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In $ΔABC$ , the sides BC=5, CA=4 and AB=3. If $A \equiv (0, 0)$ and the internal bisector of angle A meets BC in $D (\frac{12}{7}, \frac{12}{7})$ , then incentre of $ΔABC$ is

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(2,2)

(3,2)

(2,3)

(1,1)

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detailed solution

Correct option is D

We have a=5,b=4,c=3Incentre I divides AD in the ration b+c:a. Therefore, incetre coincide with centroid.Hence, coordinates of I are (1,1)

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In ΔABC, the sides BC=5, CA=4 and AB=3. If A≡(0,0) and the internal bisector of angle A meets BC in D(127,127), then incentre of ΔABC is

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In $ΔABC$ , the sides BC=5, CA=4 and AB=3. If $A \equiv (0, 0)$ and the internal bisector of angle A meets BC in $D (\frac{12}{7}, \frac{12}{7})$ , then incentre of $ΔABC$ is