A Question from Straight lines

Question

In $\triangle ABC$, the coordinates of vertex $A$ are $(4,-1)$, and lines $x-y-1=0$ and $2x-y=3$ are the internal angle bisectors of angles $B$ and $C$. Then, the radius of the incircle of triangle $ABC$ is?

I found out the point of intersection of the angle bisectors because that's the centre of the circle.

Now can someone please suggest how to proceed further with this problem? Thanks in advance!

Answer 1

Suppose you have a $\triangle ABC$ with incenter $I$ and $IF \perp AB$, so the inradius is the length of $IF$. In $\triangle AIF$, which is a right-angled triangle, $AI$ being the angle bisector of $\angle BAC$, you have $\angle IAF = \dfrac{\angle BAC}2$ and $$\dfrac{IF}{AI} = \sin(\angle IAF) = \sin\left(\dfrac{\angle BAC}2\right) \ \ \textbf{(keep this noted, this will be useful)}\\ \implies \boxed{r=AI\sin\left(\dfrac{\angle BAC}2\right)}$$

Clearly $BI,CI$ are the internal angle-bisectors of $\angle ABC, \angle ACB$ respectively. This implies that $$\begin{aligned}\angle IBC = \dfrac{\angle ABC}2&, \ \angle ICB= \dfrac{\angle ACB}2\\ \implies \angle BIC &= 180^\circ-\angle IBC - \angle ICB\\&=180^\circ-\left(\dfrac{\angle ABC+\angle ACB}2\right)\\&=180^\circ - \left(\dfrac{180^\circ - \angle BAC}2\right)\\ \implies\angle BIC &=90^\circ + \dfrac{\angle BAC}2\end{aligned} \tag1$$

You are already given the equations of the angle bisectors ($BI$ and $CI$), so

their intersection point is the coordinate of $I$, which you can find by solving those two equations, and from the coordinates of $I,A$, you can find out the distance $AI$

the only thing in the above boxed formula you need is the $\sin\left(\dfrac{\angle BAC}2\right)$ which you can find using

the $\tan\left(90^\circ +\dfrac{\angle BAC}2\right)$ angle between the two given lines from the derivation in $(1)$

and using, with $\theta=\dfrac{\angle BAC}2$, $$\tan\left(90^\circ+\theta\right)=-\cot\theta=-\dfrac{\cos\theta}{\sin\theta}= -\dfrac{\sqrt{1-\sin^2\theta}}{\sin\theta}$$