Question

14. If vertices of a AABC are A(0, ), B(1, 0) and c(0,
1), then prove that AABC is right-angled triangle.
Also, find the coordinates of its orthocentre.​

Answer 1

Given :-

Vertices of triangle ABC are

• A= (0,1)
• B = (1, 0)
• C = (0, 1)

To prove :-

The given vertices are the vertices of right -angle triangle

Also To find :-

• Co-ordinates of Orthocentre

Concept Implemented:-

Firstly we prove these are the vertices of right angle triangle .

How to prove ?

First we have to find the distance between the points . If these values satisfies the Pythagoras theorem then these were the vertices of right angle triangle

Distance formulae :-

$\sqrt{(x_1 - x_2) {}^{2} + (y_1 - y_2) {}^{2} }$

Distance between A and B = AB

$= \sqrt{(0 -1 ) {}^{2} + (1 -0 ) {}^{2} }$

$= \sqrt{( - 1) {}^{2} + (1) {}^{2} }$

$= \sqrt{1 + 1}$

$= \sqrt{2}$

$\red {\: AB = \sqrt{2} }$

Distance between B and C is BC

$= \sqrt{(1 - 0) {}^{2} + (0 - 1) {}^{2} }$

$= \sqrt{(1) {}^{2} + ( - 1) {}^{2} }$

$= \sqrt{1 + 1}$

$= \sqrt{2}$

$\red{BC = \sqrt{2} }$

Distance between A and C is AC

$= \sqrt{(0 - 0) {}^{2} +(1 - 1) {}^{2} }$

$= \sqrt{0 + 0}$

$= 0$

$\red {AC = 0}$

Now checking the Pythagoras theorem,

$AB {}^{2} + AC {}^{2} = BC {}^{2}$

$AB {}^{2} + BC {}^{2} = AC {}^{2}$

$AC {}^{2} + BC {}^{2} = AB {}^{2}$

$( \sqrt{2} ) {}^{2} + (0) {}^{2} = ( \sqrt{2) {}^{2} }$

$2 + 0 = 2$

$2 = 2$

Satisfied 1st condition [AB²+AC²=BC²]

$( \sqrt{2} ) {}^{2} + ( \sqrt{2} ) {}^{2} = (0) {}^{2}$

$2 + 2 = 0$

$4 \neq0$

2nd condition not satisfied [AB² + BC² =/ AC²]

$(0) {}^{2} + ( \sqrt{2} ) {}^{2} = ( \sqrt{2} ) {}^{2}$

$0 + 2 = 2$

$2 = 2$

Satisfied 3rd condition [AC² + BC² = AB²]

So, If any one condition satisfied , It is a Right angled triangle .

Hence proved !!!!!____________________________

Finding the Orthocentre:-

In a right angle triangle Orthocentre lies at the vertex of right angle triangle [90°]

Now refer the attachment!

The right angle triangle at B =(1,0) So, the co-ordinates of Orthocentre are (1,0)

_________________________

Know more :-

☆In a right angle triangle circumcentre is midpoint of hypotenuse

☆ In a right angle triangle Orthocentre lies vertex of right angle triangle

☆Orthocentre, Ninepoint centre, Centroid , Circumcentre lies on same line [O,N,G,S]

☆ Ninepoint centre is the midpoint of Orthocentre and Circumcentre

☆

Centroid divides Orthocentre and circumcentre in ratio 2: 1

Answer 2

Given :-

Vertices of triangle ABC are

A= (0,1)

B = (1, 0)

C = (0, 1)

To prove :-

The given vertices are the vertices of right -angle triangle

Also To find :-

Co-ordinates of Orthocentre

Concept Implemented:-

Firstly we prove these are the vertices of right angle triangle .

How to prove ?

First we have to find the distance between the points . If these values satisfies the Pythagoras theorem then these were the vertices of right angle triangle

Distance formulae :-

$\sqrt{(x_1 - x_2) {}^{2} + (y_1 - y_2) {}^{2} }$

Distance between A and B = AB

$= \sqrt{(0 -1 ) {}^{2} + (1 -0 ) {}^{2} }$

$= \sqrt{( - 1) {}^{2} + (1) {}^{2} }$

$= \sqrt{1 + 1}$

$= \sqrt{2}$

$\red {\: AB = \sqrt{2} }$

Distance between B and C is BC

$= \sqrt{(1 - 0) {}^{2} + (0 - 1) {}^{2} }$

$= \sqrt{(1) {}^{2} + ( - 1) {}^{2} }$

$= \sqrt{1 + 1}$

$= \sqrt{2}$

$\red{BC = \sqrt{2} }$

Distance between A and C is AC

$= \sqrt{(0 - 0) {}^{2} +(1 - 1) {}^{2} }$

$= \sqrt{0 + 0}$

$= 0$

$\red {AC = 0}$

Now checking the Pythagoras theorem,

$AB {}^{2} + AC {}^{2} = BC {}^{2}$

$AB {}^{2} + BC {}^{2} = AC {}^{2}$

$AC {}^{2} + BC {}^{2} = AB {}^{2}$

$( \sqrt{2} ) {}^{2} + (0) {}^{2} = ( \sqrt{2) {}^{2} }$

$2 + 0 = 2$

$2 = 2$

Satisfied 1st condition [AB²+AC²=BC²]

$( \sqrt{2} ) {}^{2} + ( \sqrt{2} ) {}^{2} = (0) {}^{2}$

$2 + 2 = 0$

$4 \neq0$

2nd condition not satisfied [AB² + BC² =/ AC²]

$(0) {}^{2} + ( \sqrt{2} ) {}^{2} = ( \sqrt{2} ) {}^{2}$

$0 + 2 = 2$

$2 = 2$

Satisfied 3rd condition [AC² + BC² = AB²]

So, If any one condition satisfied , It is a Right angled triangle .

Hence proved !!!!!____________________________

Finding the Orthocentre:-

In a right angle triangle Orthocentre lies at the vertex of right angle triangle [90°]

Now refer the attachment!

The right angle triangle at B =(1,0) So, the co-ordinates of Orthocentre are (1,0)

_________________________

Know more :-

☆In a right angle triangle circumcentre is midpoint of hypotenuse

☆ In a right angle triangle Orthocentre lies vertex of right angle triangle

☆Orthocentre, Ninepoint centre, Centroid , Circumcentre lies on same line [O,N,G,S]

☆ Ninepoint centre is the midpoint of Orthocentre and Circumcentre

☆

Centroid divides Orthocentre and circumcentre in ratio 2: 1