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Question
ABCD is rectangle formed by the points A(-1, -1), B(-1, 4), C(5, 4) and D(5, -1). If P,Q,R and S be the midpoints of AB, BC, CD and DA respectively, Show that PQRS is a rhombus.
Solution
Here, the points P ,Q, Rand S are the midpoint of ,AB ,BC, CD and DA respectively. Then
`"Coordinates of P = ((-1-1)/2 , (-1+4)/2) = (-1,3/2)`
`"Coordinates of Q = ((-1+5)/2 , (4+4)/2) = (2,.4)`
`"Coordinates of R = ((5+5)/2 , (4-1)/2)= (5,3/2)`
`"Coordinates of " S = ((-1+5)/2 ,(-1-1)/2) = (2,-1)`
Now,
`PQ = sqrt((2+1)^2 +(4-3/2)^2) = sqrt(9+25/4) = sqrt(61/2)`
`QR = sqrt((5-2)^2 +(3/2-4)^2 )= sqrt(9+25/4) = sqrt(61/2)`
`RS = sqrt((5-2)^2 +(3/2+1)^2 )= sqrt(9+25/4) = sqrt(61/2)`
`SP = sqrt((2+1)^2 +(-1-3/2)^2 )= sqrt(9+25/4) = sqrt(61/2)`
` PR = sqrt((5-1)^2 +(3/2-3/2)^2) = sqrt(36) = 6`
`QS = sqrt((2-2)^2 +(-1-4)^2) = sqrt(25) =5`
Thus, PQ = QR = RS = SP and PR ≠ QS therefore PQRS is a rhombus
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