Question

if the mid points BC,CA ,AB of the triangle ABC are (+6, - 1),( - 4, 3),( 2, - 5) respectively then find the vertices ABC

Answer 1

Heya !!!

Let A ( X1 , Y1) , B( X2 , Y2) and C ( X3 , Y3) be the vertices of the GIVEN TRIANGLE ABC , and let D(6,-1) , E(-4,3) and F ( 2,-5) be the midpoints of BC , CA and AB respectively.

B ( X2 , Y2) and C ( X3 , Y3)

Here,

X1 = X2 , Y1 = Y2 and X2 = X3 , Y2 = Y3

D is the midpoint of BC => ( X1+X2/2 , Y1+Y2/2) => ( X2 + X3/2 , Y2 , Y3/2)

But coordinates of D is ( 6 , -1)

So,

X2 + X3 / 2 = 6 , Y2 + Y3/2 = -1

=> X2 + X3 = 12 ---------(1)

And,

Y2 + Y3 = -2 --------(2)

********************************************

C ( X3 , Y3 ) and A ( X1 + Y1)

Here,

X1 = X3 , Y1 = Y3 and X2 = X1 , Y2 = Y1

E is the midpoint of CA .

Therefore,

Coordinates of E = ( X1+X2/2 , Y1+Y2/2) = ( X3 + X1/2 , Y3 + Y1/2 )

But the coordinates of E is E ( -4 , 3)

Therefore,

( X3 + X1/2 ) = -4

=> X1 + X3 = -8 --------(3)

And,

Y3 + Y1/2 = 3

Y1 + Y3 = 6 ----------(4)

**********************************************

A ( X1 , Y1) and B ( X2 , Y2)

Here,

X1 = X1 , Y1 = Y1 and X2 = X2 , Y2 = Y2

F is the midpoint of AB .

Therefore,

Coordinates of F = ( X1+X2/2 , Y1 + Y2/2) = ( X1 + X2/ 2 , Y1+ Y2/2 )

But the coordinates of F is F(2,-5)

Therefore,

X1+X2/2 = 2

X1 + X2 = 4 --------(5)

And,

Y1 + Y2/2 = -5

Y1 + Y2 = -10 --------(6)

Adding equation (1) , (3) and (5) , we get

X2 + X3 + X1+X3 + X1 + X2 = 12 + (-8) + 4

2X1 + 2X2 + 2X3 = 12-8+4

2 ( X1 + X2 + X3) = 8

( X1+X2+X3) = 8/2

( X1 + X2 + X3) = 4 ---------(7)

Using (1) , (3) and (5) with (7) , we get

( X1 + X2 + X3) = 4

X1 + 12 = 4

X1 = 4 -12

X1 = -8

**************************

( X1+ X2 + X3) = 4

X2 + X1 + X3 = 4

X2 + (-8) = 4

X2 - 8 = 4

X2 = 4 +8

X2 = 12

**********************

( X1+ X2 + X3) = 4

( X3 + X1 + X2) = 4

X3 + 4 = 4

X3 = 4-4

X3 = 0

Adding (2) , (4) And (6) we get,

2 ( Y1 + Y2 + Y3) = -8

( Y1 + Y2 + Y3) = -4 ---------(8)

Using (2) , (4) and (6) with (8) we get,

( Y1 + Y2 + Y3) = -4

Y1 + (-2) = -4

Y1 = -4 +2

Y1 = -2

************************
( Y1 + Y2 + Y3) = -4

Y2 + Y1 + Y3 = -4

Y2 + 6 = -4

Y2 = -4 -6

Y2 = -10

***********************

( Y1 + Y2 + Y3) = -4

Y3 + Y1 + Y2 = -4

Y3 + (-10) = -4

Y3 = -4 + 10

Y3 = 6

Hence,

The vertices of ∆ABC are A(X1 + Y1 ) , B ( X2 + Y2) and C(X3,Y3)

=> A ( -8 , -2 ) , B ( 12 , -10 ) and C ( 0 , 6).

HOPE IT WILL HELP YOU....... :-)