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Centroid, Incentre, Circumcentre and Ex-centre of Triangle What do you mean by the centroid of a triangle? In order to understand the term centroid, we first need to know what do we mean by a median. A median is the line joining the mid-points of the sides and the opposite vertices. A centroid is the point of intersection of the medians of the triangle. A centroid divides the median in the ratio 2:1. Result: Find the centroid of the triangle the coordinates of whose vertices are given by A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) respectively. Solution: A centroid divides the median in the ratio 2:1. Hence, since ‘G’ is the median so AG/AD = 2/1. Since D is the midpoint of BC, coordinates of D are ((x_{2 }+ x_{3)}/2, (y_{2 }+ y_{3)}/2) Using the section formula, the coordinates of G are (2(x_{2}+x_{3})/2) +1.x_{1}/2+1, (2(y_{2}+y_{3})/2) +1.y_{1}/2+1) ⇒ Coordinates of G are (x_{1}+x_{2}+x_{3}/3, y_{1}+y_{2}+y_{3}/3). If the coordinates of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the coordinates of the centroid (which is generally denoted by G) are given by What do you mean by the incentre of a triangle? The incentre of a triangle is the point of intersection of the angle bisectors of angles of the triangle. An incentre is also the centre of the circle touching all the sides of the triangle. Note:
In order to understand the term centroid, we first need to know what do we mean by a median. A median is the line joining the mid-points of the sides and the opposite vertices. A centroid is the point of intersection of the medians of the triangle. A centroid divides the median in the ratio 2:1.
Find the centroid of the triangle the coordinates of whose vertices are given by A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) respectively.
A centroid divides the median in the ratio 2:1. Hence, since ‘G’ is the median so AG/AD = 2/1.
Since D is the midpoint of BC, coordinates of D are ((x_{2 }+ x_{3)}/2, (y_{2 }+ y_{3)}/2)
Using the section formula, the coordinates of G are
(2(x_{2}+x_{3})/2) +1.x_{1}/2+1, (2(y_{2}+y_{3})/2) +1.y_{1}/2+1)
⇒ Coordinates of G are (x_{1}+x_{2}+x_{3}/3, y_{1}+y_{2}+y_{3}/3).
If the coordinates of a triangle are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}), then the coordinates of the centroid (which is generally denoted by G) are given by
The incentre of a triangle is the point of intersection of the angle bisectors of angles of the triangle. An incentre is also the centre of the circle touching all the sides of the triangle.
Angle bisector divides the oppsoite sides in the ratio of remaining sides i.e. BD/DC = AB/AC = c/b.
Incentre divides the angle bisectors in the ratio (b+c):a, (c+a):b and (a+b):c.
By geometry, we know that BD/DC = AB/AC (since AD bisects ÐA).
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 (bx_{2}+cx_{3}/b+c, by_{2}+cy_{3}/b+c)
IB bisects DB. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.
Let the coordinates of I be (x, y).
Then x = ax_{1}+bx_{2}+cx_{3}/a+b+c, y = ay_{1}+by_{2}+cy_{3}/a+b+c.
The point of intersection of perpendicualr bisectors of the sides of a triangle is called the circumcentre of triangle. This is also the centre of the circle, passing through the vertices of the given triangle.
The coordinates of circumcentre are given by
{(x_{1} sin 2A + x_{2} sin 2B + x_{3} sin 2C)/ (sin 2A + sin 2B + sin 2C), (y_{1} sin 2A + y_{2} sin 2B + y_{3} sin 2C)/ (sin 2A + sin 2B + sin 2C)}.
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.
Coordinates of centre of ex-circle opposite to vertex A are given as
I_{1}(x, y) = (–ax_{1}+bx_{2}+cx_{3}/a+b+c/–a+b+c, –ay_{1}+by_{2}+cy_{3}/–a+b+c).
Similarly co-ordinates of centre of I_{2}(x, y) and I_{3}(x, y) are
I_{2}(x, y) = (ax_{1}–bx_{2}+cx_{3}/a–b+c, ay_{1}–by_{2}+cy_{3}/a–b+c)
I_{3}(x, y) = (ax_{1}+bx_{2}–cx_{3}/a+b–c, ay_{1}+by_{2}–cy_{3}/a+b–c)
The coordinates of the excentre are given by
I_{1} = (-ax_{1} + bx_{2 }+ cx_{3})/(-a + b + c), (-ay_{1 }+ by_{2 }+ cy_{3})/(-a + b + c)}
Similarly, we have I_{2} = (ax_{1} - bx_{2 }+ cx_{3})/(a - b + c), (ay_{1 }- by_{2 }+ cy_{3})/(a - b + c)}
I_{3} = (ax_{1} + bx_{2 }- cx_{3})/(a + b - c), (ay_{1 }+ by_{2 }- cy_{3})/(a + b - c)}
In a right angled triangle, orthocentre is the point where right angle is formed.
In an isosceles triangle, all of centroid, orthocentre, incentre and circumcentre lie on the same line.
Orthocentre, centroid and circumcentre are always collinear and centroid divides the line joining orthocentre and circumcentre in the ratio 2:1.
If (0, 1), (1, 1) and (1, 0) are middle points of the sides of a triangle, find its incentre.
Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3})be teh vertices of a triangle.
x_{1} + x_{2} = 0, x_{2} + x_{3 }= 0, x_{3} + x_{1} = 0
y_{1} + y_{2} = 0, y_{2} + y_{3} = 0, y_{3} + y_{1} = 0.
Solving these equations, we get A(0, 0), B(0, 2) and C(2, 0).
Now, a = BC = 2√ 2, b = CA = 2 and c = AB = 2.
Thus, incentre of the triangle ABC is (2-√ 2, 2-√ 2).
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