If vertices of the triangle are A(0,0),B(1,0) and C(1/2,√3/2) respectively, then the centre of the circle for which the lines AB,BC,CA are tangents is
A
(1/2,1/4)
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B
(3/2,1/2√3)
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C
(1/2,1/2√3)
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D
(1/2,−1/√3)
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Solution
The correct option is C(1/2,1/2√3) As we know that sides of the triangle are always tangential to its incirle .
Hence Incentre is I≡(ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c) BC=a=√14+34=1 AC=b=1,AB=c=1 AB=BC=CA=1 is equilateral.
So, its centre is ⎛⎜
⎜
⎜
⎜⎝1+12+03,0+√32+03⎞⎟
⎟
⎟
⎟⎠≡(12,12√3)