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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 square calculator
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 1. Show that the points `A(1,2), B(5,4), C(3,8), D(-1,6)` are vertices of a square
2. Show that the points `A(2,3), B(-2,2), C(-1,-2), D(3,-1)` are vertices of a square
3. Show that the points `A(1,7), B(4,2), C(-1,-1), D(-4,4)` are vertices of a square
5. Show that the points `A(3,2), B(0,5), C(-3,2), D(0,-1)` are vertices of a square
6. Show that the points `A(5,6), B(1,5), C(2,1), D(6,2)` are vertices of a square
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Example1. Show that the points `A(1,2), B(5,4), C(3,8), D(-1,6)` are vertices of a squareSolution: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 and also the diagonals are equal, is a square.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC=BD` The given points are `A(1,2),B(5,4),C(3,8),D(-1,6)` Length of sides:`AB=sqrt((5-1)^2+(4-2)^2)` `=sqrt((4)^2+(2)^2)` `=sqrt(16+4)` `=sqrt(20)` `:. AB=2sqrt(5)` `BC=sqrt((3-5)^2+(8-4)^2)` `=sqrt((-2)^2+(4)^2)` `=sqrt(4+16)` `=sqrt(20)` `:. BC=2sqrt(5)` `CD=sqrt((-1-3)^2+(6-8)^2)` `=sqrt((-4)^2+(-2)^2)` `=sqrt(16+4)` `=sqrt(20)` `:. CD=2sqrt(5)` `AD=sqrt((-1-1)^2+(6-2)^2)` `=sqrt((-2)^2+(4)^2)` `=sqrt(4+16)` `=sqrt(20)` `:. AD=2sqrt(5)` Length of diagonals:`AC=sqrt((3-1)^2+(8-2)^2)` `=sqrt((2)^2+(6)^2)` `=sqrt(4+36)` `=sqrt(40)` `:. AC=2sqrt(10)` `BD=sqrt((-1-5)^2+(6-4)^2)` `=sqrt((-6)^2+(2)^2)` `=sqrt(36+4)` `=sqrt(40)` `:. BD=2sqrt(10)` Here, all sides `AB=BC=CD=AD` and both diagonals `AC=BD` Since, all the sides are equal and both the diagonals are equal Hence, ABCD is a square 
2. Show that the points `A(2,3), B(-2,2), C(-1,-2), D(3,-1)` are vertices of a squareSolution: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 and also the diagonals are equal, is a square.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC=BD` The given points are `A(2,3),B(-2,2),C(-1,-2),D(3,-1)` Length of sides:`AB=sqrt((-2-2)^2+(2-3)^2)` `=sqrt((-4)^2+(-1)^2)` `=sqrt(16+1)` `=sqrt(17)` `:. AB=sqrt(17)` `BC=sqrt((-1+2)^2+(-2-2)^2)` `=sqrt((1)^2+(-4)^2)` `=sqrt(1+16)` `=sqrt(17)` `:. BC=sqrt(17)` `CD=sqrt((3+1)^2+(-1+2)^2)` `=sqrt((4)^2+(1)^2)` `=sqrt(16+1)` `=sqrt(17)` `:. CD=sqrt(17)` `AD=sqrt((3-2)^2+(-1-3)^2)` `=sqrt((1)^2+(-4)^2)` `=sqrt(1+16)` `=sqrt(17)` `:. AD=sqrt(17)` Length of diagonals:`AC=sqrt((-1-2)^2+(-2-3)^2)` `=sqrt((-3)^2+(-5)^2)` `=sqrt(9+25)` `=sqrt(34)` `:. AC=sqrt(34)` `BD=sqrt((3+2)^2+(-1-2)^2)` `=sqrt((5)^2+(-3)^2)` `=sqrt(25+9)` `=sqrt(34)` `:. BD=sqrt(34)` Here, all sides `AB=BC=CD=AD` and both diagonals `AC=BD` Since, all the sides are equal and both the diagonals are equal Hence, ABCD is a square 
3. Show that the points `A(1,7), B(4,2), C(-1,-1), D(-4,4)` are vertices of a squareSolution: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 and also the diagonals are equal, is a square.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC=BD` The given points are `A(1,7),B(4,2),C(-1,-1),D(-4,4)` Length of sides:`AB=sqrt((4-1)^2+(2-7)^2)` `=sqrt((3)^2+(-5)^2)` `=sqrt(9+25)` `=sqrt(34)` `:. AB=sqrt(34)` `BC=sqrt((-1-4)^2+(-1-2)^2)` `=sqrt((-5)^2+(-3)^2)` `=sqrt(25+9)` `=sqrt(34)` `:. BC=sqrt(34)` `CD=sqrt((-4+1)^2+(4+1)^2)` `=sqrt((-3)^2+(5)^2)` `=sqrt(9+25)` `=sqrt(34)` `:. CD=sqrt(34)` `AD=sqrt((-4-1)^2+(4-7)^2)` `=sqrt((-5)^2+(-3)^2)` `=sqrt(25+9)` `=sqrt(34)` `:. AD=sqrt(34)` Length of diagonals:`AC=sqrt((-1-1)^2+(-1-7)^2)` `=sqrt((-2)^2+(-8)^2)` `=sqrt(4+64)` `=sqrt(68)` `:. AC=2sqrt(17)` `BD=sqrt((-4-4)^2+(4-2)^2)` `=sqrt((-8)^2+(2)^2)` `=sqrt(64+4)` `=sqrt(68)` `:. BD=2sqrt(17)` Here, all sides `AB=BC=CD=AD` and both diagonals `AC=BD` Since, all the sides are equal and both the diagonals are equal Hence, ABCD is a square 
4. Show that the points `A(3,2), B(0,5), C(-3,2), D(0,-1)` are vertices of a squareSolution: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 and also the diagonals are equal, is a square.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC=BD` The given points are `A(3,2),B(0,5),C(-3,2),D(0,-1)` Length of sides:`AB=sqrt((0-3)^2+(5-2)^2)` `=sqrt((-3)^2+(3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. AB=3sqrt(2)` `BC=sqrt((-3-0)^2+(2-5)^2)` `=sqrt((-3)^2+(-3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. BC=3sqrt(2)` `CD=sqrt((0+3)^2+(-1-2)^2)` `=sqrt((3)^2+(-3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. CD=3sqrt(2)` `AD=sqrt((0-3)^2+(-1-2)^2)` `=sqrt((-3)^2+(-3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. AD=3sqrt(2)` Length of diagonals:`AC=sqrt((-3-3)^2+(2-2)^2)` `=sqrt((-6)^2+(0)^2)` `=sqrt(36+0)` `=sqrt(36)` `:. AC=6` `BD=sqrt((0-0)^2+(-1-5)^2)` `=sqrt((0)^2+(-6)^2)` `=sqrt(0+36)` `=sqrt(36)` `:. BD=6` Here, all sides `AB=BC=CD=AD` and both diagonals `AC=BD` Since, all the sides are equal and both the diagonals are equal Hence, ABCD is a square 
5. Show that the points `A(5,6), B(1,5), C(2,1), D(6,2)` are vertices of a squareSolution: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 and also the diagonals are equal, is a square.
So, we have to prove all sides `AB=BC=CD=AD` and both diagonals `AC=BD` The given points are `A(5,6),B(1,5),C(2,1),D(6,2)` Length of sides:`AB=sqrt((1-5)^2+(5-6)^2)` `=sqrt((-4)^2+(-1)^2)` `=sqrt(16+1)` `=sqrt(17)` `:. AB=sqrt(17)` `BC=sqrt((2-1)^2+(1-5)^2)` `=sqrt((1)^2+(-4)^2)` `=sqrt(1+16)` `=sqrt(17)` `:. BC=sqrt(17)` `CD=sqrt((6-2)^2+(2-1)^2)` `=sqrt((4)^2+(1)^2)` `=sqrt(16+1)` `=sqrt(17)` `:. CD=sqrt(17)` `AD=sqrt((6-5)^2+(2-6)^2)` `=sqrt((1)^2+(-4)^2)` `=sqrt(1+16)` `=sqrt(17)` `:. AD=sqrt(17)` Length of diagonals:`AC=sqrt((2-5)^2+(1-6)^2)` `=sqrt((-3)^2+(-5)^2)` `=sqrt(9+25)` `=sqrt(34)` `:. AC=sqrt(34)` `BD=sqrt((6-1)^2+(2-5)^2)` `=sqrt((5)^2+(-3)^2)` `=sqrt(25+9)` `=sqrt(34)` `:. BD=sqrt(34)` Here, all sides `AB=BC=CD=AD` and both diagonals `AC=BD` Since, all the sides are equal and both the diagonals are equal Hence, ABCD is a square 
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