Question

The coordinates of incentre of the triangle formed by the lines $x=0;y=0$ and $3x+4y=12$ is

• A. $(2,\frac{3}{2})$
• B. $(4,3)$
• C. $(1,1)$
• D. $(2,1)$

Answer 1

The correct option is C $(1,1)$


Clearly we get $a=3,b=5,c=4$
$I\equiv (\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$
$\equiv (1,1)$