Q1. [2]
$A(3, 0)$, $B(6, 4)$ and $C(-1, 3)$ are vertices of a triangle ABC. Find length of its median BE.
Previously asked in CBSE board exam
2024 30/2/1 Q25(b) (OR-2)
Generated by claude-sonnet-4-6 · 2026-06-15 10:28 · grounding rag
Model Answer
BE is a median, so E is the midpoint of AC.
$$E = \left(\frac{3+(-1)}{2},\ \frac{0+3}{2}\right) = \left(1,\ \frac{3}{2}\right)$$
B = (6, 4), E = (1, 3/2)
$$BE = \sqrt{(6-1)^2 + \left(4-\frac{3}{2}\right)^2} = \sqrt{25 + \frac{25}{4}} = \sqrt{\frac{125}{4}} = \frac{5\sqrt{5}}{2} \text{ units}$$
Source: Chapter 7, Section 7.2 (Distance Formula) and Section 7.4 (Midpoint Formula)
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Explanation
- A median from B goes to the midpoint of the opposite side AC, not any arbitrary point. First find E using the midpoint formula, then apply the distance formula.
- Key step students miss: correctly identifying which side the median bisects (opposite side to vertex B is AC).
- Simplification: $\sqrt{125/4} = \sqrt{125}/2 = 5\sqrt{5}/2$. Remember $\sqrt{125} = 5\sqrt{5}$.
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