Question

# For an equilateral triangle, one side lies on x+y=6 and the 3rd vertex is mirror image of the origin in the mirror x+y=6. Then the coordinates of other two vertices are

A
(3+3,33)
B
(3+3,3+3)
C
(33,3+3)
D
(33,33)

Solution

## The correct options are A (3+√3,3−√3) C (3−√3,3+√3)​​​​Let P(x,y) be the image of origin about x+y=6 The line perpendicular to x+y=6 and passes through (0,0) will be y=x So, intersection of x=y and x+y=6 will give the midpoint of QR i.e., T ⇒T≡(3,3) Now, OT=TP=3√2  ∴QRsin60∘=3√2    (∵RP=QR=PQ) ∴QR=2√6 Now, x+y=6 So any point on the line will be  (3±rcos135°,3±rsin135°)=(3∓r√2,3±r√2) where r=TQ=√6 So, from the above condition  x=3∓√3 and y=3±√3

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