# The intersection of three lines $x-y=0,x+2y=3 \; \text {and} \; 2x+y=6$ is a :   Option: 1 None of the above Option: 2 Equilateral triangle Option: 3 Isosceles triangle   Option: 4 Right angled triangle

$\\\text{Let}\\ \mathrm{L}_{1}: \mathrm{x}-\mathrm{y}=0 \\ \mathrm{L}_{2}: \mathrm{x}+2 \mathrm{y}=3 \\\mathrm{L}_{3}: \mathrm{2x}+ \mathrm{y}=6$

Let point A be the point of intersection of L1 and L2, point B be the point of intersection of L1 and L3, and point C be the point of intersection of L3 and L2.

A = (1, 1)

B = (2, 2)

C = (3, 0)

$\\\mathrm{AC}=\sqrt{(1-3)^2+(1-0)^2}=\sqrt{4+1}=\sqrt{5} \\ \mathrm{BC}=\sqrt{(2-3)^2+(2-0)^2}=\sqrt{1+4}=\sqrt{5} \\ \mathrm{AB}=\sqrt{(1-2)^2+(1-2)^2}=\sqrt{1+1}=\sqrt{2}$

so its an isosceles triangle

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