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
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 $\displaystyle a^2 = b^2 + c^2$.
If two lines are perpendicular, the product of their gradients is -1.
$\displaystyle \text{Gradient of T1} = \frac{\Delta y}{\Delta x} = \frac{c}{a+b}$
$\displaystyle \text{Gradient of T2} = \frac{\Delta y}{\Delta x} = \frac{-c}{a-b}$
Product of them $\displaystyle = \frac{c}{a+b} \frac{-c}{a-b} = -\frac{c^2}{a^2-b^2}$
Remember we found $\displaystyle a^2 = b^2 + c^2$ before. Then, $\displaystyle c^2 = a^2 - b^2$
$\displaystyle -\frac{c^2}{a^2-b^2} = -\frac{c^2}{c^2} = -1$