Question 2 The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.
Class X1 - Maths -Straight Lines Page 211
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∆ABC is an equilateral, then AB = BC
AB² = BC²
use distance formula,
distance between (x₁,y₁) and (x₁,y₂) = √{(x₂-x₁)² +(y₂-y₁)²}
( h - 0 )² + ( 0 - a )² = (2a)²
h² + a² = 4a²
h² = 3a²
h = ±√3a
hence, the vertices of traingle are ( √3a, 0) , (0,a) and ( 0, -a) .
AB² = BC²
use distance formula,
distance between (x₁,y₁) and (x₁,y₂) = √{(x₂-x₁)² +(y₂-y₁)²}
( h - 0 )² + ( 0 - a )² = (2a)²
h² + a² = 4a²
h² = 3a²
h = ±√3a
hence, the vertices of traingle are ( √3a, 0) , (0,a) and ( 0, -a) .
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