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)
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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).