Question

find the coordinates of points which trisects the line segment joining (1,-2) and (-3,4)

Answer 1

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Given line segment joining the points A(1,-2)and B(-3,4)

Let P and Q be the points of trisection of AB i.e., AP = PQ = QB

Therefore, P divides AB internally in the ratio 1 : 2.

Therefore, the coordinates of P, by applying the section formula, are  m x 2 +n x 1 m + n , m y 2 + n y 1 m + n   . 

= [{1(-3) + 2(1)} / (1 + 2), {1(4) + 2(-2)} / (1 + 2)], = (–1 / 3, 0)

Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are

= [{2(-3) + 1(1)} / (1 + 2), {2(4) + 1(-2)} / (1 + 2)], = (–5/3, 4/3)

Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1 / 3, 0) and (–5/3, 4/3)

Answer 2

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

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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 $\frac{-1 }{3}$ , $2$