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Solution
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Solution provided by AtoZmath.com
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Show that the points are the vertices of a rhombus calculator
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 1. Show that the points `A(3,0), B(4,5), C(-1,4), D(-2,-1)` are vertices of a rhombus
2. Show that the points `A(-3,2), B(-5,-5), C(2,-3), D(4,4)` are vertices of a rhombus
3. Show that the points `A(4,-1), B(6,0), C(7,2), D(5,1)` are vertices of a rhombus
4. Show that the points `A(1,0), B(5,3), C(2,7), D(-2,4)` are vertices of a rhombus
5. Show that the points `A(7,3), B(3,0), C(0,-4), D(4,-1)` are vertices of a rhombus
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Example1. Show that the points `A(3,0), B(4,5), C(-1,4), D(-2,-1)` are vertices of a rhombusSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is
`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)` A quadrilateral, in which all sides are equal, is a rhombus.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` The given points are `A(3,0),B(4,5),C(-1,4),D(-2,-1)` Length of sides:`AB=sqrt((4-3)^2+(5-0)^2)` `=sqrt((1)^2+(5)^2)` `=sqrt(1+25)` `=sqrt(26)` `:. AB=sqrt(26)` `BC=sqrt((-1-4)^2+(4-5)^2)` `=sqrt((-5)^2+(-1)^2)` `=sqrt(25+1)` `=sqrt(26)` `:. BC=sqrt(26)` `CD=sqrt((-2+1)^2+(-1-4)^2)` `=sqrt((-1)^2+(-5)^2)` `=sqrt(1+25)` `=sqrt(26)` `:. CD=sqrt(26)` `AD=sqrt((-2-3)^2+(-1-0)^2)` `=sqrt((-5)^2+(-1)^2)` `=sqrt(25+1)` `=sqrt(26)` `:. AD=sqrt(26)` Length of diagonals:`AC=sqrt((-1-3)^2+(4-0)^2)` `=sqrt((-4)^2+(4)^2)` `=sqrt(16+16)` `=sqrt(32)` `:. AC=4sqrt(2)` `BD=sqrt((-2-4)^2+(-1-5)^2)` `=sqrt((-6)^2+(-6)^2)` `=sqrt(36+36)` `=sqrt(72)` `:. BD=6sqrt(2)` Here all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` Since, all the sides are equal and both the diagonals are not equal Hence, ABCD is a rhombus Area `=1/2 xx AC xx BD` `=1/2 xx 4sqrt(2) xx 6sqrt(2)` `=24` Hence, the area of the rhombus is `24` square units 
2. Show that the points `A(-3,2), B(-5,-5), C(2,-3), D(4,4)` are vertices of a rhombusSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is
`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)` A quadrilateral, in which all sides are equal, is a rhombus.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` The given points are `A(-3,2),B(-5,-5),C(2,-3),D(4,4)` Length of sides:`AB=sqrt((-5+3)^2+(-5-2)^2)` `=sqrt((-2)^2+(-7)^2)` `=sqrt(4+49)` `=sqrt(53)` `:. AB=sqrt(53)` `BC=sqrt((2+5)^2+(-3+5)^2)` `=sqrt((7)^2+(2)^2)` `=sqrt(49+4)` `=sqrt(53)` `:. BC=sqrt(53)` `CD=sqrt((4-2)^2+(4+3)^2)` `=sqrt((2)^2+(7)^2)` `=sqrt(4+49)` `=sqrt(53)` `:. CD=sqrt(53)` `AD=sqrt((4+3)^2+(4-2)^2)` `=sqrt((7)^2+(2)^2)` `=sqrt(49+4)` `=sqrt(53)` `:. AD=sqrt(53)` Length of diagonals:`AC=sqrt((2+3)^2+(-3-2)^2)` `=sqrt((5)^2+(-5)^2)` `=sqrt(25+25)` `=sqrt(50)` `:. AC=5sqrt(2)` `BD=sqrt((4+5)^2+(4+5)^2)` `=sqrt((9)^2+(9)^2)` `=sqrt(81+81)` `=sqrt(162)` `:. BD=9sqrt(2)` Here all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` Since, all the sides are equal and both the diagonals are not equal Hence, ABCD is a rhombus Area `=1/2 xx AC xx BD` `=1/2 xx 5sqrt(2) xx 9sqrt(2)` `=45` Hence, the area of the rhombus is `45` square units 
3. Show that the points `A(4,-1), B(6,0), C(7,2), D(5,1)` are vertices of a rhombusSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is
`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)` A quadrilateral, in which all sides are equal, is a rhombus.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` The given points are `A(4,-1),B(6,0),C(7,2),D(5,1)` Length of sides:`AB=sqrt((6-4)^2+(0+1)^2)` `=sqrt((2)^2+(1)^2)` `=sqrt(4+1)` `=sqrt(5)` `:. AB=sqrt(5)` `BC=sqrt((7-6)^2+(2-0)^2)` `=sqrt((1)^2+(2)^2)` `=sqrt(1+4)` `=sqrt(5)` `:. BC=sqrt(5)` `CD=sqrt((5-7)^2+(1-2)^2)` `=sqrt((-2)^2+(-1)^2)` `=sqrt(4+1)` `=sqrt(5)` `:. CD=sqrt(5)` `AD=sqrt((5-4)^2+(1+1)^2)` `=sqrt((1)^2+(2)^2)` `=sqrt(1+4)` `=sqrt(5)` `:. AD=sqrt(5)` Length of diagonals:`AC=sqrt((7-4)^2+(2+1)^2)` `=sqrt((3)^2+(3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. AC=3sqrt(2)` `BD=sqrt((5-6)^2+(1-0)^2)` `=sqrt((-1)^2+(1)^2)` `=sqrt(1+1)` `=sqrt(2)` `:. BD=sqrt(2)` Here all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` Since, all the sides are equal and both the diagonals are not equal Hence, ABCD is a rhombus Area `=1/2 xx AC xx BD` `=1/2 xx 3sqrt(2) xx sqrt(2)` `=3` Hence, the area of the rhombus is `3` square units 
4. Show that the points `A(1,0), B(5,3), C(2,7), D(-2,4)` are vertices of a rhombusSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is
`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)` A quadrilateral, in which all sides are equal, is a rhombus.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` The given points are `A(1,0),B(5,3),C(2,7),D(-2,4)` Length of sides:`AB=sqrt((5-1)^2+(3-0)^2)` `=sqrt((4)^2+(3)^2)` `=sqrt(16+9)` `=sqrt(25)` `:. AB=5` `BC=sqrt((2-5)^2+(7-3)^2)` `=sqrt((-3)^2+(4)^2)` `=sqrt(9+16)` `=sqrt(25)` `:. BC=5` `CD=sqrt((-2-2)^2+(4-7)^2)` `=sqrt((-4)^2+(-3)^2)` `=sqrt(16+9)` `=sqrt(25)` `:. CD=5` `AD=sqrt((-2-1)^2+(4-0)^2)` `=sqrt((-3)^2+(4)^2)` `=sqrt(9+16)` `=sqrt(25)` `:. AD=5` Length of diagonals:`AC=sqrt((2-1)^2+(7-0)^2)` `=sqrt((1)^2+(7)^2)` `=sqrt(1+49)` `=sqrt(50)` `:. AC=5sqrt(2)` `BD=sqrt((-2-5)^2+(4-3)^2)` `=sqrt((-7)^2+(1)^2)` `=sqrt(49+1)` `=sqrt(50)` `:. BD=5sqrt(2)` Here `!=` `:.` ABCD is not a rhombus 
5. Show that the points `A(7,3), B(3,0), C(0,-4), D(4,-1)` are vertices of a rhombusSolution:We know that the distance between the two points `(x_1,y_1)` and `(x_2,y_2)` is
`d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)` A quadrilateral, in which all sides are equal, is a rhombus.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` The given points are `A(7,3),B(3,0),C(0,-4),D(4,-1)` Length of sides:`AB=sqrt((3-7)^2+(0-3)^2)` `=sqrt((-4)^2+(-3)^2)` `=sqrt(16+9)` `=sqrt(25)` `:. AB=5` `BC=sqrt((0-3)^2+(-4-0)^2)` `=sqrt((-3)^2+(-4)^2)` `=sqrt(9+16)` `=sqrt(25)` `:. BC=5` `CD=sqrt((4-0)^2+(-1+4)^2)` `=sqrt((4)^2+(3)^2)` `=sqrt(16+9)` `=sqrt(25)` `:. CD=5` `AD=sqrt((4-7)^2+(-1-3)^2)` `=sqrt((-3)^2+(-4)^2)` `=sqrt(9+16)` `=sqrt(25)` `:. AD=5` Length of diagonals:`AC=sqrt((0-7)^2+(-4-3)^2)` `=sqrt((-7)^2+(-7)^2)` `=sqrt(49+49)` `=sqrt(98)` `:. AC=7sqrt(2)` `BD=sqrt((4-3)^2+(-1-0)^2)` `=sqrt((1)^2+(-1)^2)` `=sqrt(1+1)` `=sqrt(2)` `:. BD=sqrt(2)` Here all sides `AB=BC=CD=AD` and both diagonals `AC!=BD` Since, all the sides are equal and both the diagonals are not equal Hence, ABCD is a rhombus Area `=1/2 xx AC xx BD` `=1/2 xx 7sqrt(2) xx sqrt(2)` `=7` Hence, the area of the rhombus is `7` square units 
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