Question

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

Answer 1

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.

Some more questions :  

Prove that the points (3, 0), (6, 4) and (- 1, 3) are vertices of a right-angled isosceles triangle.

https://brainly.in/question/15937740

 

Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus.

https://brainly.in/question/15937739

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption

P(0, 0), Q(5, 5) and R(-5, 5)

Now,

Using distance formula,

PQ = √(x2 - x1)² + (y2 - y1)²

PQ= √(5 - 0)² + (5 - 0)²

PQ = √5² + (5)²

PQ = √25 + 25

PQ = √50 units

Now,

QR = √(-5 - 5)² + ( 5 - 5)²

QR = √(-10)² + (0)²

QR = √100 + 0

QR = √100 units

Also,

PR = √(- 5 - 0)² + (5 - 0)²

PR = √(-5)² + (5)²

PR = √25 + 25

PR = √50 units

Here we get,

PQ = PR = √50

Hence,

Triangle is an isosceles.

Now,

In ∆PQR,

Using Pythagoras theorem

QR² = PQ² + PR²

(√100)² = (√50)² + (√50)²

100 = 50 + 50

100 = 100

Hence,

QR² = PQ² + PR²

Therefore,

Vertices of a ∆ is a right isosceles triangle.