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Question
Prove that the points A(-4,-1), B(-2, 4), C(4, 0) and D(2, 3) are the vertices of a rectangle.
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Solution 1
Let A (-4,-1); B (-2,-4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle.
So we should find the lengths of opposite sides of quadrilateral ABCD.
`AB = sqrt((-2+4)^2) + (-4 + 1)^2)`
`= sqrt(4 + 9)`
`= sqrt13`
`CD = sqrt((4 - 2)^2 + (0 - 3)^2)`
`= sqrt(4 +9)`
`= sqrt13`
Opposite sides are equal. So now we will check the lengths of the diagonals.
`AC = sqrt((4 + 4)^2 + (0 + 1)^2)`
`= sqrt(64 + 1)`
`= sqrt(65)`
`BD = sqrt((2 + 2)^2 + (3 + 4)^2)`
`= sqrt(16 + 49)`
`= sqrt65`
Opposite sides are equal as well as the diagonals are equal. Hence ABCD is a rectangle.
Solution 2
The given points are A (-4,-1); B (-2,-4); C (4, 0) and D (2, 3) .
`AB = sqrt({-2-(-4)}^2 + { -4-(-1)}^2) = sqrt ((2)^2+(-3)^2) = sqrt(4+9) = sqrt(13) ` units
` BC = sqrt({ 4-(-2)}^2+{0-(-4)}^2) = sqrt((6)^2 +(4)^2) = sqrt(36+16) = sqrt(52) = 2 sqrt(13) units`
`CD = sqrt((2-4)^2 +(3-0)^2) = sqrt((-2)^2 +(3)^2) = sqrt(4+9) = sqrt(13) units`
`AD = sqrt({2-(-4)}^2 + {3-(-1)}^2) = sqrt((6)^2 +(4)^2) = sqrt(36+16) = sqrt(52) = 2 sqrt(13) units`
`Thus , AB = CD = sqrt(13) units and BC = AD = 2 sqrt(13) units`
Also , `AC = sqrt({4-(-4)}^2+{0-(-1)}^2) = sqrt ((8)^2+(1)^2 ) = sqrt(64+1) = sqrt(65) units`
`BD = sqrt({2-(-2)}^2 +{3-(-4)}^2) = sqrt((4)^2 +(7)^2) = sqrt(16+49) = sqrt(65) units`
Also, diagonal AC = diagonal BD
Hence, the given points form a rectanglr
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