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

Let $(h, k)$ be the

incentre of the $△ O A B$

From figure, it is clear that

$h = k .$

Now, perpendicular distance

From $(h, h)$ to the line $3 x + 4 y = 2$

is the radius

$h .$

$∴ \frac{| 3 h + 4 h - 24 |}{\sqrt{3^{2} + 4^{2}}} = h$

$\Rightarrow 7 h - 24 - \pm 5 h \Rightarrow h = 2$ $[∵ h \neq 12]$

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