Question

the midpoints of the sides of the triangle AB BC and CA are (3,1) (5,6) (-3,2) respectively find tge vertices if the traingle ABC​

Answer 1

Answer:

Hence, the coordinates of A are (-5,-3)

coordinates of B are (11,5)

and coordinates of C are (-1,7)

Step-by-step explanation:

the midpoints of the sides of the triangle AB BC and CA are (3,1) (5,6) (-3,2).

Let the vertices of A are (a,a')

B are (b,b')

and C are (c,c')

The mid point of AB is (3,1)

i.e.

$(3,1)=(\dfrac{1}{2}(a+b),\dfrac{1}{2}(a'+b'))\\\\3=\dfrac{1}{2}(a+b) , 1=\dfrac{1}{2}(a'+b')\\\\a+b=6,a'+b'=2$

similarly the midpoint of BC is (5,6)

i.e.

$(5,6)=(\dfrac{1}{2}(b+c),\dfrac{1}{2}(b'+c'))\\\\5=\dfrac{1}{2}(b+c),6=\dfrac{1}{2}(b'+c')\\\\b+c=10,b'+c'=12$

similarly the midpoint of CA is (-3,2)

i.e.

$(-3,2)=(\dfrac{1}{2}(a+c),\dfrac{1}{2}(a'+c'))\\\\-3=\dfrac{1}{2}(a+c),2=\dfrac{1}{2}(a'+c')\\\\a+c=-6,a'+c'=4$

Hence, we get the equations as:

a+b=6      ;    a'+b'=2------------(1)

b+c=10     ;   b'+c'=12-------------(2)

a+c= -6      ;  a'+c'=4---------------(3)

Hence, on subtracting equation (2) by (1) we get:

a-c= -4       ;   a'-c'= -10------(4)

Now on adding equation (3) and (4) we get:

2a= -10   ;  2a'=-6

a= -5    ; a'= -3

on putting the value of a and a' in equation (1) and (3) we get:

b=11     ;   b'= 5

c= -1     ;  c'=7

Hence, the coordinates of A are (-5,-3)

coordinates of B are (11,5)

and coordinates of C are (-1,7)