Three lines x + 2y + 3 = 0,x + 2y - 7 = 0,and | vedclass

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(D) The two parallel lines are $L_1: x + 2y + 3 = 0$ and $L_2: x + 2y - 7 = 0$.The distance $d$ between these two parallel lines is given by $d = \fra

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$2x - y - 14 = 0$ and $2x - y + 6 = 0$

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(D) The two parallel lines are $L_1: x + 2y + 3 = 0$ and $L_2: x + 2y - 7 = 0$.The distance $d$ between these two parallel lines is given by $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|3 - (-7)|}{\sqrt{1^2 + 2^2}} = \frac{10}{\sqrt{5}}$.Since these lines form two sides of a square,the distance between the other two parallel sides must also be $d = \frac{10}{\sqrt{5}}$.The third side is $L_3: 2x - y - 4 = 0$. The fourth side $L_4$ must be parallel to $L_3$,so let $L_4: 2x - y + \lambda = 0$.The distance between $L_3$ and $L_4$ is $\frac{|\lambda - (-4)|}{\sqrt{2^2 + (-1)^2}} = \frac{|\lambda + 4|}{\sqrt{5}}$.Equating the distances: $\frac{|\lambda + 4|}{\sqrt{5}} = \frac{10}{\sqrt{5}}$,which implies $|\lambda + 4| = 10$.This gives $\lambda + 4 = 10$ or $\lambda + 4 = -10$.So,$\lambda = 6$ or $\lambda = -14$.The equations of the fourth sides are $2x - y + 6 = 0$ and $2x - y - 14 = 0$.

Three lines $x + 2y + 3 = 0$,$x + 2y - 7 = 0$,and $2x - y - 4 = 0$ form three sides of two squares. Find the equation of the fourth side of each square.

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