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Q.

The vertices of a triangle are (3, 2), (7, 2) (7, 5) then the in-centre(I)

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a

$(3, 6)$

b

$(6, 3)$

c

$(- 3, - 6)$

d

$(- 6, - 3)$

answer is A.

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Detailed Solution

$\begin{array}{l} A (3, 2), B (7, 2), C (7, 5) a = B C = 3, b = C A = 5, c = A B = 4 \\ I = (\frac{a x_{1} + b x_{2} + c x_{3}}{a + b + c}, \frac{a y_{1} + b y_{2} + c y_{3}}{a + b + c}) = (\frac{9 + 35 + 28}{12}, \frac{6 + 10 + 20}{12}) = (6, 3) \end{array}$

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