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Question

(A) The given equation is $4x^2 + 20xy + 25y^2 + 2x + 5y - 12 = 0$. This can be written as $(2x + 5y)^2 + (2x + 5y) - 12 = 0$. Let $t = 2x + 5y$. Then

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$\frac{7}{\sqrt{29}}$

Vedclass · Answer

(A) The given equation is $4x^2 + 20xy + 25y^2 + 2x + 5y - 12 = 0$. This can be written as $(2x + 5y)^2 + (2x + 5y) - 12 = 0$. Let $t = 2x + 5y$. Then the equation becomes $t^2 + t - 12 = 0$. Factoring the quadratic: $(t + 4)(t - 3) = 0$. So,$t = -4$ or $t = 3$. This gives us two parallel lines: $2x + 5y + 4 = 0$ and $2x + 5y - 3 = 0$. The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by $d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$. Here,$A = 2, B = 5, C_1 = 4, C_2 = -3$. $d = \frac{|4 - (-3)|}{\sqrt{2^2 + 5^2}} = \frac{|7|}{\sqrt{4 + 25}} = \frac{7}{\sqrt{29}}$.

The distance between the lines represented by $4x^2 + 20xy + 25y^2 + 2x + 5y - 12 = 0$ is equal to

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