if A(x₁,y₁), B(x₂,y₂), and C(x₃,y₃) are the vertices of a triangle, then find the coordinates of the orthocentre, circumcentre, centroid and incentre of the triangle.

By geometry, we know that BD/DC = AB/AC (since AD bisects �A).

 

If the lengths of the sides AB, BC and AC are c, a and b respectively, then BD/DC = AB/AC = c/b.

 

Coordinates of D are (bx₂+cx₃/b+c, by₂+cy₃/b+c)

 

IB bisects �B. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.

Let the coordinates of I be (x, y).

_{– –}

Then x = ax₁+bx₂+cx₃/a+b+c, y = ay₁+by₂+cy₃/a+b+c.

 

 

This the point of concurrency of the perpendicular bisectors of the sides of the triangle. This is also the centre of the circle, passing through the vertices of the given triangle.

 

 

This is the point of concurrency of the altitudes of the triangle.

 

 

Excentre of a triangle is the point of concurrency of bisectors of two exterior and third interior angle. Hence there are three excentres I₁, I₂ and I₃ opposite to three vertices of a triangle.

 

If A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of a triangle ABC,

coordinates of centre of ex-circle opposite to vertex A are given as

 

I₁(x, y) = (–ax₁+bx₂+cx₃/a+b+c/–a+b+c, –ay₁+by₂+cy₃/–a+b+c).

 

Similarly co-ordinates of centre of I₂(x, y) and I₃(x, y) are

 

I₂(x, y) = (ax₁–bx₂+cx₃/a–b+c, ay₁–by₂+cy₃/a–b+c)