# area of a parallelogram defined by the equations of its borders

Let $k$ be a real number between $0$ and $1$.

What is the area of the quadrilateral formed by the lines $y = kx, y = kx + 1, x = ky$ and $x = ky + 1$?

I tried replacing $k$ with $0.5$, however it was hard to convert back. So what is an easy method that I can use to solve this problem? The method I used is so long. I tried graphing, however it is just a parallelogram.

## 1 Answer

Hint:

Plot the lines. Note the lines $y=kx$, $x=ky$ are symmetric w.r.t. the first bissectrix, and similarly for the lines $y=kx+1$, $x=ky+1$. Thus it is easy to have the coordinates of the vertices. Then remember the area of a parallelogram is (the absolute value of) a determinant.

Added: a plot of the figure • If you have the coordinates of the vertices, you have the lengths of the diagonals (and if you can use the determinant, it's even simpler). – Bernard Jan 27 '17 at 17:34
• @bernard the problem is i never learned the determinant or anything regarding matrices, so im unable to use that methos...... hence im using the the coordinates, which in this case r unknown since when i try to solve for them i get a undefined answer – exchangehelpforuni Jan 27 '17 at 17:36
• Did you plot the $4$ lines? Don't forget the symmetry. – Bernard Jan 27 '17 at 17:37
• yes i did @bernard – exchangehelpforuni Jan 27 '17 at 17:43
• @bernard the poi of the lines r not numbers on the graph ie.: poi of y=kx and y=(x-1)/k is............ (1/1-k^2, k/1-k^2) which is exxtreeemlllt hard to work with and then use pythagorean theorm with:(((( – exchangehelpforuni Jan 27 '17 at 17:44