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Question
The base QR of a n 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.
Solution
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) ).`
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