Question

The base QR of an equilateral triangle PQR lies on x-axis. The coordinates of the point Q are (–4, 0) and origin is the midpoint of the base. Find the coordinates of the points P and R.

Answer 1

Let(x ,0) be the coordinates of R. Then

`0= (-4+x)/2 ⇒ x =4`

Thus, the coordinates of R are (4,0)  . Here, PQ = QR =  PR  and the coordinates of P lies on . y  - axis Let the coordinates of P be

(0,y) . Then ,

`PQ = QR ⇒ PQ^2 = QR^2`

`⇒ (0+4)^2 +(y-0)^2 = 8^2`

`⇒ y^2 = 64-16=48`

`⇒ y = +- 4 sqrt(3)`

`"Hence, the required coordinates are " R (4,0) and P (0,4 sqrt(3) ) or P (0, -4 sqrt(3) ).`