Question

Show that the points (0, 3), (-2,1) and (-1, 4) are the vertices of a right angled triangle.

Answer 1

Answer:

The points ( 0, 3 ), ( - 2, 1 ) & ( - 1, 4 ) are the vertices of the right-angled triangle.

Step-by-step-explanation:

NOTE: Refer to the attachment for the diagram.

Let the points be A, B and C.

A ≡ ( 0, 3 ) ≡ ( x₁, y₁ )

B ≡ ( - 2, 1 ) ≡ ( x₂, y₂ )

C ≡ ( - 1, 4 ) ≡ ( x₃, y₃ )

Now, we know that,

d ( A, B ) = √[ ( x₁ - x₂ )² + ( y₁ - y₂ )² ] - - [ Distance formula ]

⇒ d ( A, B ) = √{ [ 0 - ( - 2 ) ]² + ( 3 - 1 )² }

⇒ d ( A, B ) = √[ ( 0 + 2 )² + ( 2 )² ]

⇒ d ( A, B ) = √[ ( 2 )² + 4 ]

⇒ d ( A, B ) = √( 4 + 4 )

⇒ d ( A, B ) = √8

Squaring both sides we get,

∴ ( AB )² = 8 - - ( 1 )

Now,

d ( B, C ) = √[ ( x₂ - x₃ )³ + ( y₂ - y₃ )² ]

⇒ d ( B, C ) = √{ [ - 2 - ( 1 ) ]² + ( 1 - 4 )² }

⇒ d ( B, C ) = √[ ( - 2 + 1 )² + ( - 3 )² ]

⇒ d ( B, C ) = √[ ( - 1 )² + 9 ]

⇒ d ( B, C ) = √( 1 + 9 )

⇒ d ( B, C ) = √10

Squaring both sides, we get,

∴ ( BC )² = 10 - - ( 2 )

Now,

d ( A, C ) = √[ ( x₁ - x₃ )² + ( y₁ - y₃ )² ]

⇒ d ( A, C ) = √{ [ 0 - ( - 1 ) ]² + ( 3 - 4 )² }

⇒ d ( A, C ) = √[ ( 0 + 1 )² + ( - 1 )² ]

⇒ d ( A, C ) = √[ ( 1 )² + 1 ]

⇒ d ( A, C ) = √( 1 + 1 )

⇒ d ( A, C ) = √2

Squaring both sides, we get,

∴ ( AC )² = 2 - - ( 3 )

From ( 1 ), ( 2 ) & ( 3 ),

∴ BC > AB > AC

In △ABC, BC is the greatest side.

( BC )² = 10 - - [ From ( 2 ) ]

Adding equations ( 1 ) & ( 3 ), we get,

⇒ ( AB )² + ( AC )² = 8 + 2

⇒ ( AB )² + ( AC )² = 10 - - ( 4 )

From ( 2 ) & ( 4 ), in △ABC,

The square of the greatest side is equal to the sum of squares of the remaining two sides.

∴ By converse of Pythagors theorem,

△ABC is a right-angled triangle.

∴ The points ( 0, 3 ), ( - 2, 1 ) & ( - 1, 4 ) are the vertices of the right-angled triangle.

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Additional Information:

1. Distance Formula:

The formula which is used to find the distance between two points using their coordinates is called distance formula.

• d ( A, B ) = √[ ( x₁ - x₂ )² + ( y₁ - y₂ )² ]

2. Section Formula:

The formula which is used to find the coordinates of a point which divides a line segment in a particular ratio is called section formula.

• x = ( mx₂ + nx₁ ) / ( m + n )

• y = ( my₂ + ny₁ ) / ( m + n )

Answer 2

Let , A(0, 3) , B(-2,1) and C(-1, 4) are the vertices of triangle

As we know that , the distance formula is given by

$\boxed{ \tt{Distance = \sqrt{ {( x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} } }}$

Thus ,

$\tt \implies AB = \sqrt{ {( - 2 - 0)}^{2} + {(1 - 3)}^{2} }$

$\tt \implies AB = \sqrt{4 + 4}$

$\tt \implies AB = \sqrt{8}$

$\tt \implies AB = 2 \sqrt{2}$

Similarly ,

$\tt \implies BC = \sqrt{ {( - 1 - \{ - 2\})}^{2} + {(4 - 1)}^{2} }$

$\tt \implies BC = \sqrt{1 + 9}$

$\tt \implies BC = \sqrt{10}$

Now ,

$\tt \implies AC = \sqrt{ {( - 1 - 0)}^{2} + {(4 - 3)}^{2} }$

$\tt \implies AC = \sqrt{1 + 1}$

$\tt \implies AC = \sqrt{2}$

It is observed that ,

(AB)² + (AC)² = (BC)²

Hence , the given points are the vertices of a right angled triangle