Find the co-ordinates of the incentre of a triangle formed by the lines 3x-4y=0, 4x+3y-8=0 and 24x-7y-12=0.

 

Hey!

this is a easy but lengthy problem.

solve the equations to get the coordinates of the vertices.

use distance formula to get the length of sides: a,b,c.

Coordinates of incenter is given by formula:

 

(ax1+bx2+cx3 / a+b+c , ay1+by2+cy3 /a+b+c )

 

Any shorter method????

where A (x1,y1)  B(x2,y2)  C(x3,y3) are vertices of triangle that is found by solving equations two at a time to find the point of intersection of the lines ie vertices

First  you need to find the coordinates of the three vertices of the given triangle.

 

On solving we get the coordinates as

 

A(16/25 , 12/25)

 

B(32/25 , 24/25)

 

C(23/25 , 36/25)

 

Now equations of the angle bisector between AB and AC will be:-

 

(3x - 4y) / (25) =  (4x + 3y - 8)

 

We will get two equations as

 

x + 7y - 8 = 0.................1

 

7x - y - 8  = 0.................2

 

Now solve the above equations to get the incentre.

 

Cheers !!!!!!!!!!!!!!!!!