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

# If the line $3x+4y-24=0$ intersects the-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

1. A

(4, 4)

2. B

(2, 2)

3. C

(3, 4)

4. D

(4, 3)

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### Solution:

Let $\left(h,k\right)$ be the

incentre of the $\mathrm{△}OAB$

From figure, it is clear that

Now, perpendicular distance

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