# Thread: coordinate geometry rhombus

1. ## coordinate geometry rhombus

prove diagonals of rhombus are perpendicular by using coordinate geometry.

it says to let the vertices be (0,0), (a,0), (a+b,0), (b,c)

how do i use the pythagorean theorem to show this? a^2+b^2=c^2

2. In the diagram, the vertices are (0,0), (a,0), (b,c), (a+b,c). All edges measure a. The diagonals are T1 and T2.

As you see, there's a right triangle with edges a, b and c. From pythagorean theorem, we find $a^2 = b^2 + c^2$.

If two lines are perpendicular, the product of their gradients is -1.

$\text{Gradient of T1} = \frac{\Delta y}{\Delta x} = \frac{c}{a+b}$

$\text{Gradient of T2} = \frac{\Delta y}{\Delta x} = \frac{-c}{a-b}$

Product of them $= \frac{c}{a+b} \frac{-c}{a-b} = -\frac{c^2}{a^2-b^2}$

Remember we found $a^2 = b^2 + c^2$ before. Then, $c^2 = a^2 - b^2$

$-\frac{c^2}{a^2-b^2} = -\frac{c^2}{c^2} = -1$

3. i understand everything up until this point:

Remember we found before. Then,
$
-\frac{c^2}{a^2-b^2} = -\frac{c^2}{c^2} = -1
$

how did you get -c^2/c^2 ?

4. Originally Posted by wingless
[IMG]
In the diagram, the vertices are (0,0), (a,0), (b,c), (a+b,c). All edges measure a. The diagonals are T1 and T2.

As you see, there's a right triangle with edges a, b and c. From pythagorean theorem, we find $\boxed{a^2 = b^2 + c^2}$.
From, $a^2 = b^2 + c^2$, we get to $c^2 = a^2 - b^2$

Then, in $-\frac{c^2}{a^2 - b^2}$, we just replaced $a^2 - b^2$ with $c^2$.

5. thanks.