The intersection of three lines x - y = 0, x + 2y = 3 and 2x + y =...

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${L}_{1}: x - y = 0$

${L}_{2}: x + 2y = 3$

${L}_{3}: 2x + y = 6$

On solving ${L}_{1}$ and ${L}_{2}: y = 1$ and $x = 1$

On solving ${L}_{1}$ and ${L}_{3}$, we get

$x = 2$

$y = 2$

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

$x + 2y = 3$

$2x + y = 6$

On solving ${L}_{2} \text{ and } {L}_{3}$, we get

$x = 3$

$y = 0$

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

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

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

so it is an isosceles triangle