Question

find the trisectional points which the line joining 2,6
and -4,8​

Answer 1

Answer:

(-2,-10/3) and (0,4/3) are the trisection points

Answer 2

Answer:

Trisectional Points :-

(0 , 20/3)

(-2 , 22/3)

Step-by-step explanation:

Given,

A = (2 , 6)

B = (-4 , 8)

To Find :-

Trisectional points.

How To Do :-

As we know that the ratio that trisectional points divides the line segment we need apply the ratio and the value of co-ordinates in the in the section(internal division) formula to get the value of Point of trisection.

Formula Required :-

Section(Internal Division) Formula :-

$(x,y)=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

Solution :-

The ratio of point of trisection = 1 : 2 and 2 : 1

Let, 'P' be the point that divides the line segment AB in the ratio '1 : 2'

'Q' be the point that divides the line segment AB in the ratio  '2 : 1'

Finding co-ordinates of 'P' :-

Let,

m : n = 1 : 2

A = (2 , 6)

x_1 = 2 , y_1 = 6

B = (-4 , 8)

x_2 = - 4 , y_2 = 8

$\implies P=\left(\dfrac{1(-4)+2(2)}{1+2},\dfrac{1(8)+2(6)}{1+2}\right)$

$=\left(\dfrac{-4+4}{3},\dfrac{8+12}{3}\right)$

= (0/3 , 20/3)

= (0 , 20/3)

∴ Co-ordinates of 'P' = (0 , 20/3).

Finding Co-ordinates of 'Q' :-

Let,

m : n = 2 : 1

A = (2 , 6)

x_1 = 2 , y_1 = 6

B = (-4 , 8)

x_2 = - 4 , y_2 = 8

$Q=\left(\dfrac{2(-4)+1(2)}{2+1},\dfrac{2(8)+1(6)}{2+1}\right)$

$=\left(\dfrac{-8+2}{3},\dfrac{16+6}{3}\right)$

= ( -6/3 , 22/3)

= (-2 , 22/3)

∴ Co-ordinates of 'Q' = (-2 , 22/3).