The origin is a corner of a square and two of its sides are y + 2x = 0 and y + 2x = 3. Find the equation of the other two sides.

Question

The origin is a corner of a square and two of its sides are y + 2x = 0 and y + 2x = 3. Find the equation of the other two sides.

I tried by finding one of the points in one line

(1,1) for equation y + 2x = 3 and found the distance between origin

d = √1 + 1

= √2

y + 2x = 0

y = -2x

As it is square the distance will be same

√2 = √x^2 + y^2

5x^2 = 2

x = √2/5

y = 2√2/5

What after this?

Answer 1

Since one of the sides going through the origin is $y+2x=0$, the other must be perpendicular with that line, from that we know its slope. Also we know that this line goes through the origin, so we can use the point slop form of the line:

$$y-y_1=a(x-x_1); a=\frac{-1}{-2}=1/2; y_1=0, x_1=0$$ $$y=\frac{1}{2}x$$

The last line will include a point that's a 90 degrees rotation of the point (6/5,3/5), which is the intersection of $y=\frac{1}{2}x$ and $y+2x=3$. So it will include the point (-3/5,6/5). Since this line must be parallel with $y=\frac{1}{2}x$, it must have the same slope. Using point slope again: $$y-y_1=a(x-x_1); a=1/2; y_1=6/5, x_1=3/5$$ $$y-\frac{6}{5}=\frac{1}{2}(x+\frac{3}{5})$$ $$y=\frac{1}{2}x+\frac{3}{2}$$

Answer 2

Notice the slopes of the two sides are both $-2$. $(y + 2x =0\implies y = -2x$ and $y+2x = 3\implies y=-2x +3$ so slope of both are $-2$). So the two lines given are parellel and opposite sides.

The other two sides must be perpendicular so their slopes must but be $-\frac 1{-2} = \frac 12$. So the equation of the lines must be $y = \frac 12x + b_1$ and $y = \frac 12x + b_2$ where $b_1, b_2$ are the $ y$ intercepts of the two sides.

Now a corner is $0,0$ so one of them has a $y$ intercept $0$ so $y = \frac 12x$ is a third line.

Now.... this is a square and not just a rectangle so distances between the parallel sides must be equal.

Let $(x_0, y_0)$ be the point where $y+2x=0$ and $y=\frac 12 x$ intersect.

$y_0 +2x_0 = 0$ and $y_0=\frac 12 x_0$ so $(x_0,y_0) = (0,0)$

Let $(x_1, y_1)$ be the point where $y+2x=3$ and $y=\frac 12 x$ itnersect.

$y_1 + 2x_1 = 3$ and $y_1=\frac 12x_1$ so $\frac 12 x_0 + 2x_0 = 3$ so $x_0=\frac 3{\frac 52}=\frac 65$ and $y_1 = \frac 12\frac 65 = \frac 35$.

And the side of the square is:

$s = \sqrt{(x_1 - x_0)^2 + (y_1-y_2)^2}=\sqrt{(\frac 65)^2 + (\frac 35)^2}=\sqrt{\frac{45}{25}}=\frac 3{\sqrt 5}$.

Now if $(x_2,y_2)$ is the point where $y+2x=0$ and $y=\frac 12x + b_2$ intersect then:

• $y_2 +2x_2 = 0$
• $y_2 =\frac 12x_2 + b_2$
• $\sqrt{(x_2-x_0)^2 + (y_2-x_0)^2} = \sqrt{x_2^2 + y_2^2} = \frac 3{\sqrt 5}$.

That's three equations with three unknowns:

$y_2 = -2x_2$ so $\sqrt{x_2^2 + 4x_2^2} = \frac 3{\sqrt 5}$ so $\sqrt{5x_2^2}=\sqrt{5}|x_2|=\frac 3{\sqrt 5}$ so $|x_2|=\frac 35$. Hmmm, can we assume $x_2$ is postive? Yes, The line $y+2x = 3$ is to the right of $y+2x = 0$ so the side $y+2x =0$ is going down to the right and the $x$ values are increases from $0$ to something positive while the $y$ values are decreasing to something negative. So $x_2 = \frac 35$ and $y_2 = -\frac 65$

And $-\frac 65 = \frac 12 \frac 35 +b_2$ so $b_2= -\frac 65 -\frac 3{10} = -\frac {15}{10}=-\frac 32$.

So the fourth line is $y = \frac 12 x-\frac 32$