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How do I prove that a rhombus has perpendicular diagonals if the vertices are at the points:

A (0,0)

B (b,c)

C (b+ sqrt(bsq + csq), c)

D (sqrt(bsq + csq), 0)

 Apr 10, 2015

Best Answer 

 #1
avatar+130516 
+5

A (0,0)

B (b,c)

C (b+ sqrt(bsq + csq), c)

D (sqrt(bsq + csq), 0)

 

First......let's consider the slope of AC, one of the diagonals....we have

 ([b + √b^2 + c^2] - 0 ) / {c - 0)  =   ([b + √b^2 + c^2] / c  

 

Next, let's consider the slope of the other diagonal, BD.....we have

[ 0 - c ] / (  ([b + √b^2 + c^2 ] - b ) =  -c / ([b + √b^2 + c^2]

 

And since the slopes are negative reciprocals, the diagonals are perpendicular to each other

 

  

 Apr 10, 2015
 #1
avatar+130516 
+5
Best Answer

A (0,0)

B (b,c)

C (b+ sqrt(bsq + csq), c)

D (sqrt(bsq + csq), 0)

 

First......let's consider the slope of AC, one of the diagonals....we have

 ([b + √b^2 + c^2] - 0 ) / {c - 0)  =   ([b + √b^2 + c^2] / c  

 

Next, let's consider the slope of the other diagonal, BD.....we have

[ 0 - c ] / (  ([b + √b^2 + c^2 ] - b ) =  -c / ([b + √b^2 + c^2]

 

And since the slopes are negative reciprocals, the diagonals are perpendicular to each other

 

  

CPhill Apr 10, 2015

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