Question

The line segment joining the points (3,2),(-6,5) is trisected.The coordinates of point of trisection is .....​

Answer 1

HELLO GUYS THIS IS THE SOLUTION

Answer 2

$Let \:P \:and \: Q \: be \:the\: points \:of \\trisection \:of \: the \:line \: segment \: joining \:the\\points \: A(3,2) \:and \:B(-6,5) .$

$\therefore P \:divides \: AB \: internally \: in \\the \: ratio \: 1 : 2$

$\blue { ( By \: using \: section \: formula )}$

$\boxed { \pink { P(x,y) = \Big(\frac{ m_{1}x_{2} + m_{2}x_{1}}{m_{1}+m_{2}} , \frac{ m_{1}y_{2} + m_{2}y_{1}}{m_{1}+m_{2}} \Big)}}$

$\implies P = \Big( \frac{1\times (-6)+2\times 3}{1+2} , \frac{1\times 5+2\times 2}{1+2} \Big)$

$= \Big(\frac{-6+6}{3} , \frac{5+4}{3}\Big)$

$= \Big( 0 , \frac{9}{3}\Big) \\= (0,3)$

$Now, Q \: also \: divides \: AB \: internally \:in \\the\:ratio \: 2:1$

$Q = \Big( \frac{2\times (-6)+1\times 3}{2+1} , \frac{2\times 5+1\times 2}{2+1} \Big)$

$= \Big( \frac{-12+3}{3} , \frac{10+2}{3}\Big) \\= \Big( \frac{-9}{3} , \frac{12}{3}\Big) \\= ( -3,4)$

Therefore.,

$The \: Coordinates \:of \: the \: points \:of \\trisection \: of \:the \:line \: segment \:are \\\green {P(0,3) \: and \: Q(-3,4) }$

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