The centroid of an equilateral triangle is (0,0). If two vertices of the triangle lie on x+y `= 2sqrt(2)`, then find all the possible vertices fo triangle.

Answer

`(-2sqrt(2), -2sqrt(2)), (sqrt(2)-sqrt(6), sqrt(2)+sqrt(6)), (sqrt(2)+sqrt(6), sqrt(2)-sqrt(6))`

Explanation

As shown in the figure, equilateral triangle ABC is such that B and C lie on the given line `x+y-2sqrt(2) = 0`.
AD is perpendicular to BC, which passes through centroid.
`OD = (|0+0-2sqrt(2)|)/(sqrt(2)) = 2`
`therefore " OA=4"("As centroid divides median in the ratio "2:1)`
Slope of AD is 1, i.e., tan `45^(@), "where" 45^(@)` is the inclination of AD with x-axis.
`therefore A-=(0-4"cos" 45^(@), 0-"sin" 45^(@))-=(-2sqrt(2),-2sqrt(2))`
`"and " D-=(0+2"cos" 45^(@), 0+2"sin" 45^(@))-=(sqrt(2),sqrt(2))`
`"Now, "DC =DB=2 "cot" 30^(@) = 2sqrt(3)`
Therefore, B `-=(sqrt(2) +2sqrt(3) "cos" 135^(@), sqrt(2) + 2sqrt(3) "cos" 135^(@))`
`-=(sqrt(2) -sqrt(6), sqrt(2) + sqrt(6))`
`"And "C-=(sqrt(2) -2sqrt(3)"cos" 135^(@), sqrt(2) -2sqrt(3)"cos" 135^(@))`
`-=(sqrt(2) +sqrt(6),sqrt(2) -sqrt(6))`