The straight lines x+y=0, 5x+y=4, and x+5y=4 | vedclass

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(A) The given lines are $L_1: x+y=0$, $L_2: 5x+y=4$, and $L_3: x+5y=4$.Solving $L_1$ and $L_2$: $x+y=0 \implies y=-x$. Substituting into $L_2$: $5x-x=

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an isosceles triangle

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(A) The given lines are $L_1: x+y=0$, $L_2: 5x+y=4$, and $L_3: x+5y=4$.Solving $L_1$ and $L_2$: $x+y=0 \implies y=-x$. Substituting into $L_2$: $5x-x=4 \implies 4x=4 \implies x=1, y=-1$. So, $A = (1, -1)$.Solving $L_1$ and $L_3$: $x+y=0 \implies x=-y$. Substituting into $L_3$: $-y+5y=4 \implies 4y=4 \implies y=1, x=-1$. So, $C = (-1, 1)$.Solving $L_2$ and $L_3$: $5x+y=4$ and $x+5y=4$. Subtracting: $4x-4y=0 \implies x=y$. Substituting into $L_2$: $5x+x=4 \implies 6x=4 \implies x=2/3, y=2/3$. So, $B = (2/3, 2/3)$.Calculating side lengths:$AB = \sqrt{(2/3 - 1)^2 + (2/3 + 1)^2} = \sqrt{(-1/3)^2 + (5/3)^2} = \sqrt{1/9 + 25/9} = \sqrt{26/9} = \frac{\sqrt{26}}{3}$.$BC = \sqrt{(-1 - 2/3)^2 + (1 - 2/3)^2} = \sqrt{(-5/3)^2 + (1/3)^2} = \sqrt{25/9 + 1/9} = \sqrt{26/9} = \frac{\sqrt{26}}{3}$.$CA = \sqrt{(-1 - 1)^2 + (1 - (-1))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.Since $AB = BC$, the triangle is an isosceles triangle.

The straight lines $x+y=0$, $5x+y=4$, and $x+5y=4$ form

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