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

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

Im sorry i cant answer it : ( im so sorry

Iknow one of the answers is a and maybe b but not 100% sure about b