Question

Using coordinates geometry prove that diagonals of a rhombus are perpendicular to each other

Answer 1

Let s be the length of the sides of the rhombus ABCD.
Orient the rhombus so that its bottom side is AB where A = (0, 0) and B = (s, 0)
and let D = (a,b) lie in the first quadrant.
Then a>0, b>0, and a^2 + b^2 = s^2 (because the length of AD = length of AB = s).
The final point must be C = (a+s,b).

Summary so far
A = (0,0)
B = (s,0)
C = (a+s,b)
D = (a,b)
where a>0, b>0,s>0, and a² + b² = s² and s is the length of the sides of the rhombus.

We can check that it really is a rhombus by checking that the lengths of the four sides is really s.
length of AB = √((s-0)²+(0-0)²) = s
length of BC = √((a+s-s)²+(b-0)²) = √(a²+b²) = √(s²) = s
length of CD = √((a-a-s)²+(b-b)²) = √(s²) = s
length of DA = √((a-0)²+(b-0)²) = √(a²+b²) = √(s²) = s

The diagonals are AC and BD.

The slope of AC = (b-0) / (a+s-0) = b/(a+s)

The slope of BD = (b-0) / (a-s) = b/(a-s)

(slope of AC)*(slope of BD) =
b/(a+s) * b/(a-s) = b²/(a²-s²) = b²/(a²-(a²+b²)) = b²/(-b²) = -1

Since the product of the slopes of the diagonals is -1, then the diagonals of a rhombus are perpendicular to each other.