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 with base on the x-axis and vertex on the positive y-axis:
The x-coordinates of the base vertices are always $\pm \frac{h}{\sqrt{3}}$, where $h$ is the altitude.
Here, $h = \frac{5\sqrt{3}}{2}$.
So the x-coordinates are:
\[ \pm \frac{\frac{5\sqrt{3}}{2}}{\sqrt{3}} = \pm \frac{5}{2} \]
This quick shortcut directly yields the base coordinates in one line!
Step 1: Represent B and C with a single unknown t, and use the distance formula instead of the altitude-to-side formula.
Since $D(0,0)$ is the midpoint of side $BC$ and $BC$ lies along the x-axis (because $AD$, along the y-axis, is perpendicular to it), we can write $B = (-t, 0)$ and $C = (t, 0)$ for some positive number $t$, so that $D$ is automatically their midpoint.
Step 2: Write down the equilateral condition using the distance formula.
For the triangle to be equilateral, side $AB$ must equal side $BC$. Using the distance formula for $AB$, where $A = \left(0, \frac{5\sqrt{3}}{2}\right)$ and $B = (-t, 0)$:
\[ AB = \sqrt{(0-(-t))^2 + \left(\frac{5\sqrt{3}}{2} - 0\right)^2} = \sqrt{t^2 + \frac{75}{4}} \]
And $BC$, the distance between $(-t,0)$ and $(t,0)$, is simply:
\[ BC = 2t \]
Step 3: Set AB equal to BC and solve for t.
\[ \sqrt{t^2 + \frac{75}{4}} = 2t \]
Square both sides:
\[ t^2 + \frac{75}{4} = 4t^2 \]
\[ \frac{75}{4} = 3t^2 \]
\[ t^2 = \frac{25}{4} \]
\[ t = \frac{5}{2} \]
Step 4: Write down the coordinates of B and C using this value of t.
\[ B = \left(-\frac{5}{2}, 0\right), \qquad C = \left(\frac{5}{2}, 0\right) \]
Final Answer:
The coordinates of $B$ and $C$ are $\left(-\frac{5}{2}, 0\right)$ and $\left(\frac{5}{2}, 0\right)$, which matches Option (B).
\[ \boxed{B\left(-\tfrac{5}{2},0\right),\ C\left(\tfrac{5}{2},0\right)} \]
