question_answer

The incentre of triangle formed by the lines \[x=0,\] \[y=0\] and \[3x+4y=12\] is   [RPET 1990]

A) \[\left( \frac{1}{2},\,\frac{1}{2} \right)\]

B) (1, 1)

C) \[\left( 1,\,\frac{1}{2} \right)\]

D) \[\left( \frac{11}{2},\,1 \right)\]

Correct Answer: B

Solution :

Here \[a=BC=5,\,\,b=AC=4,\,\,\,c=AB=3\] Hence incentre is \[\left( \frac{0+0+3\times 4}{5+4+3},\frac{0+4\times 3+0}{5+4+3} \right)\]= (1, 1).