One vertex of an equilateral triangle with side 2 units is at origin. Another vertex lies on the line X=root 3y, and is in the first quadrant. Please find all the coordinates.
Answers
Given : One vertex of an equilateral triangle with side 2 units is at origin
Another vertex lies on the line x=√3y, and is in the first quadrant
To Find : all the coordinates.
Solution:
One vertex of an equilateral triangle with side 2 units is at origin
A = ( 0 , 0)
B lies on x = √3y
B = ( √3y₁ , y₁)
2 =√(√3y₁ - 0)² + (y₁-0)²
=> 4 = 3y₁² + y₁²
=> 4 = 4y₁²
=> 1 = y₁²
=> y₁ = ± 1
but its in 1st Quadrant
Hence B = ( √3 , 1)
A = ( 0 , 0) , B = ( √3 , 1)
C = (x , y)
x² + y² = 4
=> x² + y² = (x - √3)² + (y - 1)²
=> x² + y² = x² -2√3x + 3 + y² - 2y + 1
=> 2√3x + 2y = 4
=> √3x + y = 2
=> y = 2 - √3x
x² + y² = 4
=> x² + (2 - √3x)² = 4
=> x² + 4 + 3x² - 4√3x = 4
=> 4x² - 4√3x = 0
=> x² - √3x = 0
=> x(x - √3) = 0
=> x = 0 , x = √3
=> y = 2 , y = - 1
( 0 , 2) and ( √3 , - 1) are the points
all the coordinates.
A = ( 0 , 0) , B = ( √3 , 1)
C = ( 0 , 2) or ( √3 , - 1)
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