show that point A ( 1,2), B ( 5,2), ( 3,1), D ( 1,5) are the vertices of a parallelogram or not
Answers
Answer:
We know that the distance between the two points (x
1
,y
1
) and (x
2
,y
2
) is
d=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
Let the given vertices be A=(−3,−2), B=(5,−2), C=(9,3) and D=(1,3)
We first find the distance between A=(−3,−2) and B=(5,−2) as follows:
AB=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(1−(−5))
2
+(−11−(−3))
2
=
(1+5)
2
+(−11+3)
2
=
6
2
+(−8)
2
=
36+64
=
100
=
10
2
=10
Similarly, the distance between B=(5,−2) and C=(9,3) is:
BC=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(7−1)
2
+(−6−(−11))
2
=
6
2
+(−6+11)
2
=
6
2
+5
2
=
36+25
=
61
Now, the distance between C=(9,3) and D=(1,3) is:
CD=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
=
(1−7)
2
+(2−(−6))
2
=
(−6)
2
+(2+6)
2
=
(−6)
2
+8
2
=
36+64
=
100
=
10
2
=10
Now, the distance between D=(1,3) and A=(−3,−2) is:
Answer:
By using distance formula
D=√[ (x2-x1)²+(y2-y1)² ]
thus,
AB=√[ (5-1)² + (2-2)² ]
AB=√[4²+0]
AB=√16
AB=4
CD=√[ (-2)²+4²]
CD= √20.