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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 parallelogram calculator
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 1. Show that the points `A(-3,-2), B(5,-2), C(9,3), D(1,3)` are vertices of a parallelogram
2. Show that the points `A(7,3), B(6,1), C(8,2), D(9,4)` are vertices of a parallelogram
3. Show that the points `A(1,-2), B(3,6), C(5,10), D(3,2)` are vertices of a parallelogram
4. Show that the points `A(6,8), B(3,7), C(-2,-2), D(1,-1)` are vertices of a parallelogram
5. Show that the points `A(3,1), B(0,-2), C(1,1), D(4,4)` are vertices of a parallelogram
6. Show that the points `A(2,1), B(5,2), C(6,4), D(3,3)` are vertices of a parallelogram
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Example1. Show that the points `A(-3,-2), B(5,-2), C(9,3), D(1,3)` are vertices of a parallelogramSolution: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, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD` The given points are `A(-3,-2),B(5,-2),C(9,3),D(1,3)` Length of sides:`AB=sqrt((5+3)^2+(-2+2)^2)` `=sqrt((8)^2+(0)^2)` `=sqrt(64+0)` `=sqrt(64)` `:. AB=8` `CD=sqrt((1-9)^2+(3-3)^2)` `=sqrt((-8)^2+(0)^2)` `=sqrt(64+0)` `=sqrt(64)` `:. CD=8` `BC=sqrt((9-5)^2+(3+2)^2)` `=sqrt((4)^2+(5)^2)` `=sqrt(16+25)` `=sqrt(41)` `:. BC=sqrt(41)` `AD=sqrt((1+3)^2+(3+2)^2)` `=sqrt((4)^2+(5)^2)` `=sqrt(16+25)` `=sqrt(41)` `:. AD=sqrt(41)` Here, opposite sides `AB=CD` and `BC=AD` Since, the lengths of the opposite sides are equal Hence, ABCD is a parallelogram 
2. Show that the points `A(7,3), B(6,1), C(8,2), D(9,4)` are vertices of a parallelogramSolution: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, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD` The given points are `A(7,3),B(6,1),C(8,2),D(9,4)` Length of sides:`AB=sqrt((6-7)^2+(1-3)^2)` `=sqrt((-1)^2+(-2)^2)` `=sqrt(1+4)` `=sqrt(5)` `:. AB=sqrt(5)` `CD=sqrt((9-8)^2+(4-2)^2)` `=sqrt((1)^2+(2)^2)` `=sqrt(1+4)` `=sqrt(5)` `:. CD=sqrt(5)` `BC=sqrt((8-6)^2+(2-1)^2)` `=sqrt((2)^2+(1)^2)` `=sqrt(4+1)` `=sqrt(5)` `:. BC=sqrt(5)` `AD=sqrt((9-7)^2+(4-3)^2)` `=sqrt((2)^2+(1)^2)` `=sqrt(4+1)` `=sqrt(5)` `:. AD=sqrt(5)` Here, opposite sides `AB=CD` and `BC=AD` Since, the lengths of the opposite sides are equal Hence, ABCD is a parallelogram 
3. Show that the points `A(1,-2), B(3,6), C(5,10), D(3,2)` are vertices of a parallelogramSolution: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, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD` The given points are `A(1,-2),B(3,6),C(5,10),D(3,2)` Length of sides:`AB=sqrt((3-1)^2+(6+2)^2)` `=sqrt((2)^2+(8)^2)` `=sqrt(4+64)` `=sqrt(68)` `:. AB=2sqrt(17)` `CD=sqrt((3-5)^2+(2-10)^2)` `=sqrt((-2)^2+(-8)^2)` `=sqrt(4+64)` `=sqrt(68)` `:. CD=2sqrt(17)` `BC=sqrt((5-3)^2+(10-6)^2)` `=sqrt((2)^2+(4)^2)` `=sqrt(4+16)` `=sqrt(20)` `:. BC=2sqrt(5)` `AD=sqrt((3-1)^2+(2+2)^2)` `=sqrt((2)^2+(4)^2)` `=sqrt(4+16)` `=sqrt(20)` `:. AD=2sqrt(5)` Here, opposite sides `AB=CD` and `BC=AD` Since, the lengths of the opposite sides are equal Hence, ABCD is a parallelogram 
4. Show that the points `A(6,8), B(3,7), C(-2,-2), D(1,-1)` are vertices of a parallelogramSolution: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, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD` The given points are `A(6,8),B(3,7),C(-2,-2),D(1,-1)` Length of sides:`AB=sqrt((3-6)^2+(7-8)^2)` `=sqrt((-3)^2+(-1)^2)` `=sqrt(9+1)` `=sqrt(10)` `:. AB=sqrt(10)` `CD=sqrt((1+2)^2+(-1+2)^2)` `=sqrt((3)^2+(1)^2)` `=sqrt(9+1)` `=sqrt(10)` `:. CD=sqrt(10)` `BC=sqrt((-2-3)^2+(-2-7)^2)` `=sqrt((-5)^2+(-9)^2)` `=sqrt(25+81)` `=sqrt(106)` `:. BC=sqrt(106)` `AD=sqrt((1-6)^2+(-1-8)^2)` `=sqrt((-5)^2+(-9)^2)` `=sqrt(25+81)` `=sqrt(106)` `:. AD=sqrt(106)` Here, opposite sides `AB=CD` and `BC=AD` Since, the lengths of the opposite sides are equal Hence, ABCD is a parallelogram 
5. Show that the points `A(3,1), B(0,-2), C(1,1), D(4,4)` are vertices of a parallelogramSolution: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, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD` The given points are `A(3,1),B(0,-2),C(1,1),D(4,4)` Length of sides:`AB=sqrt((0-3)^2+(-2-1)^2)` `=sqrt((-3)^2+(-3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. AB=3sqrt(2)` `CD=sqrt((4-1)^2+(4-1)^2)` `=sqrt((3)^2+(3)^2)` `=sqrt(9+9)` `=sqrt(18)` `:. CD=3sqrt(2)` `BC=sqrt((1-0)^2+(1+2)^2)` `=sqrt((1)^2+(3)^2)` `=sqrt(1+9)` `=sqrt(10)` `:. BC=sqrt(10)` `AD=sqrt((4-3)^2+(4-1)^2)` `=sqrt((1)^2+(3)^2)` `=sqrt(1+9)` `=sqrt(10)` `:. AD=sqrt(10)` Here, opposite sides `AB=CD` and `BC=AD` Since, the lengths of the opposite sides are equal Hence, ABCD is a parallelogram 
6. Show that the points `A(2,1), B(5,2), C(6,4), D(3,3)` are vertices of a parallelogramSolution: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, is a parallelogram.
So, we have to prove opposite sides `AB=CD` and `BC=AD` The given points are `A(2,1),B(5,2),C(6,4),D(3,3)` Length of sides:`AB=sqrt((5-2)^2+(2-1)^2)` `=sqrt((3)^2+(1)^2)` `=sqrt(9+1)` `=sqrt(10)` `:. AB=sqrt(10)` `CD=sqrt((3-6)^2+(3-4)^2)` `=sqrt((-3)^2+(-1)^2)` `=sqrt(9+1)` `=sqrt(10)` `:. CD=sqrt(10)` `BC=sqrt((6-5)^2+(4-2)^2)` `=sqrt((1)^2+(2)^2)` `=sqrt(1+4)` `=sqrt(5)` `:. BC=sqrt(5)` `AD=sqrt((3-2)^2+(3-1)^2)` `=sqrt((1)^2+(2)^2)` `=sqrt(1+4)` `=sqrt(5)` `:. AD=sqrt(5)` Here, opposite sides `AB=CD` and `BC=AD` Since, the lengths of the opposite sides are equal Hence, ABCD is a parallelogram 
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