The straight lines x+3y-4=0,x+y-4=0,and 3x+y- | vedclass

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(A) To find the vertices of the triangle,we solve the equations of the lines in pairs: $1$. For vertex $A$,solve $x+3y-4=0$ and $3x+y-4=0$: Multiplyin

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form an isosceles triangle

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(A) To find the vertices of the triangle,we solve the equations of the lines in pairs: $1$. For vertex $A$,solve $x+3y-4=0$ and $3x+y-4=0$: Multiplying the first equation by $3$,we get $3x+9y-12=0$. Subtracting $3x+y-4=0$ from this,we get $8y-8=0$,so $y=1$. Substituting $y=1$ into $x+3(1)-4=0$,we get $x=1$. Thus,$A = (1, 1)$. $2$. For vertex $B$,solve $x+3y-4=0$ and $x+y-4=0$: Subtracting the second from the first,we get $2y=0$,so $y=0$. Substituting $y=0$ into $x+y-4=0$,we get $x=4$. Thus,$B = (4, 0)$. $3$. For vertex $C$,solve $3x+y-4=0$ and $x+y-4=0$: Subtracting the second from the first,we get $2x=0$,so $x=0$. Substituting $x=0$ into $x+y-4=0$,we get $y=4$. Thus,$C = (0, 4)$. Now,calculate the lengths of the sides: $AB = \sqrt{(4-1)^2 + (0-1)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10}$ $BC = \sqrt{(0-4)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$ $CA = \sqrt{(1-0)^2 + (1-4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1+9} = \sqrt{10}$ Since $AB = CA = \sqrt{10}$,the triangle is an isosceles triangle.

The straight lines $x+3y-4=0$,$x+y-4=0$,and $3x+y-4=0$

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