Question

The incenter of the triangle formed by co-ordinate axes and the line x/a+y/b=1 is

Answer 1

A(a,0),B(0,b)

C(0,0) be the vertices

AB=

$\sqrt{ {a}^{2} + {b}^{2} }$

BC=b

AC=a

incentre Ix=(BCx1+ACx2+ABx3)÷(BC+AC+AB)

=(ba+0+0)÷[b+a+root(a^2+b^2)]

$\frac{ab}{ \sqrt{a + b + \sqrt{ {a}^{2} + {b}^{2} } } }$

similarly y coordinate also same