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 = √([tex]y_{2}[/tex] - [tex]y_{1}[/tex])² + ([tex]x_{2}[/tex] - [tex]x_{1}[/tex])²
Distance AB
A(4, 6), B(-1, 5)
[tex]x_{1}[/tex] =4 [tex]y_{1}[/tex]=6 [tex]x_{2}[/tex] =-1 [tex]y_{2}[/tex] =5
|AB|=√([tex]y_{2}[/tex] - [tex]y_{1}[/tex])² + ([tex]x_{2}[/tex] - [tex]x_{1}[/tex])²
=√(5- 6)² + (-1- 4)²
=√(-1)² + (-5)²
=√1+ 25
=√26
Distance BC
B(-1, 5), C(-2, 0)
[tex]x_{1}[/tex] =-1 [tex]y_{1}[/tex]=5 [tex]x_{2}[/tex] =-2 [tex]y_{2}[/tex] =0
|BC|=√([tex]y_{2}[/tex] - [tex]y_{1}[/tex])² + ([tex]x_{2}[/tex] - [tex]x_{1}[/tex])²
=√(0 - 5)² + (-2+1)²
=√(-5)² + (-1)²
=√25+1
=√26
Distance CD
C(-2, 0), D(3,
1)
[tex]x_{1}[/tex] =-2 [tex]y_{1}[/tex]=0 [tex]x_{2}[/tex] =3 [tex]y_{2}[/tex] =1
|CD|=√([tex]y_{2}[/tex] - [tex]y_{1}[/tex])² + ([tex]x_{2}[/tex] - [tex]x_{1}[/tex])²
=√(1 - 0)² + (3+2)²
=√(1)² + (5)²
=√1+25
=√26
Distance DA
D(3,
1) A(4,6)
[tex]x_{1}[/tex] =3 [tex]y_{1}[/tex]=1 [tex]x_{2}[/tex] =4 [tex]y_{2}[/tex] =6
|DA|=√([tex]y_{2}[/tex] - [tex]y_{1}[/tex])² + ([tex]x_{2}[/tex] - [tex]x_{1}[/tex])²
=√(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.
