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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,

\Rightarrow (\frac{m 1 x 2 + m 2 x 1}{m 1 + m 2}, \frac{m 1 y 2 + m 2 y 1}{m 1 + m 2})

(x1,y1)=(-3,0)
(x2,y2) =(6,6)
m1:m2 = 1:2
Substituting the values in the formula

\Rightarrow (\frac{1 (6) + 2 (- 3)}{1 + 2}, \frac{1 (6) + 2 (0)}{1 + 2})

= (\frac{6 - 6}{3}, \frac{6 + (0)}{3})

= (\frac{0}{3}, \frac{6}{3})

⇒

(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

\Rightarrow (\frac{m 1 x 2 + m 2 x 1}{m 1 + m 2}, \frac{m 1 y 2 + m 2 y 1}{m 1 + m 2})

(x1,y1) = (-3,0)
(x2,y2) = (6,6)
m1:m2= 2:1

\Rightarrow (\frac{(2 (6) + 1 (- 3))}{2 + 1}, \frac{2 (6) + 1 (0)}{2 + 1})

\Rightarrow (\frac{12 - 3}{3}, \frac{12 + 0}{3}) \Rightarrow (\frac{9}{3}, \frac{12}{3}) = (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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[[1]] is the coordinates of the points of trisection of the line joining the points (-3,0) and (6,6).