The points $(-5, 0)$, $(5, 0)$ and $(0, 4)$ are the vertices of a triangle which is a/an :

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Accepted Answer

## Model Answer **Option B: Isosceles triangle** Let A(–5, 0), B(5, 0), C(0, 4). $AB = \sqrt{(10)^2+0^2} = 10$ $AC = \sqrt{25+16} = \sqrt{41}$ $BC = \sqrt{25+16} = \sqrt{41}$ Since AC = BC ≠ AB, it is an **isosceles triangle**. Source: Chapter 7, Section 7.2 (Distance Formula) --- ## Explanation The examiner expects you to apply the distance formula to all three sides and observe that two sides are equal (AC = BC = √41) while the third is different (AB = 10). Equal two sides → isosceles. Also verify it is not equilateral (all sides equal) or right-angled (no Pythagoras triplet holds here: 41 + 41 ≠ 100).

Mathematics — Standard (041) — CBSE Class 10 board question