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QJEEM23Ma47cf: JEE Main 2023 Mathematics - Class 11 JEEM23Ma47cf Solution: Let and be the two points on the line such that and are…

Let BB and CC be the two points on the line y+x=0y+x=0 such that BB and CC are symmetric with respect to the origin. Suppose AA is a point on y−2x=2y-2 x=2 such that △ABC\triangle A B C is an equilateral triangle. Then, the area of the △ABC\triangle A B C is :

Multiple Choice Options

  • A 103\frac{10}{\sqrt{3}}
  • B 232 \sqrt{3}
  • C 333 \sqrt{3}
  • D 83\frac{8}{\sqrt{3}}

Step-by-Step Verified Solution

Origin (O)(O) is mid-point of BC(x+y=0)B C(x+y=0) .

AA lies on perpendicular bisector of BCB C , which is x−y=0x-y=0

A is point of intersection of x−y=0x-y=0 and y−2x=2y-2 x=2

∴A≡(−2,−2)\therefore A \equiv(-2,-2)

Let h=AO=−2−212+12=22h=A O=\frac{-2-2}{\sqrt{1^{2}+1^{2}}}=2 \sqrt{2}

 Area =h23=83\text { Area }=\frac{h^{2}}{\sqrt{3}}=\frac{8}{\sqrt{3}}

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