The lines represented by the equation 3x^2 + | vedclass

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(B) First,factorize the pair of lines $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$. Let the equation be $3x^2 + (8y + 2)x - (3y^2 + 4y + 1) = 0$. Using the q

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form a right-angled triangle

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(B) First,factorize the pair of lines $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$. Let the equation be $3x^2 + (8y + 2)x - (3y^2 + 4y + 1) = 0$. Using the quadratic formula for $x$,$x = \frac{-(8y+2) \pm \sqrt{(8y+2)^2 + 4(3)(3y^2+4y+1)}}{2(3)}$. Simplifying the discriminant: $(64y^2 + 32y + 4) + (36y^2 + 48y + 12) = 100y^2 + 80y + 16 = (10y+4)^2$. Thus,$x = \frac{-8y-2 \pm (10y+4)}{6}$. This gives two lines: $x = \frac{2y+2}{6} \Rightarrow 3x - y - 1 = 0$ and $x = \frac{-18y-6}{6} \Rightarrow x + 3y + 1 = 0$. The slopes of these lines are $m_1 = 3$ and $m_2 = -1/3$. Since $m_1 	imes m_2 = -1$,these two lines are perpendicular. The third line is $4x - 3y - 2 = 0$,with slope $m_3 = 4/3$. Since the first two lines are perpendicular,they form a right-angled triangle with any third line that is not parallel to them. Thus,the lines form a right-angled triangle.

The lines represented by the equation $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$ and the line $4x - 3y - 2 = 0$:

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