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A triangle has corners A(0, 0), B(21, 0) and C(6, 8)…

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A triangle has corners $A(0, 0)$ , $B(21, 0)$ and $C(6, 8)$ . Find its incentre, and the distance from the incentre to the corner $A$ . Give the distance to 3 decimal places.

Need a reminder? The key ideaShow

$G$ is the average of the corners. $H = 3G - 2O$ , so $O$ , $G$ , $H$ lie on one line with $GH = 2\,OG$ . $O$ is equally far from the corners. The incentre is $I = \dfrac{aA + bB + cC}{a + b + c}$ , with $a$ , $b$ , $c$ the sides opposite $A$ , $B$ , $C$ . The area is $\dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ . For $ax^2 + 2hxy + by^2 = 0$ , $\tan\theta = \dfrac{2\sqrt{h^2 - ab}}{|a + b|}$ , and the slopes $m$ solve $bm^2 + 2hm + a = 0$ .

The worked solution

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