show that the straight lines represented by (X+2a)^2-3y^2=0 and x=a form an eqiulateral triangle
Answers
Given : straight lines represented by (X+2a)^2-3y^2=0 and x=a form an equilateral triangle
To Find : Prove that
Solution:
(X+2a)²-3y²=0
=> (X+2a)²-(√3 y)²=0
=> ( x + 2a + √3y)( x + 2a - √3y) = 0
x + 2a + √3y = 0
x + 2a - √3y = 0
intersection point : ( - 2a , 0 )
x + 2a + √3y = 0 , x = a
=> intersection point : (a , - √3a)
x + 2a - √3y = 0 , x = a
=> intersection point : (a , √3a)
points of triangle are
( - 2a , 0 ) , (a , - √3a) , (a , √3a)
Side length = √(-2a - a)² + (0 - (-√3a))² = √12a² = 2√3a
= √(-2a - a)² + (0 - (√3a))² = √12a² = 2√3a
= √(a - a)² + (-√3 a - √3a)² = 2√3a
all three sides = 2√3a
Hence Equilateral Triangle
QED
Hence proved
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