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 $L_3: x+y-1=0$. The slopes of the lines are $m_1 = \frac{2}{3}$,$m_2 = -\frac{1}{2}$,and $m_3 = -1$. Check for perpendicularity: $m_1 	imes m_2 = \frac{2}{3} 	imes (-\frac{1}{2}) = -\frac{1}{3} 
eq -1$. $m_1 	imes m_3 = \frac{2}{3} 	imes (-1) = -\frac{2}{3} 
eq -1$. $m_2 	imes m_3 = -\frac{1}{2} 	imes (-1) = \frac{1}{2} 
eq -1$. Calculate the lengths of the sides: The vertices are the intersection points: $V_1 (L_1 \cap L_2) = (0,0)$. $V_2 (L_1 \cap L_3): 2x-3y=0 \implies x=\frac{3}{2}y$. Substituting in $x+y=1 \implies \frac{3}{2}y+y=1 \implies \frac{5}{2}y=1 \implies y=\frac{2}{5}, x=\frac{3}{5}$. So $V_2 = (\frac{3}{5}, \frac{2}{5})$. $V_3 (L_2 \cap L_3): x+2y=0 \implies x=-2y$. Substituting in $x+y=1 \implies -2y+y=1 \implies y=-1, x=2$. So $V_3 = (2, -1)$. Side lengths: $d_1 (V_1V_2) = \sqrt{(\frac{3}{5})^2 + (\frac{2}{5})^2} = \sqrt{\frac{9+4}{25}} = \frac{\sqrt{13}}{5}$. $d_2 (V_1V_3) = \sqrt{2^2 + (-1)^2} = \sqrt{5}$. $d_3 (V_2V_3) = \sqrt{(2-\frac{3}{5})^2 + (-1-\frac{2}{5})^2} = \sqrt{(\frac{7}{5})^2 + (-\frac{7}{5})^2} = \sqrt{\frac{49+49}{25}} = \frac{7\sqrt{2}}{5}$. Since all sides are of different lengths,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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