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 $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.For this to represent a pair of lines,the condition i

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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 $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.For this to represent a pair of lines,the condition is $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.Comparing $(2p-3)x^2 + 2pxy - y^2 = 0$ with the standard form,we have $a = 2p-3$,$h = p$,$b = -1$,$g = 0$,$f = 0$,and $c = 0$.Substituting these values into the condition: $(2p-3)(0)(-1) + 2(0)(0)(p) - (2p-3)(0)^2 - (-1)(0)^2 - (0)(p)^2 = 0$.This simplifies to $0 = 0$,which is always true for any $p \in R$.For the lines to be distinct,the condition $h^2 - ab > 0$ must be satisfied.Here,$h^2 - ab = p^2 - (2p-3)(-1) = p^2 + 2p - 3$.We require $p^2 + 2p - 3 > 0$.Factoring the quadratic: $(p+3)(p-1) > 0$.The inequality holds when $p  1$.Thus,the lines are distinct for all $p \in R - [-3, 1]$.

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

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