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# An equilateral triangle has vertex at the point (3,4) and at another point (-2,3). Find the coordinates of the third vertex.

Asked by Udit Lilhare 29th January 2014, 4:46 PM
Let the equilateral triangle be ABC where A = (3,4) and B=(-2,3) and C=(x,y) Now, AB=BC=AC (equilateral triangle) Thus, AB^2 = BC^2 = AC^2 Now, AB^2 = [3-(-2)]^2 + [4-3]^2 = 25+1=26 AC^2 = [3-x]^2 + [4-y]^2 = 9 - 6x + x^2 + 16 -8y + y^2 = x^2 + y^2 - 6x - 8y + 25 And, BC^2 = [-2-x]^2 + [3-y]^2 = 4 + 4x +x^2 + 9 -6y + y^2 =x^2 + y^2 + 4x - 6y + 13 Now, AC^2 = BC^2 i.e. x^2 + y^2 - 6x - 8y + 25 =x^2 + y^2 + 4x - 6y + 13 i.e. -10x - 2y = -12 i.e. 5x + y = 6 i.e. y = 6 - 5x Now, AB^2 = AC^2 26 = = x^2 + y^2 - 6x - 8y + 25 26 = x^2 + (6-5x)^2 -6x -8(6-5x) + 25 26 = x^2 + (36 -60x + 25x^2) - 6x -48 + 40x + 25 26 = x^2 + 36 - 60x + 25x^2 - 6x - 48 + 40x + 25 26 = 26x^2 - 26x + 13 26x^2 - 26x - 13 = 0 2x^2 - 2x - 1 = 0 For getting further solution in detail, please refer to our Ask the Expert Section.
Answered by Expert 29th January 2014, 5:15 PM
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