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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 rectangle calculator
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 1. Show that the points `A(-4,-1), B(-2,-4), C(4,0), D(2,3)` are vertices of a rectangle
2. Show that the points `A(2,-2), B(8,4), C(5,7), D(-1,1)` are vertices of a rectangle
3. Show that the points `A(0,-1), B(-2,3), C(6,7), D(8,3)` are vertices of a rectangle
4. Show that the points `A(2,-2), B(14,10), C(11,13), D(-1,1)` are vertices of a rectangle
5. Show that the points `A(0,-1), B(-2,3), C(6,7), D(8,3)` are vertices of a rectangle
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Example1. Show that the points `A(-4,-1), B(-2,-4), C(4,0), D(2,3)` are vertices of a rectangleSolution: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 opposites sides are equal and also the diagonals are equal, is a rectangle.
So, we have to prove opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` The given points are `A(-4,-1),B(-2,-4),C(4,0),D(2,3)` Length of sides:`AB=sqrt((-2+4)^2+(-4+1)^2)` `=sqrt((2)^2+(-3)^2)` `=sqrt(4+9)` `=sqrt(13)` `:. AB=sqrt(13)` `CD=sqrt((2-4)^2+(3-0)^2)` `=sqrt((-2)^2+(3)^2)` `=sqrt(4+9)` `=sqrt(13)` `:. CD=sqrt(13)` `BC=sqrt((4+2)^2+(0+4)^2)` `=sqrt((6)^2+(4)^2)` `=sqrt(36+16)` `=sqrt(52)` `:. BC=2sqrt(13)` `AD=sqrt((2+4)^2+(3+1)^2)` `=sqrt((6)^2+(4)^2)` `=sqrt(36+16)` `=sqrt(52)` `:. AD=2sqrt(13)` Length of diagonals:`AC=sqrt((4+4)^2+(0+1)^2)` `=sqrt((8)^2+(1)^2)` `=sqrt(64+1)` `=sqrt(65)` `:. AC=sqrt(65)` `BD=sqrt((2+2)^2+(3+4)^2)` `=sqrt((4)^2+(7)^2)` `=sqrt(16+49)` `=sqrt(65)` `:. BD=sqrt(65)` Here opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` Since, all the opposite sides are equal and both the diagonals are equal Hence, ABCD is a rectangle 
2. Show that the points `A(2,-2), B(8,4), C(5,7), D(-1,1)` are vertices of a rectangleSolution: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 opposites sides are equal and also the diagonals are equal, is a rectangle.
So, we have to prove opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` The given points are `A(2,-2),B(8,4),C(5,7),D(-1,1)` Length of sides:`AB=sqrt((8-2)^2+(4+2)^2)` `=sqrt((6)^2+(6)^2)` `=sqrt(36+36)` `=sqrt(72)` `:. AB=6sqrt(2)` `CD=sqrt((-1-5)^2+(1-7)^2)` `=sqrt((-6)^2+(-6)^2)` `=sqrt(36+36)` `=sqrt(72)` `:. CD=6sqrt(2)` `BC=sqrt((5-8)^2+(7-4)^2)` `=sqrt((-3)^2+(3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. BC=3sqrt(2)` `AD=sqrt((-1-2)^2+(1+2)^2)` `=sqrt((-3)^2+(3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. AD=3sqrt(2)` Length of diagonals:`AC=sqrt((5-2)^2+(7+2)^2)` `=sqrt((3)^2+(9)^2)` `=sqrt(9+81)` `=sqrt(90)` `:. AC=3sqrt(10)` `BD=sqrt((-1-8)^2+(1-4)^2)` `=sqrt((-9)^2+(-3)^2)` `=sqrt(81+9)` `=sqrt(90)` `:. BD=3sqrt(10)` Here opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` Since, all the opposite sides are equal and both the diagonals are equal Hence, ABCD is a rectangle 
3. Show that the points `A(0,-1), B(-2,3), C(6,7), D(8,3)` are vertices of a rectangleSolution: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 opposites sides are equal and also the diagonals are equal, is a rectangle.
So, we have to prove opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` The given points are `A(0,-1),B(-2,3),C(6,7),D(8,3)` Length of sides:`AB=sqrt((-2-0)^2+(3+1)^2)` `=sqrt((-2)^2+(4)^2)` `=sqrt(4+16)` `=sqrt(20)` `:. AB=2sqrt(5)` `CD=sqrt((8-6)^2+(3-7)^2)` `=sqrt((2)^2+(-4)^2)` `=sqrt(4+16)` `=sqrt(20)` `:. CD=2sqrt(5)` `BC=sqrt((6+2)^2+(7-3)^2)` `=sqrt((8)^2+(4)^2)` `=sqrt(64+16)` `=sqrt(80)` `:. BC=4sqrt(5)` `AD=sqrt((8-0)^2+(3+1)^2)` `=sqrt((8)^2+(4)^2)` `=sqrt(64+16)` `=sqrt(80)` `:. AD=4sqrt(5)` Length of diagonals:`AC=sqrt((6-0)^2+(7+1)^2)` `=sqrt((6)^2+(8)^2)` `=sqrt(36+64)` `=sqrt(100)` `:. AC=10` `BD=sqrt((8+2)^2+(3-3)^2)` `=sqrt((10)^2+(0)^2)` `=sqrt(100+0)` `=sqrt(100)` `:. BD=10` Here opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` Since, all the opposite sides are equal and both the diagonals are equal Hence, ABCD is a rectangle 
4. Show that the points `A(2,-2), B(14,10), C(11,13), D(-1,1)` are vertices of a rectangleSolution: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 opposites sides are equal and also the diagonals are equal, is a rectangle.
So, we have to prove opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` The given points are `A(2,-2),B(14,10),C(11,13),D(-1,1)` Length of sides:`AB=sqrt((14-2)^2+(10+2)^2)` `=sqrt((12)^2+(12)^2)` `=sqrt(144+144)` `=sqrt(288)` `:. AB=12sqrt(2)` `CD=sqrt((-1-11)^2+(1-13)^2)` `=sqrt((-12)^2+(-12)^2)` `=sqrt(144+144)` `=sqrt(288)` `:. CD=12sqrt(2)` `BC=sqrt((11-14)^2+(13-10)^2)` `=sqrt((-3)^2+(3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. BC=3sqrt(2)` `AD=sqrt((-1-2)^2+(1+2)^2)` `=sqrt((-3)^2+(3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. AD=3sqrt(2)` Length of diagonals:`AC=sqrt((11-2)^2+(13+2)^2)` `=sqrt((9)^2+(15)^2)` `=sqrt(81+225)` `=sqrt(306)` `:. AC=3sqrt(34)` `BD=sqrt((-1-14)^2+(1-10)^2)` `=sqrt((-15)^2+(-9)^2)` `=sqrt(225+81)` `=sqrt(306)` `:. BD=3sqrt(34)` Here opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` Since, all the opposite sides are equal and both the diagonals are equal Hence, ABCD is a rectangle 
5. Show that the points `A(0,-1), B(-2,3), C(6,7), D(8,3)` are vertices of a rectangleSolution: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 opposites sides are equal and also the diagonals are equal, is a rectangle.
So, we have to prove opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` The given points are `A(0,-1),B(-2,3),C(6,7),D(8,3)` Length of sides:`AB=sqrt((-2-0)^2+(3+1)^2)` `=sqrt((-2)^2+(4)^2)` `=sqrt(4+16)` `=sqrt(20)` `:. AB=2sqrt(5)` `CD=sqrt((8-6)^2+(3-7)^2)` `=sqrt((2)^2+(-4)^2)` `=sqrt(4+16)` `=sqrt(20)` `:. CD=2sqrt(5)` `BC=sqrt((6+2)^2+(7-3)^2)` `=sqrt((8)^2+(4)^2)` `=sqrt(64+16)` `=sqrt(80)` `:. BC=4sqrt(5)` `AD=sqrt((8-0)^2+(3+1)^2)` `=sqrt((8)^2+(4)^2)` `=sqrt(64+16)` `=sqrt(80)` `:. AD=4sqrt(5)` Length of diagonals:`AC=sqrt((6-0)^2+(7+1)^2)` `=sqrt((6)^2+(8)^2)` `=sqrt(36+64)` `=sqrt(100)` `:. AC=10` `BD=sqrt((8+2)^2+(3-3)^2)` `=sqrt((10)^2+(0)^2)` `=sqrt(100+0)` `=sqrt(100)` `:. BD=10` Here opposite sides `AB=CD` and `BC=AD` and both diagonals `AC=BD` Since, all the opposite sides are equal and both the diagonals are equal Hence, ABCD is a rectangle
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