In the given figure, \(\Delta ABC\) is an equilateral triangle. \(AD\) is a median of the triangle joining the points \(A\left(0, \frac{5\sqrt{3}}{2}\right)\), \(D(0, 0)\). Points \(B\) and \(C\) are (in same order) :
Show Hint
For any equilateral triangle symmetric about an axis, the height and half-base form a \(30^\circ - 60^\circ - 90^\circ\) right-angled triangle.
The ratio of sides in such a triangle is \(1 : \sqrt{3} : 2\).
Since the height is \(\frac{5}{2}\sqrt{3}\), the half-base is simply \(\frac{5}{2}\), which instantly gives the x-coordinates.
Step 1: Place $B$ and $C$ symmetrically on the x-axis and use the distance formula instead of the height formula.
Since $A$ and $D$ both have x-coordinate 0, the median $AD$ lies along the y-axis, so $BC$ must lie along the x-axis with $D(0,0)$ as its midpoint. Let $B=(-p,0)$ and $C=(p,0)$ for some $p>0$, so that $BC = 2p$.
Step 2: Use the fact that all three sides of an equilateral triangle are equal.
Side $AB$ must equal side $BC$. Using the distance formula for $AB$, where $A=\left(0,\frac{5\sqrt{3}}{2}\right)$ and $B=(-p,0)$:
\[ AB^{2} = (0-(-p))^{2} + \left(\frac{5\sqrt{3}}{2}-0\right)^{2} = p^{2} + \frac{75}{4} \]
Step 3: Set $AB^{2}=BC^{2}$ and solve for $p$.
\[ p^{2} + \frac{75}{4} = (2p)^{2} = 4p^{2} \]
\[ \frac{75}{4} = 3p^{2} \]
\[ p^{2} = \frac{25}{4} \]
\[ p = \frac{5}{2} \]
Final Answer:
So $B=\left(-\frac{5}{2},0\right)$ and $C=\left(\frac{5}{2},0\right)$, matching option (B).
\[ \boxed{B=\left(-\tfrac{5}{2},0\right),\ C=\left(\tfrac{5}{2},0\right)} \]
