- Thread starter
- #1

#### DrunkenOldFool

##### New member

- Feb 6, 2012

- 20

- Thread starter DrunkenOldFool
- Start date

- Thread starter
- #1

- Feb 6, 2012

- 20

- Feb 29, 2012

- 342

- Jan 27, 2012

- 95

Let one of the side be $y=x+k$ (Note that the line $y=x+k$ is always paralell to $y=x-7$) . The half of the length of the side of square will be equal to the perpendicular distance from the origin to the line $x+k=y$.

\[\frac{l}{2}= \Bigg| \frac{0-0+k}{\sqrt{2}}\Bigg|=\frac{|k|}{\sqrt{2}}\]

The length of each side is $l=\sqrt{2}|k|$ and the diameter of the circle is $\frac{4}{\sqrt{3}}$. Also, the diagonal of a square is $\sqrt{2}$ times its length of side. Therefore

\[\sqrt{2} \times \sqrt{2}|k|=\frac{4}{\sqrt{3}}\]

From here we get two values of $k$ i.e $2/\sqrt{3},-2/\sqrt{3}$. So two sides are $y=x+\frac{2}{\sqrt{3}},y=x-\frac{2}{\sqrt{3}}$.

\[\frac{l}{2}= \Bigg| \frac{0-0+k}{\sqrt{2}}\Bigg|=\frac{|k|}{\sqrt{2}}\]

The length of each side is $l=\sqrt{2}|k|$ and the diameter of the circle is $\frac{4}{\sqrt{3}}$. Also, the diagonal of a square is $\sqrt{2}$ times its length of side. Therefore

\[\sqrt{2} \times \sqrt{2}|k|=\frac{4}{\sqrt{3}}\]

From here we get two values of $k$ i.e $2/\sqrt{3},-2/\sqrt{3}$. So two sides are $y=x+\frac{2}{\sqrt{3}},y=x-\frac{2}{\sqrt{3}}$.

Last edited:

- Moderator
- #4

- Feb 7, 2012

- 2,703

[graph]psuayl2bwh[/graph]

The line $x-y=7$ has gradient 1, so the sides of the square will have gradient 1 and $-1.$ The circle is centred at the origin and has radius $2/\sqrt3$, so the vertices of the square will be at the points $(\pm2/\sqrt3,0)$ and $(0,\pm2/\sqrt3).$ You want the equations of the lines with gradient $\pm1$ going through those points, and with that information you can write down the answers.

- Admin
- #5

- Jan 26, 2012

- 4,191

Great word!(click on the diagram to embiggen it).

- Moderator
- #6

- Feb 7, 2012

- 2,703

It comes fromGreat word!

- Thread starter
- #7

- Feb 6, 2012

- 20

Thank You! This was extremely helpful.

[graph]psuayl2bwh[/graph]

The line $x-y=7$ has gradient 1, so the sides of the square will have gradient 1 and $-1.$ The circle is centred at the origin and has radius $2/\sqrt3$, so the vertices of the square will be at the points $(\pm2/\sqrt3,0)$ and $(0,\pm2/\sqrt3).$ You want the equations of the lines with gradient $\pm1$ going through those points, and with that information you can write down the answers.