If the vertices of a triangle are (0,0),(6,0) | vedclass

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(C) Let the vertices be $A(0,0)$,$B(6,0)$,and $C(6,8)$.The lengths of the sides are:$c = AB = \sqrt{(6-0)^2 + (0-0)^2} = 6$$a = BC = \sqrt{(6-6)^2 + (

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$(4,2)$

Vedclass · Answer

(C) Let the vertices be $A(0,0)$,$B(6,0)$,and $C(6,8)$.The lengths of the sides are:$c = AB = \sqrt{(6-0)^2 + (0-0)^2} = 6$$a = BC = \sqrt{(6-6)^2 + (8-0)^2} = 8$$b = AC = \sqrt{(6-0)^2 + (8-0)^2} = \sqrt{36+64} = 10$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{8(0) + 10(6) + 6(6)}{8+10+6}, \frac{8(0) + 10(0) + 6(8)}{8+10+6} \right)$$I = \left( \frac{0 + 60 + 36}{24}, \frac{0 + 0 + 48}{24} \right)$$I = \left( \frac{96}{24}, \frac{48}{24} \right) = (4, 2)$.

If the vertices of a triangle are $(0,0)$,$(6,0)$,and $(6,8)$,then its incentre will be:

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