the coordinates of the vertices of parellelogram ABCD are A (-2,-1);B (2,-1);C (2,-5)D (x,y) i) find mid point of AC ii) find x and y
Answers
Answer:
Given sides of parallelogram A(−2,1),B(a,0),C(4,b),D(1,2)
We know that diagonals of Parallelogram bisect each other.
Mid-point let say O of diagonal AC is given by
x=(
2
x
1
+x
2
)and y=(
2
y
1
+y
2
)
O (
2
−2+4
,
2
1+b
) .(1)
Mid-point let say P of diagonal BD is given by
P (
2
a+1
,
2
0+2
) . .(2)
Points O and P are same
Equating the corresponding co-ordinates of both midpoints, we get
2
−2+4
=
2
a+1
⇒a=1
and
2
1+b
=
2
0+2
⇒b=1
Now the Given co-ordinates of the parallelogram are written as
A(−2,1),B(1,0),C(4,1),D(1,2)
By distance formula,
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
we can find the length of each side
AB=
(−2−1)
2
+(1−0)
2
AB=
(3)
2
+(1)
2
=
10
AB=CD .(pair of opposite sides of the parallelogram are parallel and equal)
BC=
(4−1)
2
+(1−0)
2
BC=
(3)
2
+(1)
2
=
10
BC=AD .(pair of opposite sides of the parallelogram are parallel and equal )
⇒AB=BC=CD=AD=
10
⇒ABCD is a Rhombus