show that the following points are the vertices of right angle isosceles triangle is 1,2 .4,2,1,5
Answers
Answer:
The given points are vertices of isosceles right triangle.
Step-by-step-explanation:
Let the triangle be ABC.
- A ( 1, 2 ) ≡ ( x₁, y₁ )
- B ( 4, 2 ) ≡ ( x₂, y₂ )
- C ( 1, 5 ) ≡ ( x₃, y₃ )
Now, by distance formula,
d ( A, B ) = √[ ( x₁ - x₂ )² + ( y₁ - y₂ )² ]
⇒ d ( A, B ) = √[ ( 1 - 4 )² + ( 2 - 2 )² ]
⇒ d ( A, B ) = √[ ( - 3 )² + 0² ]
⇒ d ( A, B ) = √( 9 + 0 )
⇒ d ( A, B ) = √( 9 )
⇒ d ( A, B ) = 3
∴ Length of AB = 3 units - - - ( 1 )
Now,
d ( B, C ) = √[ ( x₂ - x₃ )² + ( y₂ - y₃ )² ]
⇒ d ( B, C ) = √[ ( 4 - 1 )² + ( 2 - 5 )² ]
⇒ d ( B, C ) = √[ ( 3 )² + ( - 3 )² ]
⇒ d ( B, C ) = √( 9 + 9 )
⇒ d ( B, C ) = √18
∴ Length of BC = √18 units
∴ BC² = ( √18 )²
⇒ BC² = 18 units - - - ( 2 )
Now,
d ( A, C ) = √[ ( x₁ - x₃ )² + ( y₁ - y₃ )² ]
⇒ d ( A, C ) = √[ ( 1 - 1 )² + ( 2 - 5 )² ]
⇒ d ( A, C ) = √[ 0² + ( - 3 )² ]
⇒ d ( A, C ) = √( 0 + 9 )
⇒ d ( A, C ) = √( 9 )
⇒ d ( A, C ) = 3
∴ Length of AC = 3 units - - - ( 3 )
From ( 1 ) & ( 3 )
Length of AB = Length of AC
∴ △ABC is an isosceles triangle.
Now,
AB² + AC² = ( 3 )² + ( 3 )²
⇒ AB² + AC² = 9 + 9
⇒ AB² + AC² = 18 units - - - ( 4 )
From ( 2 ) & ( 4 ),
AB² + AC² = BC²
By converse of Pythagoras theorem,
m∠A = 90°
∴ △ABC is an isosceles right triangle.
∴ The given points are vertices of isosceles right triangle.