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$\begin{array}{l} Let A, B, C be angles of triangle ABC with vertexx A (4, - 1) \\ and x - 1 = 0 and x - y - 1 = 0 are internal angle bisectors through \\ B and C respectively. Let D, E, F be the points of contact of sides \\ BC, CA and AB with incircle of triangle ABC \\ images of D, E, F through the internal angle bisectors A, B, C respectively \end{array}$

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$\begin{array}{l} Equatin of incircle of triangle ABC is {(x - 1)}^{2} + y^{2} = 5 \end{array}$

$c i r c u m c i r c l e o f t h e t r i a n g l e D^{1} E^{1} F^{1} i s {(x - 1)}^{2} + y^{2} = 5$

$\begin{array}{l} Area of traignle ABC is 23 square units \end{array}$

$\begin{array}{l} Equation of incircle of triangle ABC is x^{2} + y^{2} = 25 \end{array}$

answer is A, B.

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Detailed Solution

Mirror images of $A (4, - 1)$ About the two bisectors x – y – 1 = 0 and x – 1 = 0 are respectively (0, 3) and (-2, -1) which lie in BC

Equation of BC is 2x – y + 3 = 0 Inradiur $r = \frac{5}{\sqrt{5}} = \sqrt{5}$

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