The triangle formed by the lines 2x^2+xy-6y^2 | vedclass

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(D) The given pair of lines is $2x^2+xy-6y^2=0$. Factoring the quadratic expression: $2x^2+4xy-3xy-6y^2=0 \implies 2x(x+2y)-3y(x+2y)=0 \implies (2x-3y

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scalene

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(D) The given pair of lines is $2x^2+xy-6y^2=0$. Factoring the quadratic expression: $2x^2+4xy-3xy-6y^2=0 \implies 2x(x+2y)-3y(x+2y)=0 \implies (2x-3y)(x+2y)=0$. So,the two lines are $L_1: 2x-3y=0$ and $L_2: x+2y=0$. The third line is $L_3: x+y-1=0$. To find the vertices,we solve the systems of equations: $1$. Intersection of $L_1$ and $L_2$: $(0,0)$. $2$. Intersection of $L_1$ and $L_3$: $2x-3y=0 \implies x=3y/2$. Substituting into $L_3$: $3y/2+y=1 \implies 5y/2=1 \implies y=2/5, x=3/5$. Vertex $A = (3/5, 2/5)$. $3$. Intersection of $L_2$ and $L_3$: $x+2y=0 \implies x=-2y$. Substituting into $L_3$: $-2y+y=1 \implies -y=1 \implies y=-1, x=2$. Vertex $B = (2, -1)$. Now,calculate the lengths of the sides: $OA = \sqrt{(3/5)^2+(2/5)^2} = \sqrt{9/25+4/25} = \sqrt{13/25} = \sqrt{13}/5$. $OB = \sqrt{2^2+(-1)^2} = \sqrt{4+1} = \sqrt{5}$. $AB = \sqrt{(2-3/5)^2+(-1-2/5)^2} = \sqrt{(7/5)^2+(-7/5)^2} = \sqrt{49/25+49/25} = \sqrt{98/25} = 7\sqrt{2}/5$. Since $OA^2+OB^2 = 13/25 + 5 = 138/25$ and $AB^2 = 98/25$,the triangle is not right-angled. Since $OA 
eq OB 
eq AB$,the triangle is scalene.

The triangle formed by the lines $2x^2+xy-6y^2=0$ and $x+y-1=0$ is

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