Question

the midpoint of the sides of the triangle AB,BC and CA are(3,1)(5,6) and(-3,2) respectively.find the vertices of the triangle​

Answer 1

Answer:

Hence, the vertices of the triangle are:

A(4,-3)  , B(2,5)  , C(8,7)

Step-by-step explanation:

We are given that:

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

Let the vertices of the triangle are:

A(a,a') , B(b,b') , C(c,c') .

• As (3,1) is mid point of AB.

This means that:

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

This means:

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

and

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

• Similarly,

(5,6) is mid point of BC.

This means that:

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

This means:

$5=\dfrac{1}{2}(b+c)\\\\b+c=10$

and

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

• Similarly,

(-3,2) is midpoint of CA.

This means that:

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

This means:

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

and

$2=\dfrac{1}{2}(a'+c')\\\\a'+c'=4$

Hence we get six equations as:

a+b=6     , a'+b'=2-------(1)

b+c=10    , b'+c'=12------(2)

a+c= -6     , a'+c'=4------(3)

on subtracting first and second equation we get:

a-c= -4        ; a'-c'= -10-----(4)

Now on adding this equation to equation 3 we have:

2a= -10    ;  2a'= -6

a= -5       ; a'= -3

Putting the value of a and a' in equation (4) we have:

c= -1     ; c'=7

Also on putting value of c and c' in equation (2) we have:

b=11   ; b'=5

Hence, the vertices of the triangle are:

A(-5,-3)  , B(11,5)  , C(-1,7)