For l in R,the equation (2 l-3) x^2+2 l x y-y | vedclass

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(B) The general second-degree equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines if $h^2 - ab \geq 0$.Comparing the given equation

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for all values of $l \in R-(-3,1)$

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(B) The general second-degree equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines if $h^2 - ab \geq 0$.Comparing the given equation $(2l-3)x^2 + 2lxy - y^2 = 0$ with the standard form,we have $a = (2l-3)$,$h = l$,and $b = -1$.The condition for representing a pair of lines is $h^2 - ab \geq 0$.Substituting the values,we get $l^2 - (2l-3)(-1) \geq 0$.$l^2 + (2l-3) \geq 0$$l^2 + 2l - 3 \geq 0$$(l+3)(l-1) \geq 0$.Solving this inequality,we find that the expression is non-negative when $l \in (-\infty, -3] \cup [1, \infty)$,which can be written as $l \in R - (-3, 1)$.

For $l \in R$,the equation $(2 l-3) x^2+2 l x y-y^2=0$ represents a pair of lines

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