Question

Find the coordinates of the points of trisection of the line segment joining (1, -2) and (-3, 4)​

Answer 1

Answer:

let , A = (1, -2) and B = (-3, 4)

let the other three points are C, D, E.

assume the as following,

A(1, -2)_____C(?)_____D(?)_____E(?)____B(-3, 4)

point C divides the linesegment in the ratio 1 : 2

$section \: formula = (\frac{m1x1 + m2x2}{m1 + m2} ) \: \: \: \: ( \frac{m1y1 + m2y2}{m1 + m2} )$

a/c to formula,to find the co-ordinates of the point C , m1 = 1 , m2 = 2 , x1 = 1 , x2 = -3 , y1 = -2 , y2 = 4

as per the formula,

= [1 (1) + 2(-3)/ 1 + 2 , 1(-2)+2(4)/ 1 + 2 ]

= [ 1 - 6 / 3 , -2 + 8 / 3 ]

= [ -5/3 , 6 / 3] = [ -5/3 , 2]

therefore, the co-ordinates of point C = (-5/3, 2)

now we have to find the co-ordinates of point E,

point E divides the linesegments in 2 : 1 ratio.

as per formula, m1 = 2 , m2 = 1 , x1 = 1 , x2 = -3

y1 = -2 , y2 = 4

substituting in formula,

= [ 2(1) + 1(-3)/2+1 , 2(-2) + 1(4)/2+1 ]

= [ 2 - 3/3 , -4 + 4 / 3 ]

= [ - 1/3 , 0/3] = [ -1/3 , 0]

therefore, the co-ordinates of point E = (-1/3 , 0)

now point D is the midpoint of the line segment,

$mid \: point \: formula = ( \frac{x1 + x2}{2} ) \: ( \frac{y1 + y2}{2} )$

as per formula, x1 = 1 , x2 = -3 , y1 = -2 , y2 = 4

now insert in the formula ,

= [ 1 - 3/2 , -2 + 4 / 2 ]

= [ -2/2 , 2/2 ]

= ( -1 , 1 )

therefore, the co-ordinates of the point D = (-1 , 1)

Answer 2

$\mathbb{HOPE \: THIS \: HELPS \:U }$  

$\huge{\mathcal{\pink{A}\green{N}\red{S}\blue{W} {E}\green{R}}$

we will use this formula $\frac{m_1x_2+m_2x_1}{m_1+m_2}$ , $\frac{m_1y_2+m_2y_1}{m_1+m_2}$

D divides line segment ab into ratio of 1:2 $m_1=1,m_2= 2$ $x_1=1,x_2= -3$ $y_1=-2,y_2= 4$  

D = $\frac{1(-3)+2(1)}{1+2}$ , $\frac{1(4)+2(-2)}{1+2}$

D = $\frac{-3+2 }{3}$ , $\frac{4-4}{3}$

D = $\frac{-1 }{3}$ , $\frac{0}{3}$

D = $\frac{-1 }{3}$ , $0$

D coordinates are $\frac{-1 }{3}$ , $0$

C divides line segment ab into ration of 2:1 $m_1=2,m_2= 1$ $x_1=1,x_2= -3$ $y_1=-2,y_2= 4$  

c = $\frac{2(-3)+1(1)}{1+2}$ , $\frac{2(4)+1(-2)}{1+2}$

c = $\frac{-6+1 }{3}$ , $\frac{8-2}{3}$

c = $\frac{-5 }{3}$ , $\frac{6}{3}$

c = $\frac{-5 }{3}$ , $2$

c coordinates are $\frac{-1 }{3}$ , $2$