Show that the points (4, 6), (-1, 5), (-2, 0), (3, 1) are the vertices of a rhombus.​

Question

Show that the points (4, 6), (-1, 5), (-2, 0), (3,
1) are the vertices of a rhombus.​

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Gianna 2 weeks 2021-09-09T09:00:58+00:00 1 Answer 0

Answers ( )

    0
    2021-09-09T09:02:17+00:00

    Answer:

    |AB|=|BC|=|CD|=|DA|=√26

    This shows that the points are vertices of a rhombus.

    Step-by-step explanation:

    Show that the points (4, 6), (-1, 5), (-2, 0), (3,
    1) are the vertices of a rhombus, we simply have to show that they all have equal distances.

    Let ABCD be the vertices of the rhombus (4, 6), (-1, 5), (-2, 0), (3,
    1)  respectively.  That is;

    A(4, 6), B(-1, 5), C(-2, 0), D(3,
    1)  

    We will find the distance AB, BC, CD and DA, If the distances between them are equal, then we are able to prove that it is a rhombus;

    Using the distance formula;

    D = √(y_{2}y_{1})² + (x_{2}x_{1}

    Distance AB

    A(4, 6), B(-1, 5)

    x_{1} =4        y_{1}=6      x_{2} =-1    y_{2} =5

    |AB|=√(y_{2}y_{1})² + (x_{2}x_{1}

             =√(5- 6)² + (-1- 4)²

               =√(-1)² + (-5)²

               =√1+ 25

               =√26

    Distance BC

    B(-1, 5), C(-2, 0)

    x_{1} =-1        y_{1}=5     x_{2} =-2   y_{2} =0

    |BC|=√(y_{2}y_{1})² + (x_{2}x_{1}

          =√(0 – 5)² + (-2+1)²

           =√(-5)² + (-1)²

           =√25+1

            =√26

    Distance CD

    C(-2, 0), D(3,
    1)

    x_{1} =-2       y_{1}=0     x_{2} =3  y_{2} =1

    |CD|=√(y_{2}y_{1})² + (x_{2}x_{1}

          =√(1 – 0)² + (3+2)²

           =√(1)² + (5)²

           =√1+25

            =√26

    Distance DA

    D(3,
    1)  A(4,6)

    x_{1} =3     y_{1}=1     x_{2} =4   y_{2} =6

    |DA|=√(y_{2}y_{1})² + (x_{2}x_{1}

          =√(6 – 1)² + (4-3)²

           =√(5)² + (1)²

           =√25+1

            =√26

    |AB|=|BC|=|CD|=|DA|=√26

    This shows that the points are vertices of a rhombus.

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