To prove a rhombus you need to show that the sides are congruent and the diagonals are perpendicular.
Sides:
[tex]d_{RM}=\sqrt{(-4-1)^2+(5-4)^2}[/tex]
[tex]=\sqrt{(-5)^2+(1)^2}[/tex]
[tex]=\sqrt{25+1}[/tex]
[tex]=\sqrt{26}[/tex]
[tex]d_{MB}=\sqrt{(1-2)^2+(4+1)^2}[/tex]
[tex]=\sqrt{(-1)^2+(15)^2}[/tex]
[tex]=\sqrt{1+25}[/tex]
[tex]=\sqrt{26}[/tex]
[tex]d_{BS}=\sqrt{(2+3)^2+(-1-0)^2}[/tex]
[tex]=\sqrt{(5)^2+(-1)^2}[/tex]
[tex]=\sqrt{25+1}[/tex]
[tex]=\sqrt{26}[/tex]
[tex]d_{SR}=\sqrt{(-3+4)^2+(0-5)^2}[/tex]
[tex]=\sqrt{(1)^2+(-5)^2}[/tex]
[tex]=\sqrt{1+25}[/tex]
[tex]=\sqrt{26}[/tex]
[tex]\overline{RM}[/tex] ≅ [tex]\overline{MB}[/tex] ≅ [tex]\overline{BS}[/tex] ≅ [tex]\overline{SR}[/tex]
Diagonals:
Use the slope formula: [tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
[tex]m_{RB}=\dfrac{5+1}{-4-2}[/tex]
[tex]=\dfrac{6}{-6}[/tex]
= -1
[tex]m_{MS}=\dfrac{4-0}{1+3}[/tex]
[tex]=\dfrac{4}{4}[/tex]
= 1
Slopes are opposite reciprocals so they are perpendicular.
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All of the sides are congruent and the diagonals are perpendicular so RMBS is a rhombus.