Question

If the line $3x+4y-24=0$ intersects the x-axis at the point $A$ and y-axis at the point $B$, then the incentre of the triangle $OAB$, where $O$ is the origin is :

• A. $(4,3)$
• B. $(2,2)$
• C. $(3,4)$
• D. $(4,4)$

Answer 1

The correct option is B $(2,2)$

Let $r$ be the radius of the circle $\Rightarrow (r,r)$ will be the centre of the circle.

$\therefore$ Equation of $AB$ is $3x+4y-24=0$
$\therefore$The distance of the centre of the incircle with the line $AB$ is -
$\left|\frac{3r+4r-24}{\sqrt{3^2+4^2}}\right|=r$
$\Rightarrow \left|\frac{7r-24}{5}\right|=r$
$\Rightarrow 7r-24=5r$ and $7r-24=-5r$
$\Rightarrow r=12$ or $r=2$
Hence, incentre of the circle is $(2,2)$