Question

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, -3). The origin is the midpoint of the base. Find the coordinates of the points A and B. Also, find the coordinates of another point D such that ABCD is a rhombus.

Answer 1

Let (0, y)  be the coordinates of B. Then

` 0= (-3+y)/2 ⇒ y=3`

Thus, the coordinates of B are (0,3)

Here. AB = BC = AC  and by symmetry the coordinates of A lies on x-axis Let the coordinates of A be (x, 0). Then

`AB= BC⇒AB^2 = BC^2`

`⇒ (x-0)^2 +(0-3)^2 = 6^2`

`⇒ x^2 = 36-9=27`

`⇒ x = +- 3 sqrt(3) `

`"If the coordinates of point A are "(3 sqrt(3),0)  ."then the coordinates of D are " (-3 sqrt(3), 0).`

`"If the coordinates of point A are "(-3 sqrt(3),0)  ."then the coordinates of D are " (-3 sqrt(3), 0).` 

`"Hence the required coordinates are " A(3sqrt(3),0) , B(0,3) and  D (-3 sqrt(3),0) or `

`A (-3sqrt(3),0) , B(0,3) and D (3sqrt(3),0).`