Mid points of sides of a triangle are (1/2,0) , (0,1/2) , (12,1/2) then circum centre of triangle is
Answers
Answer
Let A(x
1
,y
1
),B(x
2
,y
2
),C(x
3
,y
3
) be the vertices of △ABC
Let D(2,1) , F(-1,-3) and E(4,5) be the mid points of AB,AC and BC.
D and F are mid points pf ABC and AC
∴DF∥BE
E and F are mid points pf BC and Ac
∴EF∥BD
∴ DBEF is a parallelogram.
The diagonals of a parallelogram bisect each other i.e, both diagonals have same mid point
i.e, Midpoint BF=Midpoint of DE
(
2
x
2
+(−1)
,
2
y
2
+(−3)
)=(
2
2+4
,
2
1+5
)
∴
2
x
2
+(−1)
=
2
2+4
∴x
2
=7
Similarly
2
y
2
+(−3)
=
2
1+5
∴y
2
=9
i.e,(x
2
,y
2
)=(7,9)
D is the mid point of AB
D=(2,1)=(
2
x
1
+x
2
,
2
y
1
+y
2
)
2
x
1
+7
=2∴x
1
=−3
i.e,(x
1
,y
1
)=(−3,−7)
2
y
1
+9
=1∴y
1
=−7
F is the midpoint of AC
F=(−1,−3)=(
2
x
1
+x
3
,
2
y
1
+y
3
)
−1=
2
−3+x
3
∴x
3
=1
i.e,(x
3
,y
3
)=(1,1)
−3=
2
−7+y
3
∴y
3
=1
∴ The vertices of triangle are =(−3,−7),(7,9),(1,1)