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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.
Solution
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).`
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