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Using analytical geometry, prove that the diagonals of a rhombus areperpendicular to each other. |
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Answer» we need to prove ABCD is a rhombus where AB=BC=BD=DA A(0,0) B(S,0) C(A+S,B) D(A,B) `AD=SQRT(a^2+b^2)=s` `a^2+b^2=s^2` AB=`sqrt(s^2+0^2)=s` BC=`sqrt((a+s-s)^2+(b-0)^2)=sqrt(a^2+b^2)=S` CD=`sqrt((a+S-a)^2+(b-b)^2)=sqrt(s^2)=S` DA=`sqrt(a^2+b^2)=sqrt(s^2)=S` secondly we need to prove they are perpendicular `M_(AC)=(b-0)/(a+s-0)=b/(a+s)` `M_(BD)=(b-0)/(a-s)=b/(a-s)` `M_(AC)*M_(BD) ` `b/(a+s)*b/(a-s)` `b^2/(a^2-s^2)` `b^2/-b^2` -1 multipliplication of there slope is-1 so they are perpenticular |
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