In $ΔABC$ , the coordinates of the vertex A are $(4, - 1)$ , and lines $x - y - 1 = 0$ and $2 x - y = 3$ are the internal bisectors of angles B and C. Then, the radius of the in circle of triangle ABC is

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$4 / \sqrt{5}$

$3 / \sqrt{5}$

$6 / \sqrt{5}$

$7 / \sqrt{5}$

answer is C.

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Detailed Solution

Incenre 1 is point of intersection of two angle bisectors which is ( 2,1).

$IA = \sqrt{2^{2} + 2^{2}} = 2 \sqrt{2}$

Also, $AI = rcosec \frac{A}{2}$

$∠ BIC = \frac{π}{2} + \frac{A}{2}$

or $\tan (\frac{π}{2} + \frac{π}{2}) = |\frac{1 - 2}{1 + 2}|$

or $\tan \frac{A}{2} = 3$

So, $r = \frac{AI}{cosec \frac{A}{2}} = \frac{2 \sqrt{2}}{\sqrt{1 + \frac{1}{9}}} \frac{2 \sqrt{2} \times 3}{\sqrt{10}} = \frac{6}{\sqrt{5}}$

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