The equation (2p-3)x^2 + 2pxy - y^2 = 0 repre | vedclass

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(B) The general equation of a pair of straight lines is given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. For this to represent a pair of lines,the c

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For all values of $p \in R - [-3, 1]$

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(B) The general equation of a pair of straight lines is given by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$. For this to represent a pair of lines,the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$. Comparing the given equation $(2p-3)x^2 + 2pxy - y^2 = 0$ with the general form,we have $a = 2p-3$,$h = p$,$b = -1$,$g = 0$,$f = 0$,and $c = 0$. Substituting these values into the condition for a pair of lines,we get $0 = 0$,which is always true. For the lines to be distinct,the condition $h^2 - ab > 0$ must be satisfied. Substituting the values,we get $p^2 - (2p-3)(-1) > 0$. $p^2 + 2p - 3 > 0$. Factoring the quadratic,we get $(p+3)(p-1) > 0$. This inequality holds for $p \in (-\infty, -3) \cup (1, \infty)$,which is equivalent to $p \in R - [-3, 1]$.

The equation $(2p-3)x^2 + 2pxy - y^2 = 0$ represents a pair of distinct lines:

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