# Find the centroid, the in-centre, the circum-centre and the orthocentre of the triangle whose sides have the equations ( 3 x-4 y=0,12 y+5 x=0 ) and ( y-15=0 )

Centroid of TriangleThe centroid of a triangle is the point of concurrency of the medians. The centroid G of the triangle ABC, divides the median AD, in the ratio of 2: 1 .

Incentre of TriangleThe incentre 'I' of a triangle is the point of concurrency of the bisectors of the angles of the triangle.

Circum Centre of TriangleThis 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.

Orthocentre of TriangleThis is the point of concurrency of the altitudes of the triangle. We have to find the co-ordinates of the centroid and the incentre of the triangle which is formed by the 3 lines whose equations are( 3 x-4 y=0 )

( 12 y+5 x=0 )

( y-15=0 )

Solve the 3 linear equation taking 2 at a time to get the co-ordinates of the vertices of the triangle as( A(0,0) quad B(20,15) quad C(-36,15) )

Now use distance formula to calculate the sides of the triangle as,

( A B(c)=sqrt{(20-0)^{2}+((15-0))^{2}} )

[

=25

]

( B C(a)=sqrt{(20+36)^{2}+((15-15))^{2}} )

[

=56

]

( A C(b)=sqrt{(0+36)^{2}+((0-15))^{2}} )

( =39 )

Thus centroid,

[

begin{array}{l}

=left(frac{x_{1}+x_{2}+x_{3}}{3}, frac{y_{1}+y_{2}+y_{3}}{3}right)

=left(frac{0+20-36}{3}, frac{0+15+15}{3}right)

end{array}

]

( =left(-frac{16}{3}, 10right) )

co-ordinates of incentre are-

[

begin{array}{l}

=left(frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}right)

=left(frac{56(0)+20(39)-25(36)}{120}, frac{56(0)+15(39)+25(15)}{120}right)

end{array}

]

( =(-1,8) )