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If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a’ are (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}), then `|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|^2 = (3"a"^4)/4`

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#### Solution

The area of a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is given by

Δ = `1/2 |(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`

Also, area of an equilateral triangle with side a is given by

Δ = `sqrt(3)/2 "a"^2`

∴ `1/2 |(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)| = sqrt(3)/4 "a"^2`

Squaring both sides, we get

⇒ Δ^{2} = `1/4 |(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)| = 3/16 "a"^4`

or `|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|^2 = (3"a"^4)/4`

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