The intersection of three lines x-y=0,x+2y=3, | vedclass

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(C) Let the lines be $L_1: x-y=0$,$L_2: x+2y=3$,and $L_3: 2x+y=6$.Solving $L_1$ and $L_2$:$x-y=0 \implies x=y$$x+2x=3 \implies 3x=3 \implies x=1, y=1$

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Isosceles triangle

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(C) Let the lines be $L_1: x-y=0$,$L_2: x+2y=3$,and $L_3: 2x+y=6$.Solving $L_1$ and $L_2$:$x-y=0 \implies x=y$$x+2x=3 \implies 3x=3 \implies x=1, y=1$. Point $A = (1, 1)$.Solving $L_1$ and $L_3$:$x-y=0 \implies x=y$$2x+x=6 \implies 3x=6 \implies x=2, y=2$. Point $B = (2, 2)$.Solving $L_2$ and $L_3$:$x+2y=3 \implies x=3-2y$$2(3-2y)+y=6 \implies 6-4y+y=6 \implies -3y=0 \implies y=0, x=3$. Point $C = (3, 0)$.Now,calculate the lengths of the sides:$AB = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$$BC = \sqrt{(3-2)^2 + (0-2)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}$$AC = \sqrt{(3-1)^2 + (0-1)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4+1} = \sqrt{5}$Since $BC = AC = \sqrt{5}$,the triangle is an isosceles triangle.

The intersection of three lines $x-y=0$,$x+2y=3$,and $2x+y=6$ forms a:

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