Prove that the points $(3, 0)$, $(6, 4)$ and $(-1, 3)$ are the vertices of an isosceles triangle.
Generated by claude-sonnet-4-6 · 2026-06-15 10:29 · grounding rag
Model Answer
Let A(3, 0), B(6, 4) and C(–1, 3) be the given points.
$$AB = \sqrt{(6-3)^2 + (4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$$
$$BC = \sqrt{(-1-6)^2 + (3-4)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$$
$$CA = \sqrt{(3-(-1))^2 + (0-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$$
Since AB = CA = 5, two sides are equal.
Therefore, A, B and C are the vertices of an isosceles triangle.
Source: Chapter 7, Section 7.2 – Distance Formula
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Explanation
- Examiners expect all three distances calculated using the distance formula, with clear working.
- The conclusion must explicitly state which two sides are equal and name the triangle as isosceles.
- No need to check collinearity separately here; equal sides are sufficient proof.
- A common mistake is an arithmetic error in BC — note $(-7)^2 + (-1)^2 = 50$, not 48.