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प्रश्न
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.
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उत्तर
Given:
ΔABC is equilateral
Base BC lies on the y–axis
C = (0, –3)
Origin (0, 0) is midpoint of BC
B = (0, b), C = (0, −3)
`0 = (b+(-3))/2`
b − 3 = 0
b = 3
B = (0, 3)
C = (0, –3)
In an equilateral triangle,
height = `sqrt3/2 xx "side"`
Height = `sqrt3/2 xx 6`
`= 3sqrt3`
Base lies on y-axis, so the altitude is horizontal, meaning A lies on the x-axis but at distance `3sqrt3` from O.
`A = (+-3sqrt3, 0)`
A1 = `(3sqrt3, 0)`
A2 = (`-3sqrt3, 0`)
A rhombus has all sides equal, and ABCD is cyclic order.
A = (`3sqrt3`, 0), B = (0, 3), C = (0, −3)
To form a rhombus ABCD, the 4th point D is:
D = A + C − B
D = (`3sqrt3`, 0) + (0, −3) − (0, 3)
D = (`3sqrt3`, −6)
Base points:
B = (0, 3), C = (0, −3)
Vertices of equilateral triangle:
A = (`3sqrt3`, 0) or A = (`−3sqrt3`, 0)
Point D such that ABCD is a rhombus:
If A = (`3sqrt3`, 0):
D = (`3sqrt3`, −6)
If A = (`−3sqrt3`, 0):
D = (`-3sqrt3`, −6)
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