The base $BC$ of an equilateral triangle $ABC$ lies on the $y$-axis. The coordinates of vertex $C$ are $(0, -3)$. The origin is the midpoint of the base $BC$. Find the coordinates of vertices $A$ and $B$.
Step-by-Step Solution
Key Concept: Origin $O(0,0)$ is midpoint of $BC$. $C(0, -3) \Rightarrow B(0, 3)$. Base length $BC = 6$. Height of equilateral triangle $h = \dfrac{\sqrt{3}}{2} \times 6 = 3\sqrt{3}$. Vertex $A$ lies on $x$-axis at $(3\sqrt{3}, 0)$ or $(-3\sqrt{3}, 0)$.
Origin is midpoint of $BC \Rightarrow$ Since $C = (0, -3)$, $B = (0, 3)$. [1.0 Mark]
Length of base $BC = 3 - (-3) = 6$ units. [1.0 Mark]
In equilateral $\Delta ABC$, side length $s = 6$. Height $h = \dfrac{\sqrt{3}}{2} \times 6 = 3\sqrt{3}$. [1.5 Marks]
Vertex $A$ lies on $x$-axis at distance $3\sqrt{3}$ from origin. Thus $A = (3\sqrt{3}, 0)$ or $A = (-3\sqrt{3}, 0)$. Vertices are $B(0, 3), C(0, -3)$ and $A(\pm 3\sqrt{3}, 0)$. [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Finding $B(0, 3)$: 1.0 Mark
Calculating side length $= 6$: 1.0 Mark
Calculating altitude height $= 3\sqrt{3}$: 1.5 Marks
Evaluating vertex $A(\pm 3\sqrt{3}, 0)$: 1.5 Marks
Correct Answer: