One vertex of the equilateral triangle with centroid at the origin and one side as x + y - 2 = 0 is

Question

One vertex of the equilateral triangle with centroid at the origin and one side as x + y - 2 = 0 is (A) (-1, -1) (B) (2, 2) (C) (-2, - 2) (D) (2, -2)

Tardigrade Admin · Accepted Answer

[Hint: Let Let ABC be the equilateral triangle with vertex A(h, k) and let D (α, β) be the point on BC. Then (2 α + h/3) =0 = (2 β + k/3) Also α + β - 2 = 0 and ( ( k - 0/h -0)) × (-1) = - 1 ]

Q. One vertex of the equilateral triangle with centroid at the origin and one side as x + y - 2 = 0 is

(-1, -1)

(2, 2)

(-2, - 2)

(2, -2)

[Hint : Let Let ABC be the equilateral triangle with vertex A(h, k) and let $D (α, β)$ be the point on BC. Then $\frac{2 α + h}{3} = 0 = \frac{2 β + k}{3}$
Also $α + β - 2 = 0$ and $(\frac{k - 0}{h - 0}) \times (- 1) = - 1]$

Q. One vertex of the equilateral triangle with centroid at the origin and one side as x + y - 2 = 0 is

(-1, -1)

(2, 2)

(-2, - 2)

(2, -2)

[Hint : Let Let ABC be the equilateral triangle with vertex A(h, k) and let D(α,β) be the point on BC. Then 32α+h​=0=32β+k​ Also α+β−2=0 and (h−0k−0​)×(−1)=−1]

[Hint : Let Let ABC be the equilateral triangle with vertex A(h, k) and let $D (α, β)$ be the point on BC. Then $\frac{2 α + h}{3} = 0 = \frac{2 β + k}{3}$
Also $α + β - 2 = 0$ and $(\frac{k - 0}{h - 0}) \times (- 1) = - 1]$