Question

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Find the coordinates of the points of trisection of ths line segment joining points A ( 2,6) and B ( - 4 ,8) ? ​

Answer 1

Answer:

Co-ordinates of 'P' = (-2/3 , 22/3).

Co-ordinates of 'Q' = (0 , 20/3).

Step-by-step explanation:

Given,

A = (2 , 6)

B = (-4 , 8)

To Find :-

Co-ordinate of point of trisection.

How To Do :-

Here they gave the values of co-ordinates of 'A' and 'B' and we are asked to find the co-ordinates of point of trisection. We know that point of trisection divides the line segment in the ratios '2 : 1' and '1 : 2'. So by using these ratios we need to substitute the values in internal division formula and we need find the value of point of trisection.

Formula Required :-

Internal division :-

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

Solution :-

Let,

'P' be the co-ordinate that divide the line segment AB in he ratio '2 : 1'

'Q' be the co-ordinate that divide the line segment AB in he ratio '1 : 2'

Finding co-ordinates of 'P' :-

m : n = 2 : 1

A = (2 , 6)

Let,

x_1 = 2 , y_1 = 6

B = (-4 , 8)

Let,

x_2 = - 4 , y_2 = 8

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

$=\left(\dfrac{-4+2}{3},\dfrac{16+6}{3}\right)$

= (-2/3 , 22/3)

∴ Co-ordinates of 'P' = (-2/3 , 22/3).

Finding co-ordinates of 'Q' :-

m : n = 1 : 2

A = (2 , 6)

Let,

x_1 = 2 , y_1 = 6

B = (-4 , 8)

Let,

x_2 = - 4 , y_2 = 8

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

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

= (0/3 , 20/3)

= (0 , 20/3)

∴ Co-ordinates of 'Q' = (0 , 20/3).

Answer 2

Answer:

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