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Question
Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.
Solution
Let the given points be A (-3, 2), B (-5, -5), C (2, -3) and D (4, 4).
AB = `sqrt((-5 + 3)^2 + (-5 - 2)^2) = sqrt(4+49) = sqrt(53)`
BC = = `sqrt((2+5)^2 + (-3+5)^2) = sqrt(49+4) = sqrt(53)`
CD= = `sqrt((4 - 2)^2 + (4+3)^2) = sqrt(4 + 49) = sqrt(53)`
DA = `sqrt((-3 - 4)^2 + (2 - 4)^2) = sqrt(49+4) = sqrt(53)`
AC =`sqrt((2+3)^2 + (-3 - 2)^2) = sqrt(25+25) = 5sqrt(2)`
BD =`sqrt((4+5)^2 + (4+5)^2) = sqrt(81+ 81) = 9sqrt(2)`
Since, AB = BC = CD = DA and AC ≠ BD
The given vertices are the vertices of a rhombus.
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