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Question
Show hat A(1,2), B(4,3),C(6,6) and D(3,5) are the vertices of a parallelogram. Show that ABCD is not rectangle.
Solution
The given vertices are A(1,2), B(4,3),C(6,6) and D(3,5) .
`AB = sqrt((1-4)^2+(2-3)^2) = sqrt((-3)^2 +(-1)^2) `
`= sqrt(9+1) = sqrt(10) `
`BC = sqrt((4-6)^2 +(3-6)^2) = sqrt((-2)^2 +(-3)^2)`
`= sqrt(4+9) = sqrt(13)`
`CD = sqrt((6-3) ^2 +(6-5)^2) = sqrt((3)^2 +(1)^2) `
`= sqrt(9+1) = sqrt(10)`
`AD = sqrt((1-3)^2 +(2-5)^2 ) = sqrt((-2)^2 +(-3)^2)`
`= sqrt(4+9) = sqrt(13) `
`∵ AB = CD = sqrt(10) " units and" BC= AD = sqrt(13) units `
Therefore, ABCD is a parallelogram
`AC = sqrt((1-6)^2 +(2-6)^2 )= sqrt((-5)^2 +(-4)^2)`
`= sqrt(25+16) = sqrt(41) `
`BD = sqrt((4-3)^2 +(3-5)^2 ) = sqrt((1)^2 +(-2)^2) `
`= sqrt(1+4) = sqrt(5) `
Thus, the diagonal AC and BD are not equal and hence ABCD is not a rectangle
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