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(C) The given pair of lines is $x^2 - 4xy + y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1, h = -2, b = 1$. Th

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$\frac{3}{2}$

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(C) The given pair of lines is $x^2 - 4xy + y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1, h = -2, b = 1$. The perpendicular distance from a point $(x_1, y_1)$ to the lines $ax^2 + 2hxy + by^2 = 0$ is given by the formula: $d_1 d_2 = \frac{|ax_1^2 + 2hx_1y_1 + by_1^2|}{\sqrt{(a-b)^2 + 4h^2}}$. Substituting the values $(x_1, y_1) = (1, -1)$,$a = 1, h = -2, b = 1$: $d_1 d_2 = \frac{|1(1)^2 + 2(-2)(1)(-1) + 1(-1)^2|}{\sqrt{(1-1)^2 + 4(-2)^2}}$. $d_1 d_2 = \frac{|1 + 4 + 1|}{\sqrt{0 + 16}} = \frac{6}{4} = \frac{3}{2}$.

The product of the perpendicular distances from $(1, -1)$ to the pair of lines $x^2 - 4xy + y^2 = 0$ is:

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