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.
Let QR be the base
Since origin is mid – point O(0,0) of QR
Then the coordinates of R(x,y) is given by
(– 4 + x)/2 = 0
x = 4
(0 + y)/2 = 0
y = 0
R(4,0)
Distance of QR = √(4 + 4)2 + 0
QR = 8
∴ PR = 8
Let P(x,y)
8 = √(4 – x)2 + (0 – y)2
64 = 16 + x2 – 8x + y2
Since it will lie on x axis
∴ × = 0
64 = 16 + y2
48 = y2
y = 4√3 or – 4√3
Hence,
P(0, 4√3) or P(0, – 4√3) and R(4, 0)