If P_1 and P_2 are perpendicular distances (i | vedclass

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(D) The given equation of the pair of lines is $2x^2 - 5xy + 2y^2 = 0$. Factorizing the equation: $2x^2 - 4xy - xy + 2y^2 = 0$. $2x(x - 2y) - y(x - 2y

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$4$

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(D) The given equation of the pair of lines is $2x^2 - 5xy + 2y^2 = 0$. Factorizing the equation: $2x^2 - 4xy - xy + 2y^2 = 0$. $2x(x - 2y) - y(x - 2y) = 0$. $(2x - y)(x - 2y) = 0$. Thus,the separate equations of the lines are $L_1: 2x - y = 0$ and $L_2: x - 2y = 0$. The perpendicular distance from a point $(x_1, y_1)$ to a line $ax + by + c = 0$ is given by $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$. For point $(2, -1)$: $P_1 = \frac{|2(2) - (-1)|}{\sqrt{2^2 + (-1)^2}} = \frac{|4 + 1|}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5}$. $P_2 = \frac{|(2) - 2(-1)|}{\sqrt{1^2 + (-2)^2}} = \frac{|2 + 2|}{\sqrt{5}} = \frac{4}{\sqrt{5}}$. Therefore,$P_1 P_2 = \sqrt{5} 	imes \frac{4}{\sqrt{5}} = 4$.

If $P_1$ and $P_2$ are perpendicular distances (in units) from point $(2, -1)$ to the pair of lines $2x^2 - 5xy + 2y^2 = 0$,then the value of $P_1 P_2$ is

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