The sides of a $∆ ABC$ are $BC = 5, CA = 4, AB = 3 .$ If $A (0, 0)$ and the bisector of the internal angle $A$ meets $BC$ in $D = (\frac{12}{7}, \frac{12}{7})$ then the coordinates of in-centre is

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$(1, 1)$

$(2, 3)$

$(2, 2)$

$(3, 2)$

answer is D.

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Detailed Solution

$Given AB = 3, BC = 5, CA = 4 Now Incentre' I' divides AD in the ratio AB + AC : BC = 7 : 5 internally then I = [\frac{7 (\frac{12}{7}) + 0}{12}, \frac{7 (\frac{12}{7}) + 0}{12}] = [\frac{12}{12}, \frac{12}{12}] = [1, 1]$

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