The intersection of three lines

Solution:

$L_{1}: x-y=0$

$L_{2}: x+2 y=3$

$\mathrm{L}_{3}: \mathrm{x}+\mathrm{y}=6$

on solving $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ :

$\mathrm{y}=\mathrm{L}$ and $\mathrm{x}=1$

$\mathrm{L}_{1}$ and $\mathrm{L}_{3}$ :

$x=2$

$y=2$

$\mathrm{L}_{2}$ and $\mathrm{L}_{3}:$

$x+y=3$

$2 x+y=6$

$x=3$

$y=0$

$\mathrm{AC}=\sqrt{4+1}=\sqrt{5}$

$\mathrm{BC}=\sqrt{4+1}=\sqrt{5}$

$\mathrm{AB}=\sqrt{1+1}=\sqrt{2}$

so its an isosceles triangle