If the coordinates of the vertices of an equilateral triangle with sides of length `a` are `(x_(1),y_(1)),(x_(2),y_(2))` and `(x_(3),y_(3))` then `|(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)|^2=(3a^(4))/4`

Answer

N/a

Explanation

Since we know that area of a triangle with vertices `(x_(1),y_(1)),(x_(2),y_(2))` and `(x_(3),y_(3))` is given by
`Delta=1/2|(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)|`
`impliesDelta^(2)=1/4|(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)|^(2)`...................(i)
We know that ara fo an equilateral triangle with side `a`.
`=Delta=1/2((sqrt(3))/2)a^(2)=(sqrt(3))/4a^(2)`
`implies Delta^(2)=3/16a^(4)`.............(ii)
From eqs (i) and (ii) `3/16 a^(4)=1/4 |(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)|^(2)`
`implies |(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)|^(2)=3/4a^(4)` Hence proved