Question

find the trisectional points of line joining (2,6),(-4,8).

Answer 1

check attachment mate!!

Answer 2

Trisection  points of line joining (2,6),(-4,8) are (0 , 20/3)  and (-2 , 22/3)

Given:

• Line joining (2,6),(-4,8)

To Find:

• Transactional points of line segment

Solution:

• A point P(x,y) divided (x₁ , y₁)  and (x₂ , y₂) in m : n ratio then
• P(x , y)  = (  (mx₂ + nx₁ )/(m + n)  , ((my₂ + ny₁ )/(m + n) )
• Mid point of  (x₁ , y₁)  and (x₂ , y₂)   is ( ( x₂ +  x₁ )/2  ,  ( y₂ +  y₁ )/2 )

Step 1:

A = (2 , 6) and B = (-4 , 8)  and C , D are trisection points

Hence AC = CD = BD = AB/3

Hence C  divides AB in 1 : 2 ratio

and D divided AB in 2 : 1 ratio

Step 2:

Using the formula coordinate of C are :

C divides (2,6),(-4,8) in 1 : 2 ratio

C = ( 1 * (-4) + 2 * 2) /(1 + 2) , ( 1 * (8)  +2 * 6) /(1 + 2) )

C = ( 0/3 , 20/3 )

C = (0 , 20/3)

Step 3:

Using the formula coordinate of D are :

D divides (2,6),(-4,8) in 2 : 1 ratio

D= (2* (-4) + 1 * 2) /(2 + 1) , ( 2* (8)  +1 * 6) /(2 + 1) )

D = ( -6/3 , 22/3 )

D = (-2 , 22/3)

or D can be found as mid point of C and B

= ( 0  - 4)/2 , ( 20/3 + 8)/2

= (-2 , 22/3)

Trisection  points of line joining (2,6),(-4,8) are (0 , 20/3)  and (-2 , 22/3)