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!

The first two lines are parallel to each other, and so you are looking for a line parallel to the third one. You have


*

*$x+2y+3=0$

*$x+2y-7=0$

*$2x+(-1)y+4=0$

*$2x+(-1)y+K=0$


where $K$ is an unknown coefficient to complete the square. 
The distance between the first and second line (square side) is
$$ d_{12} = \frac{(3)-(-7)}{\sqrt{1^2+2^2}} = 2\sqrt{5} = 4.4721\ldots$$
The same distance should exist between the third and fourth line
$$ d_{34} = \frac{4-K}{\sqrt{2^2+(-1)^2}} = d_{12}$$
$$ K = 4-\sqrt{5} d_{12} = -6$$
So the equation is $$2x-y-6=0$$
this is verified with GeoGebra.

