Question

If the coordinates of the midpoints of the sides of triangle ABC are (1,1),(2,-3),(3,4).Find the coordinates of the vertices of Triangle ABC.
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Answer 1

A(4, 2) , B(2, 10) and C(0,-4)

let A(x1, x2) , B(y1, y2) and C(z1, z2)

given, midpoint of BC is D(1, 1),

midpoint of CA is E(2, -3)

and midpoint of AB is F(3, 4)

from midpoint section formula,

point, D (1,1) = [(y1 + z1)/2 , (y2 + z2)/2]

so, (y1 + z1) = 2....(1)

(y2 + z2) = 2......(2)

similarly,

point E(2,-3) = [(x1 + z1)/2, (x2, z2)/2]

x1 + z1 = 4 ......(3)

x2 + z2 = -6 .......(4)

point F (3,4) = [(x1 + y1)/2, (x2 + y2)/2]

x1 + y1 = 6.......(5)

x2 + y2 = 8 .......(6)

adding (1), (3) and (5) equations we get,

2(x1 + y1 + z1) = 2 + 4 + 6 = 12

⇒x1 + y1 + z1 = 6 .......(7)

subtracting equation (1) from equation (7) we get, x1 = 6 - 2 = 4

subtracting equation (3) from equation (7) we get, y1 = 6 - 4 = 2

subtracting equation (5) from equation (7) we get, z1 = 6 - 6 = 0

similarly doing with equations (2), (4) and (6),

x2 = 4 - 2 = 2, y2 = 4 - (-6) =10, z2 = 4 - 8 = -4

hence, A(4, 2) , B(2, 10) and C(0,-4)

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Answer 2

Step-by-step explanation:

Let A(x

1

,y

1

), B(x

2

,y

2

) and C(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, CA and AB respectively

Since D is the mid-point of BC

∴

2

x

2

+x

3

=1and

2

y

2

+y

3

=2

⇒x

2

+x

3

=2andy

2

+y

3

=4

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

=0andy

1

+y

3

=−2

and,

2

x

1

+x

2

=2and

2

y

1

+y

2

=−1

⇒x

1

+x

2

=4andy

1

+y

2

=−2

From (i), (ii) and (iii), we get

(x

2

+x

3

)+(+x

1

x

3

)+(x

1

+x

2

)=2+0+4and,(y

2

+y

3

)+(y

1

+y

3

)+(y

1

+y

2

)=4−2−2

⇒2(x

1

+x

2

+x

3

)=6and2(y

1

+y

2

+y

3

)=0

⇒x

1

+x

2

+x

3

=3andy

1

+y

2

+y

3

=0

From (i) and (Iv), we get

x

1

+2=3andy

1

+4=0

⇒x

1

=1andy

1

=−4

So, the coordinates of A are (1,-4)

From (ii) and (iv), we get

x

2

+0=3andy

2

−2=0

⇒x

2

=3andy

2

=2

So, coordinates of B are (3, 2)From (iii) and (iv), we get

x

3

+4=3andy

3

−2=0

⇒x

3

=−1andy

3

=2

So, coordinates of C are (-1, 2)

Hence, the vertices of the triangle ABC are A(1, -4), B(3, 2) and (-1, 2).