Incentre of triangle whose vertices are A(-36,7), B(20,7), C(0,-8), is

(0,-1)

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(-1,0)

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(-1,-1)

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(0,0)

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Solution

The correct option is B
(-1,0)

$∵ a = B C = \sqrt{(20 - 0)^{2} + (7 + 8)^{2}} = 25 b = C A = \sqrt{(- 36 - 0)^{2} + (7 + 8)^{2}} = 39 a n d c = A B = \sqrt{(36 + 20)^{2} + (7 + 7)^{2}} = 56 I = (\frac{25 (- 36) + 39 (20) + 56 (0)}{25 + 39 + 56}, \frac{25 (7) + 39 (7) + 56 (- 8)}{25 + 39 + 56}) I = (- 1, 0) H e n c e, (c) i s t h e c o r r e c t a n s w e r .$

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Similar questions

$Find the coordinates of the centre of the circle inscribed in a triangle whose vertices are A(-36,7),B(20,7)and C(0,-8).$

A (- 36, 7), B (20, 7)

C (0, - 8)

G I = \frac{25}{2} \sqrt{205 λ}

λ

A (- 36, 7),

B (20, 7)

C (0, - 8)

G

I

A \equiv (- 36, 7)

B \equiv (20, 7)

C \equiv (0, - 8)

G I

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