In the given figure, ` DeltaABC` is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices.

Explanation

Since B is at a distance of 3 units from A on X- axis in the positive direction. So B will be 5 units away from the origin.
So, `" "B-=(5, 0)`
Let M be the mid-point of AB
`therefore" "AM=(1)/(2)AB=(3)/(2)` units
Also `" " AC= 3`units
`therefore` In right `DeltaCMA`, by Pythagoras theorem
`" "CM^(2)=AC^(2)-AM^(2)`
`" "=9-(9)/(4)=(27)/(4)`
`therefore" " CM= sqrt((27)/(4))=(3sqrt(3))/(2)`
So, the coordinates of C are (OM, MC)
` rArr" "C-= (OA+AM,MC). `
`rArr" "C-= (2+(3)/(2), (3sqrt(3))/(2))-= ((7)/(2), (3sqrt(3))/(2))`