Math, asked by sologamer5618, 2 months ago

The mid.-points of the sides of a triangle are (-3, 2), (1, -2) and (4, 5) respectively.
Find the co-ordinates of the vertices of the triangle.

Answers

Answered by jasvindarsinghkuttan
3

Step-by-step explanation:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. Velocity is equivalent to a specification of an object's speed and direction of motion

Answered by hgautam1978
1

Answer:

Let A(x

1

y

1

),B(x

2

,y

2

)andC(x

3

,y

3

) be the vertices of △ABC.

Let D(1,2) ,E(0,-1) and F(2,-1) be the mid -points of sides BC,AC and AB

Since D is the mid point of BC.

A(x

1

y

1

),B(x

2

,y

2

)andC(x

3

,y

3

)

2

x

2

+x

3

=1and

2

y

2

+y

3

=2

⟹x

2

+x

3

=2y

2

+y

3

=4→ (i)

Similarly , E and F are the mid-points of CA and Ab respectively

2

x

1

+x

3

=0and

2

y

1

+y

3

=−1

⟹x

1

+x

3

=2andy

1

+y

3

−2→(2)

2

x

1

+x

2

=2and

2

y

1

+y

2

=−1

⟹x

1

+x

2

=4andy

1

+y

2

=−2→(3)

From (1),(2),and(3)

(x

2

+x

3

)+(x

1

+x

3

)+(x

1

+x

2

)=2+0+4

(y

2

+y

3

)+(y

1

+y

3

)+(y

1

+y

2

)=4−2−2

⟹x

1

+x

2

+x

3

=3andy

1

+y

2

+y

3

=0→(4)

From (1) and (4)

x

1

+2=3andy

1

+4=0

⟹x

1

=1andy

1

=−4∴(1,−4)

From (2) and (4)

x

2

+0=3andy

2

−2=0

⟹(x

2

,y

2

)=B(3,2)

From (3) and (4)

x

3

+4=3andy

3

−2=0

∴(x

3

,y

3

)=C(−1,2)

∴ Coordinate of △ABC are ⇒ (1,−4);B(3,2);C(−1,2)

Step-by-step explanation:

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