<s>[INST] The lines $x+2y+3=0$ , $x+2y-7=0$ and $2x-y+4=0$ are sides of a square. Equation of the remaining side is? I found out the area between parallel lines as $ \frac{10}{\sqrt{5}} $ and then I used
$ \frac{|\lambda - 4|}{\sqrt{5}} = \frac{10}{\sqrt{5}} $ to get the values as $-6$ and $14$ . I   am getting the final equations as $2x-y-6=0$ and $2x-y+14=0$ but this answer is wrong. According to my book the correct equations are $2x-y+6=0$ and $2x-y-14=0$. Please tell me where I am wrong!
 [/INST] It is clear that the lines: $x+2y+3=0$ & $x+2y-7$ are parallel hence the forth line (side of square) must be parallel to third line: $2x-y+4=0$ Hence, let the forth line: $2x-y+c=0$ having slope $2$.
Let each side of square $a$ then it is equal to the distance between the first two (first & second) parallel lines calculated as follows $$a=\frac{\left|3-(-7)\right|}{\sqrt{1^2+2^2}}=\frac{10}{\sqrt{5}}=2\sqrt{5}$$
The side $a$ is also equal to the distance between the second two (third & fourth) parallel lines calculated as follows $$a=\frac{\left|c-4\right|}{\sqrt{(2)^2+(-1)^2}}=\frac{\left|c-4\right|}{\sqrt{5}}$$ Now, equating both the values of side $a$ of the square, we have $$\frac{\left|c-4\right|}{\sqrt{5}}=2\sqrt{5}\implies \left|c-4\right|=10 $$ $$ c-4=10 \implies c=14$$ & $$ c-4=-10 \implies c=-6$$ Thus, corresponding above two values of $c$, we get two lines: $2x-y+14=0$ & $2x-y-6=0$ representing fourth unknown side of the square lying on either side of third line: $2x-y+4=0$. Obviously, the answers are same as you have obtained. There is some printing mistake in the answers provided in your book. 
According to your book, the answers are: $2x-y+6=0$ & $2x-y-14=0$ Then, note that  the sides of the square are not equal as  you have already mentioned that the figure is a square & first two parallel sides: $x+2y+3=0$ & $x+2y-7=0$ confirms that the square has its each side $2\sqrt{5}$. Hence the answers according to your book are wrong because those do not satisfy the given conditions. The correct answers are: $2x-y-6=0$ & $2x-y+14=0$        
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