \(C\) be the two points on the line \(y+x=0\) such that \(B\) and \(C\) are symmetric with respect to the origin. Suppose \(A\) is a point on \(y -2 x =2\) such that \(\triangle ABC\) is an equilateral triangle. Then, the area of the \(\triangle ABC\) is
- A \(3 \sqrt{3}\)
- B \(2 \sqrt{3}\)
- C \(\frac{8}{\sqrt{3}}\)
- D \(\frac{10}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\frac{8}{\sqrt{3}}\)
Step-by-step Solution
Detailed explanation
At A \(x=y\) \(Y-2 x=2\) \((-2,-2)\) Height from line \(x + y =0\) \(h=\frac{4}{\sqrt{2}}\) Area of \(\Delta=\frac{\sqrt{3}}{4} \frac{ h ^2}{\sin ^2 60}=\frac{8}{\sqrt{3}}\)
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