If the equation 7x^2 - 14xy + py^2 - 12x + qy | vedclass

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(C) The given equation of the pair of lines is $7x^2 - 14xy + py^2 - 12x + qy - 4 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 +

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$\sqrt{109}$

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(C) The given equation of the pair of lines is $7x^2 - 14xy + py^2 - 12x + qy - 4 = 0$. Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 7$,$h = -7$,$b = p$,$g = -6$,$f = \frac{q}{2}$,and $c = -4$. Since the lines are parallel,$h^2 = ab$. Substituting the values,$(-7)^2 = 7p$ $\Rightarrow 49 = 7p$ $\Rightarrow p = 7$. For the equation to represent a pair of lines,the condition $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$ must be satisfied. Substituting $p = 7$,$a = 7$,$h = -7$,$g = -6$,$f = \frac{q}{2}$,and $c = -4$: $7(7)(-4) + 2(\frac{q}{2})(-6)(-7) - 7(\frac{q}{2})^2 - 7(-6)^2 - (-4)(-7)^2 = 0$. $-196 + 42q - \frac{7q^2}{4} - 252 + 196 = 0$. $42q - \frac{7q^2}{4} - 252 = 0$. Multiplying by $-4/7$,we get $q^2 - 24q + 144 = 0$. $(q - 12)^2 = 0 \Rightarrow q = 12$. Finally,$\sqrt{p^2 + q^2 - pq} = \sqrt{7^2 + 12^2 - (7)(12)} = \sqrt{49 + 144 - 84} = \sqrt{109}$.

If the equation $7x^2 - 14xy + py^2 - 12x + qy - 4 = 0$ represents a pair of parallel lines,then the value of $\sqrt{p^2 + q^2 - pq}$ is

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