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+√3,3−√3)
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B
(3+√3,3+√3)
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C
(3−√3,3+√3)
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D
(3−√3,3−√3)
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Solution
The correct option is 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