Question

The base AB of the two equilateral triangles ABC and ABC' with side $2a$ lies along the X-axis such that the mid-point of AB is at the origin. Find the coordinates of the vertices C and C' of the triangles.

Answer 1

Since the mid-point of AB is at the origin O and AB=2a
$\therefore$ OA=OB=a.
Thus, the coordinates of A and B are $(a,0)$ and $(-a,0)$ respectively.
Since triangles ABC. and ABC' are equilateral. Therefore, their third vertices C and C'
lie on the perpendicular bisector of base AB. Clearly, YOY is the perpendicular bisector
of AB. Thus, C and C' lie on Y-axis. Consequently, their x-coordinates are equal to Zero.
In $△AOC,wehave$
$O{A}^2+{OC}^2={AC}^2$
$\Rightarrow a^2+{OC}^2={\left(2a\right)}^2$
$\Rightarrow {OC}^2={4a}^2-a^2$
$\Rightarrow {OC}^2={3a}^2$
$OC=\sqrt{3a}$
Similarly, by applying Pythagoras theorem in $△$AOC; we have $O{C}^{\prime }=\sqrt{3a}$
Thus, the coordinates of C and C' are $(0,\sqrt{3}a)and(0,-\sqrt{3}a)respectively.$