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Two vertices of an isosceles triangle in the coordinate plane have coordinates (0,0) and (2a,0). Where might the third vertex be located?

User Entalyan
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1 Answer

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The third vertex might be (a,0). Consider A = (0,0) B = (2a,0) and C = (a,0) For an isosceles triangle, two sides and two angles will be equal. Based on that theory, side BC = CA. So calculating the distance BC = (2a-a)^2 +(0-0)^2 = a^2 Calculating the distance CA = (a-0)^2 + (0-0)^2 = a^2 The sides are equal that is a^2
User Zmey
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