Question

prove that (4,-1) (6,0) (7,2) and (5,1) are the vertices of a rhombus . is it a square ?

Answer 1

let A (4,-1) B (6,0) C (7,2) and D (5,1)
first find length of side
AB=root {(2)^2+(1)^2}=root5
BC=root {(1)^2+(2)^2}=root5
CD=root {(2)^2+(1)^2}=root5
AD=root {(1)^2+(2)^2}=root5

here we see
AB=BC=CD=AD
hence ABCD is rhombus

now AC=root {(3)^2+(3)^2}=3root2
here AC^2=BC^2+CA^2 isn't possible so
ABCD isn't square this is only rhombus

Answer 2

Sorry for more struck in my diary