Question

If vertices of the triangle are $A(0,0),B(1,0)$ and $C(1/2,\sqrt{3}/2)$ respectively, then the centre of the circle for which the lines $AB,BC,CA$ are tangents is

• A. $(1/2,1/4)$
• B. $(3/2,1/2\sqrt{3})$
• C. $(1/2,1/2\sqrt{3})$
• D. $(1/2,-1/\sqrt{3})$

Answer 1

The correct option is C $(1/2,1/2\sqrt{3})$

As we know that sides of the triangle are always tangential to its incirle .
Hence Incentre is
$I\equiv (\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$
$BC=a=\sqrt{\frac{1}{4}+\frac{3}{4}}=1$
$AC=b=1,AB=c=1$
$AB=BC=CA=1$ is equilateral.
So, its centre is $(\frac{1+\frac{1}{2}+0}{3},\frac{0+\frac{\sqrt{3}}{2}+0}{3})\equiv (\frac{1}{2},\frac{1}{2\sqrt{3}})$