Question

In the given figure, ΔABC is an equilateral triangle. Coordinates of vertices A and B are (0, 3) and (0, −3), respectively. Find the coordinates of points C.

Answer 1

AB = 6 cm = AC

OC = `sqrt(36 − 9)` = `3sqrt3` cm

Point C is `(3sqrt3, 0)`.

Answer 2

Given:

• △ABC is an equilateral triangle.

• Coordinates of:

  • A = (0, 3)

  • B = (0, −3)

Since A and B lie on the same vertical line (x = 0), the midpoint M of AB is:

`M =((0+0)/2, (3+(-3))/2) = (0, 0)`

So, the origin is the midpoint of AB.

So triangle ABC is symmetric about the y-axis, and AC = BC = AB.

We know that triangle is equilateral ⇒ AC = AB = BC

`AB = sqrt((0-0)^2+(3-(-3)^2)` `=sqrt(0+36) = 6`

Use distance formula:

`AC = sqrt((x-0)^2 + (0-3)^2) = 6 => sqrt(x^2+9) = 6 => x^2+9=36=>x^2 = 27 =>x=+-sqrt27 = +- 3sqrt3`

`(3sqrt3, 0) or (-3sqrt3, 0)`