The product of the perpendicular distances fr | vedclass

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Question

(C) The given equation is $2x^2 - 5xy + 2y^2 = 0$. Factoring the quadratic expression: $2x^2 - 4xy - xy + 2y^2 = 0 \Rightarrow 2x(x - 2y) - y(x - 2y)

Vedclass · Accepted Answer

$4$ units

Vedclass · Answer

(C) The given equation is $2x^2 - 5xy + 2y^2 = 0$. Factoring the quadratic expression: $2x^2 - 4xy - xy + 2y^2 = 0 \Rightarrow 2x(x - 2y) - y(x - 2y) = 0$. Thus,the lines are $2x - y = 0$ and $x - 2y = 0$. 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 the point $(2, -1)$ and line $2x - y = 0$,the distance $d_1 = \frac{|2(2) - 1(-1)|}{\sqrt{2^2 + (-1)^2}} = \frac{|4 + 1|}{\sqrt{5}} = \frac{5}{\sqrt{5}}$. For the point $(2, -1)$ and line $x - 2y = 0$,the distance $d_2 = \frac{|1(2) - 2(-1)|}{\sqrt{1^2 + (-2)^2}} = \frac{|2 + 2|}{\sqrt{5}} = \frac{4}{\sqrt{5}}$. The product of the distances is $d_1 	imes d_2 = \frac{5}{\sqrt{5}} 	imes \frac{4}{\sqrt{5}} = \frac{20}{5} = 4$ units.

The product of the perpendicular distances from the point $(2, -1)$ to the pair of lines represented by $2x^2 - 5xy + 2y^2 = 0$ is:

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