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Question
Show that the points A(2,1), B(5,2), C(6,4) and D(3,3) are the angular points of a parallelogram. Is this figure a rectangle?
Solution
The given points are s A(2,1), B(5,2), C(6,4) and D(3,3)
`AB = sqrt((5-2)^2 +(2-1)^2 ) = sqrt((3)^2 +(1)^2 ) = sqrt(9+1) = sqrt(10) ` units
`BC = sqrt((6-5)^2 +(4-2)^2 )= sqrt((1)^2 +(2)^3) = sqrt(1+4) = sqrt(5) `units
`CD = sqrt((3-6)^2 +(3-4)^2) = sqrt((-3)^2 +(-1)^2) = sqrt(9+1) = sqrt(10) `units
`AD = sqrt((3-2)^2+(3-1)^2) = sqrt((1)^2 +(2)^2) = sqrt(1+4) = sqrt(5) ` units
Thus, AB = CD = `sqrt(10) "units and " BC= AD = sqrt(5) ` units
So, quadrilateral ABCD is a parallelogram
`Also , AC = sqrt((6-2)^2 +(4-1)^2) = sqrt((4)^2 +(3)^2 )= sqrt(16+9) = sqrt(25) = 5 ` units
`BD = sqrt((3-5) ^2 +(3-2)^2 ) = sqrt((-2)^2 +(1)^2) = sqrt(4+1) = sqrt(5) units `
But diagonal AC is not equal to diagonal BD. Hence, the given points do not form a rectangle.
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