The incentre of the triangle formed by the ve | vedclass

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Question

(A) Let the vertices be $A(0, 0)$,$B(5, 12)$,and $C(16, 12)$.Calculate the lengths of the sides opposite to the vertices:$a = BC = \sqrt{(16-5)^2 + (1

Vedclass · Accepted Answer

$(7, 9)$

Vedclass · Answer

(A) Let the vertices be $A(0, 0)$,$B(5, 12)$,and $C(16, 12)$.Calculate the lengths of the sides opposite to the vertices:$a = BC = \sqrt{(16-5)^2 + (12-12)^2} = \sqrt{11^2 + 0^2} = 11$$b = AC = \sqrt{(16-0)^2 + (12-0)^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$$c = AB = \sqrt{(5-0)^2 + (12-0)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$The coordinates of the incentre $(I)$ are given by $\left( \frac{ax_1 + bx_2 + cx_3}{a+b+c}, \frac{ay_1 + by_2 + cy_3}{a+b+c} \right)$.Substituting the values: $I = \left( \frac{11(0) + 20(5) + 13(16)}{11+20+13}, \frac{11(0) + 20(12) + 13(12)}{11+20+13} \right)$$I = \left( \frac{0 + 100 + 208}{44}, \frac{0 + 240 + 156}{44} \right) = \left( \frac{308}{44}, \frac{396}{44} \right) = (7, 9)$.

The incentre of the triangle formed by the vertices $(0, 0)$,$(5, 12)$,and $(16, 12)$ is:

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