The locus of the incentre of the triangle formed by the lines $x y-4 x-4 y+16=0$ and $x+y=5$ is

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Aicharya · Accepted Answer

$x y-4 x-4 y+16=0$ $\begin{aligned} & x(y-4)-4(y-4)=0 \ & (y-4)(x-4)=0\end{aligned}$ The 2 lines are $x=4$ and $y=4$ $\begin{aligned} & \mathrm{L}_1: x=4 \ & \mathrm{~L}_2: y=4 \ & \mathrm{~L}_3: x+y=5\end{aligned}$ $\begin{aligned} & \mathrm{AB}=\mathrm{c}=3 \ & \mathrm{BC}=\mathrm{a}=3 \sqrt{2} \ & \mathrm{CA}=\mathrm{b}=3\end{aligned}$ $\begin{aligned} & \mathrm{A}\left(x_1 y_1ight) \equiv(4,4) \ & \mathrm{B}\left(x_2, y_2ight) \equiv(4,1) \ & \mathrm{C}\left(x_3, y_3ight) \equiv(1,4)\end{aligned}$ Incentre $=\left(\frac{a x_1+b x_2+c x_3}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}ight)$ $\begin{aligned} & =\frac{3 \sqrt{2} \cdot 4+3 \cdot 4+3 \cdot 1}{3+3+3 \sqrt{2}}, \frac{3 \sqrt{2} \cdot 4+3.1+3 \cdot 4}{3 \sqrt{2}+3+3} \ & =\frac{15+12 \sqrt{2}}{6+3 \sqrt{2}}, \frac{15+12 \sqrt{2}}{6+3 \sqrt{2}} \ & =\frac{5+4 \sqrt{2}}{2+\sqrt{2}}, \frac{5+4 \sqrt{2}}{2+\sqrt{2}}\end{aligned}$ Co-ordinates of $x$ and $y$ satisfy $x=y$ locus of incentre is $x-y=0$

The locus of the incentre of the triangle formed by the lines $x y-4 x-4 y+16=0$ and $x+y=5$ is

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