question_answer

Line\[3x+4y-24=0\]intersect the\[x-\]axis at point A and y-axis at point B, then incentre of \[\Delta \]AOB, where 0 is the origin, is

A)  (1, 2)             

B)  (2, 2)

C)  (2, 12)            

D)  (12, 12)

Correct Answer: B

Solution :

 Given, line is\[3x+4y-24=0\]. It cuts the\[x-\]axis at A (8,0) and y-axis at B(0,6). In\[\Delta OAB,\] \[a=6,b=8\] and     \[C=\sqrt{{{6}^{2}}+{{8}^{2}}}=\sqrt{100}=10\] \[\therefore \]Coordinates of incentre are \[\left( \frac{a{{x}_{1}}+b{{x}_{2}}+c{{x}_{3}}}{a+b+c},\frac{a{{y}_{1}}+b{{y}_{2}}+c{{y}_{3}}}{a+b+c} \right)\] \[=\left( \frac{6\times 8+8\times 0+10\times 0}{6+8+10},\frac{6\times 0+8\times 6+10\times 0}{6+8+10} \right)\] \[=\left( \frac{48}{24},\frac{48}{24} \right)=(2,2)\]