How do I prove analytically using co- ordinate geometry that the diagonals of a rhombus are perpendicular to each other

Question

Suppose I have a rhombus $ABCD$ with sides $AB,BC,CD$ and $DA$ as s units. With the help of concepts of co ordinate geometry I need to prove that the diagonals $AC$ and $BD$ are perpendicular to each other. How do I prove this particular property of rhombus?

Answer 1

Say $A(0,0)$ and $C(p,q)$, then $B$ and $D$ are the points of intersection of circles $x^2+y^2=1$ and $(x-p)^2+(y-q)^2=1$. Then the coordinates of $B$ and $D$ are the solutions of this system of equations. If we substract these two we get $$2px+2qy =0 \Longrightarrow y=-{p\over q}x$$ and $B,D$ define this equation (since they are on it). The line $AC$ has equation $y={q\over p}x$ we have:

$$k_{BD}\cdot k_{AC} = -{p\over q}\cdot {q\over p} =-1 \Longrightarrow AC\bot BD$$