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(B) The given line is $3x - 2y = 1$,which can be written as $3x - 2y = 1$. To make the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ homogeneous,we multiply

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Perpendicular to each other

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(B) The given line is $3x - 2y = 1$,which can be written as $3x - 2y = 1$. To make the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ homogeneous,we multiply the terms of degree $1$ by $(3x - 2y)$ and the constant term by $(3x - 2y)^2$. $3x^2 + 5xy - 3y^2 + (2x + 3y)(3x - 2y) = 0$. Expanding this,we get $3x^2 + 5xy - 3y^2 + 6x^2 - 4xy + 9xy - 6y^2 = 0$. Combining like terms,we get $9x^2 + 10xy - 9y^2 = 0$. For a pair of lines $ax^2 + 2hxy + by^2 = 0$,the lines are perpendicular if $a + b = 0$. Here,$a = 9$ and $b = -9$. Since $a + b = 9 + (-9) = 0$,the lines are perpendicular to each other.

The lines joining the origin to the points of intersection of the line $3x - 2y = 1$ and the curve $3x^2 + 5xy - 3y^2 + 2x + 3y = 0$ are:

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