Question

Prove that the point (0,0),(5,5) and (-5,5)are the vertices of a right isosceles triangle.

Answer 1

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$let \: for \: the \: \triangle ABC \: the \: vertices \: are \: \\ A(0,0),B(5,5) \: \: \& \: \: C( - 5&#44- 5)respectively \\ &#126 \: now &#126 \\ \underline{finding \: edges \: of \: the \: \triangle ABC} \\ \rightarrow AB = \sqrt{(0 - 5) {}^{2} + (0 - 5) {}^{2} } = 5\sqrt{2} \\ \rightarrow BC= \sqrt{(5 + 5) {}^{2} + (5 + 5) {}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: =10 \sqrt{2} \\ \rightarrow CA = \sqrt{( - 5 - 0) {}^{2} + ( - 5 - 0) {}^{2} } = 5 \sqrt{2} \\ therefore \: for \: the \triangle ABC \: AB =CA \\ therefore \: it \: is \: a \: isoscels \: triangle. \\ &#95{ now } \\ \implies AB {}^{2} + CA {}^{2} = (5 \sqrt{2} ) {}^{2} + (5 \sqrt{2} ) {}^{2} \\ \implies AB {}^{2} + CA {}^{2} = 200 \\ \implies AB {}^{2} + CA {}^{2} = (10 \sqrt{2} ) {}^{2} \\ \implies \boxed{ AB {}^{2} + CA {}^{2} = BC {}^{2} } \\ \therefore \triangle ABC \: is \: an \: right \: angled \\ \: isosceles \: triangle .\: ( \red{proved)}$

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Answer 2

Answer:

Given : Points (0, 0), (5, 5) and (-5, 5) are vertices of a triangle.

 

To prove : Vertices of a right-angled isosceles triangle

 

 Solution :  

Let A(0, 0), B(5, 5) and C (- 5, 5)

By using distance formula : √(x2 - x1)² + (y2 - y1)²

Vertices : A(0, 0), B(5, 5)

Length of side AB = √(5 - 0)² + (5 - 0)²

AB = √5² + (5)²  

AB = √25 + 25

AB = √50 units

 

Vertices :  B(6, 4) and C (- 1, 3)

Length of side BC = √(- 5 - 5)² + ( 5 - 5)²

BC = √(-10)² + (0)²

BC = √100 + 0

BC = √100 units

 

Vertices : A(3, 0), C (- 1, 3)

Length of side AC = √(- 5 - 0)² + (5 - 0)²

AC = √(-5)² + (5)²

AC = √25 + 25

AC = √50 units

Since the 2 sides AB = AC = √50 .

Therefore ∆ is an isosceles.

Now, in ∆ABC, by using Pythagoras theorem

BC² = AB² + AC²

(√100)²  = (√50)². + (√50)²

100 = 50 + 50

100 = 100

Since BC² = AB² + AC²  

Hence, the given vertices of a triangle is a right isosceles triangle.

Step-by-step explanation: