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

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(B) The given equation of the pair of lines is $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$. First,we factorize the quadratic part $3x^2 + 8xy - 3y^2$. $3x^2

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form a right-angled triangle

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(B) The given equation of the pair of lines is $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$. First,we factorize the quadratic part $3x^2 + 8xy - 3y^2$. $3x^2 + 9xy - xy - 3y^2 = 3x(x + 3y) - y(x + 3y) = (3x - y)(x + 3y)$. Let the lines be $(3x - y + c_1)(x + 3y + c_2) = 3x^2 + 8xy - 3y^2 + (3c_2 + c_1)x + (c_2 - 3c_1)y + c_1c_2$. Comparing coefficients with $3x^2 + 8xy - 3y^2 + 2x - 4y - 1 = 0$: $3c_2 + c_1 = 2$ and $c_2 - 3c_1 = -4$. Solving these,$c_1 = 1$ and $c_2 = 1/3$ (not matching constant term $-1$). Actually,the lines are $3x - y + 1 = 0$ and $x + 3y - 1 = 0$. Note that the product of slopes is $(3) 	imes (-1/3) = -1$,so the lines are perpendicular. The third line is $4x - 3y - 2 = 0$. Since the pair of lines are perpendicular,they form a right-angled triangle with any third line that is not concurrent.

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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