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Q.

[[1]] is the coordinates of the points of trisection of the line joining the points (-3,0) and (6,6).


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Detailed Solution

https://www.vedantu.com/question-sets/9bc2385a-2171-4bba-8bd8-243919b3d75a2327594320610948749.png
The points of trisection P and Q divide the line segments A(-3,0) and B(6,6) into three equal parts. So, AP = PQ = QB = 1 unit.
We have to find  AP : PB = 1:2 (where PB = PQ+QB = 1+1 = 2)
As a result, P internally divides AB in the ratio 1:2. Using the section formula, the coordinates of P are thus PB.

Using section formula,
m1x2+m2x1m1+m2,m1y2+m2y1m1+m2 (x1,y1)=(-3,0)
(x2,y2) =(6,6)
m1:m2 = 1:2
Substituting the values in the formula
(16+2-31+2,16+201+2) =(6-63,6+03)
=(03,63) (0,2)
AQ:QB = 2:1
Also, Q  divides AB internally in the ratio 2:1. Therefore, the coordinates of Q, by applying section formula, are
m1x2+m2x1m1+m2,m1y2+m2y1m1+m2 (x1,y1) = (-3,0)
(x2,y2) = (6,6)
m1:m2= 2:1
 26+1-32+1,26+102+1  12-33,12+03 93,123=(3,4)
Therefore, the coordinates of the points of trisection of the line segment joining A and B are   We get   (3,4) and (0,2)
 
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