show that the points A (3,5) B(6,0) C(1,-3) and D(-2,2)are the vertices of the square ABCD
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Answer:
A(x
1 ,y
1 )=(3,5)
B(x 2 ,y 2 )=(6,0)
C(x 3y 3 )=(1,−3)
D(x
4
,y
4
)=(−2,2)
Then,
Distance between
AB=
(x
1
−x
2
)
2
+(y
1
−y
2
)
2
=
(3−6)
2
+(5−0)
2
=
9+25
AB=
34
BC=
(x
2
−x
3
)
2
+(y
2
−y
3
)
2
=
(6−1)
2
+(0+3)
2
=
25+9
BC=
34
CD=
(x
3
−x
4
)
2
+(y
3
−y
4
)
2
=
(1+2)
2
+(−3−2)
2
=
9+25
CD=
34
DA=
(x
4
−x
1
)
2
+(y
4
−y
1
)
2
=
(−2−3)
2
+(2−5)
2
=
25+9
DA=
34
Then, AB=BC=CD=DA
All sides are equal.
Hence, ABCD is a square
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