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
In an equilateral triangle, the altitude is always \(\frac{\sqrt{3}}{2}\) times the side length.
If the altitude is \(\frac{5\sqrt{3}}{2}\), then the side must be 5.
Since the base is bisected by the origin, the coordinates are simply half of the side length on either side of the origin, giving \(\pm \frac{5}{2}\).
This simple mental calculation helps you find the correct coordinates quickly.
Step 1: Understand the setup using right-triangle trigonometry instead of the altitude formula.
Since $A\left(0, \frac{5\sqrt{3}}{2}\right)$ and $D(0,0)$ both have $x = 0$, the segment $AD$ lies along the y-axis, and $D$ is the midpoint of $BC$. Because triangle $ABC$ is equilateral and $AD$ is the median from $A$, $AD$ is also perpendicular to $BC$, so $BC$ lies along the x-axis and angle $ADB = 90^{\circ}$.
Step 2: Find the length AD.
Using the distance formula between $A\left(0, \frac{5\sqrt{3}}{2}\right)$ and $D(0,0)$:
\[ AD = \sqrt{(0-0)^2 + \left(\frac{5\sqrt{3}}{2} - 0\right)^2} = \frac{5\sqrt{3}}{2} \]
Step 3: Use the base angle of the equilateral triangle in right triangle ABD.
In an equilateral triangle every interior angle is $60^{\circ}$, so angle $ABD = 60^{\circ}$. Triangle $ABD$ is right angled at $D$ (since $AD \perp BC$). In this right triangle, $AD$ is the side opposite angle $B$, and $BD$ is the side adjacent to angle $B$. So we can write the tangent ratio:
\[ \tan(\angle ABD) = \frac{AD}{BD} \]
\[ \tan 60^{\circ} = \frac{AD}{BD} \]
Step 4: Solve for BD.
We know $\tan 60^{\circ} = \sqrt{3}$, so
\[ \sqrt{3} = \frac{\frac{5\sqrt{3}}{2}}{BD} \]
\[ BD = \frac{\frac{5\sqrt{3}}{2}}{\sqrt{3}} = \frac{5}{2} \]
Step 5: Locate B and C using symmetry.
Since $D(0,0)$ is the midpoint of $BC$ and $BC$ lies on the x-axis, $B$ and $C$ are at equal distances of $\frac{5}{2}$ from $D$, on opposite sides. So
\[ B = \left(-\frac{5}{2}, 0\right), \quad C = \left(\frac{5}{2}, 0\right) \]
Final Answer:
The coordinates of B and C are $\left(-\frac{5}{2}, 0\right)$ and $\left(\frac{5}{2}, 0\right)$, so option (B) is correct.
\[ \boxed{\left(-\tfrac{5}{2}, 0\right), \left(\tfrac{5}{2}, 0\right)} \]
