show that the points (a,a),(-a,-a) and (-√3a,√3a). are the vertices of an equilateral triangle.
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For having the vertices of an equilateral triangle all the sides of the triangle should be equal.
Let points A,B,C have vertices (a,a) , (-√3a,✓3a) , (-a,a).
Distance between two points is
✓(X2-X1) + (Y2-Y1)
therefore,
AB = √(-√3a-a)^2 - (√3a-a)^2
= √3a^2+a^2+2a^2√3+3a^2+a^2-2a^2√3
= √8a^2
= 2a√2
Similarly,
AC = √(-a-a)^2 + (-a-a)^2
= √4a^2+4a^2
= √8a^2
= 2a√2
and
BC = √(-a-(-√3a))^2 + (-a-√3a)^2
= √8a^2
= 2a√2
therefore, length of AB=AC=BC
Hence proved .
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