question_answer

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

A) (3, 4)                

B)  (2, 2)

C) (4, 3)                

D) (4, 4)

Correct Answer: B

Solution :

\[I=\left( \frac{a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{a+b+c},\frac{a{{y}_{1}}+b{{y}_{2}}+c{{y}_{2}}}{a+b+c} \right)\]

\[=\left( \frac{8(60)+6(0)+10(0)}{24},\frac{10(0)+6(8)+8(0)}{24} \right)\]

\[=(2,2).\]