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(C) The product of the perpendiculars from the origin $(0,0)$ to the pair of lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by

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$p_3 < p_2 < p_1$

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(C) The product of the perpendiculars from the origin $(0,0)$ to the pair of lines represented by $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ is given by $p = \left| \frac{c}{\sqrt{(a-b)^2 + 4h^2}} \right|$.For $xy+x+y+1=0$:$a=0, b=0, h=1/2, c=1$.$p_1 = \left| \frac{1}{\sqrt{(0-0)^2 + 4(1/2)^2}} \right| = \left| \frac{1}{\sqrt{1}} \right| = 1$.For $x^2-y^2+2x+1=0$:$a=1, b=-1, h=0, c=1$.$p_2 = \left| \frac{1}{\sqrt{(1-(-1))^2 + 4(0)^2}} \right| = \left| \frac{1}{\sqrt{4}} \right| = 1/2$.For $2x^2+3xy-2y^2+2x+1=0$:$a=2, b=-2, h=3/2, c=1$.$p_3 = \left| \frac{1}{\sqrt{(2-(-2))^2 + 4(3/2)^2}} \right| = \left| \frac{1}{\sqrt{16 + 9}} \right| = \frac{1}{\sqrt{25}} = 1/5$.Comparing the values: $1/5 < 1/2 < 1$,which means $p_3 < p_2 < p_1$.

If the products of the perpendiculars from the origin to the pairs of lines $xy+x+y+1=0$,$x^2-y^2+2x+1=0$,and $2x^2+3xy-2y^2+2x+1=0$ are $p_1, p_2$,and $p_3$ respectively,then:

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