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)
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
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