The given question is a medium difficulty 2-mark question from the chapter Coordinate Geometry. Concept tested: Computing lengths of sides of a triangle using the coordinates of the three vertices of the triangle and applying Pythagoras Theorem to establish that the triangle is a right triangle. The question appeared in the 2016 CBSE class 10 board paper.

Question 8: Prove that the point (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle

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Let (3, 0), (6, 4), and (-1, 3) be the coordinates of vertices A, B, and C respectively.

AB = \\sqrt{(6 - 3)^2 + (4 - 0)^2 }) = \\sqrt{3^2 + 4^2}) = \\sqrt{25}) = 5

BC =\\sqrt{(-1 - 6)^2 + (3 - 4)^2}) = \\sqrt{7^2 + 1^2}) = \\sqrt{50})

CA =\\sqrt{(-1 - 3)^2 + (3 - 0)^2}) = \\sqrt{4^2 + 3^2}) = \\sqrt{25}) = 5

Sides of triangle ABC measure 5, 5, and √50.

**Inference 1**: 2 sides are equal and therefore, the triangle is an Isosceles triangle.

**Observation**: 5^{2} + 5^{2} = \\sqrt{50^2})

**Inference 2**: Sum of squares of two sides equal square of the third side. Satisfies Pythagoras theorem.

Hence, the given triangle is a right triangle.

∴ the given triangle is a __Right Isosceles__ triangle.

Class 10 Maths

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CBSE Class 10 Maths - 2022

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