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?

Question

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!

Answer 1

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: $2x-y-6=0$ & $2x-y+14=0$

Answer 2

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.