radius of the incircle,

$\mathrm{r}=\frac{\mathrm{\Delta }}{\mathrm{s}}=\frac{(18\cdot 24)2}{2(18+24+30)}=\frac{18\cdot 24}{72}=6$

Hence coordinates of incentre (6, 6) ; orthocentre is (0, 0)

distance  $\mathrm{OI}=\sqrt{36+36}=6\sqrt{2}$

II–part:

$\text{ centroid is }=\left(\frac{18}{3},\frac{24}{3}\right)=(6,8)$

Hence the triangle formed by I, G and circumcentre is

$\therefore \mathrm{A}=\frac{1}{2}\left|\begin{array}{ccc}6& 6& 1\\ 9& 12& 1\\ 6& 8& 1\end{array}\right|=\frac{1}{2}\cdot 2\cdot 3\cdot |\cdot |\begin{array}{ccc}2& 3& 1\\ 3& 6& 1\\ 2& 4& 1\end{array}|=3\cdot |\begin{array}{ccc}2& -1& 1\\ 3& 0& 1\\ 2& 0& 1\end{array}\mid =3\text{ sq. units }$

$\text{ Let }\frac{1}{{\mathrm{r}}_1},\frac{1}{{\mathrm{r}}_2},\frac{1}{{\mathrm{r}}_3}\text{ are in A.P. }$

or    $\frac{\mathrm{s}-\mathrm{a}}{\mathrm{\Delta }},\frac{\mathrm{s}-\mathrm{b}}{\mathrm{\Delta }},\frac{\mathrm{s}-\mathrm{c}}{\mathrm{\Delta }}\text{ in A.P. }$

or    s – a, s – b, s – c in A.P.
or    a, b, c in A.P.
       which is true for 18, 24, 30