1.

Mid points of sides of a triangle are (1/2,0) , (0,1/2) , (12,1/2) then circum centre of triangle is

Answer»

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)


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