Prove that a shape must be a rhombus using rules of coordinate geometry

Question

Quadrilateral ABCD has points A (0,0) B (a,b) C (a+b,a+b) D (b,a). Prove it is a rhombus for any values of a and b.

I have tried to solve this by saying that since points A and C both have equal x and y coordinates, that this means all 4 sides are equal etc. However, this doesn't sound correct to me. Any ideas?

Just calculate the $4$ sides and show that $,AB=BC=CD=DA=\sqrt{a^2+b^2},$. — dxiv, May 30 '17 at 06:18

Hypergeometricx · Accepted Answer · 2017-05-30T06:56:49.863

2

Define $m(PQ)$ as the slope of PQ.

By inspection, $m(AB)=m(CD)=\dfrac ba$ and $m(BC)=m(AD)=\dfrac ab$, i.e. opposite sides are parallel.

Also, by inspection, $m(AC)\cdot m(BD)=1\cdot (-1)=-1$ i.e. diagonals are perpendicular.

Hence $ABCD$ is a rhombus. $\blacksquare$

edited May 30 '17 at 06:56

answered May 30 '17 at 06:49

Hypergeometricx

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Prove that a shape must be a rhombus using rules of coordinate geometry

1 Answers1