Question

Find the coordinates of the points of trisection of the line segment AB with A (2,7)
and B(-4,-8).​

Answer 1

$\underline\blue{\bold{Given \: Question :- }}$

• Find the coordinates of the points of trisection of the line segment AB with A (2,7) and B(-4,-8).

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$\huge \blue{AηsωeR} ✍$

${ \boxed {\bf{Given}}}$

• A line segment AB joining the points A (2,7) and B(-4,-8).

${ \boxed {\bf{To Find}}}$

• The point of trisection of line segment AB.

${ \boxed {\bf{Formula \: used :- }}}$

Section Formula

Let us consider a line segment joining the points

$\sf \: ⟼A(x_1,y_1) \: and \: B(x_2,y_2)$

and Let C (x, y) be any point which divides AB internally in the ratio m: n, then coordinates of C is given by

$\bf \:( x, y) = (\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} )$

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${ \boxed {\bf{Solution}}}$

$\sf \: Let P \: and \: Q \: be \: the \: point \: of \: trisection.$

$\sf \: So \: that, AP = PQ = QB$

$\sf \:So, \: it \: implies \: P \: divides \: AB \: in \: ratio \: 1 : 2$

$\sf \:and \: Q \: divides \: AB \: in \: the \: ratio \: 2 : 1.$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{Coordinates \: of \: P \: be \: (a, b)} \\ &\sf{Coordinates \: of \: Q \: be \: (c, d)} \end{cases}\end{gathered}\end{gathered}$

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Case :- 1

$\bf \:When \: P \: divides \: AB \: in \: the \: ratio \: 1 : 2.$

☆ Using section Formula, we get

$\bf \:Coordinates \: of \: P =$

$\bf \:( a, b) = (\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} )$

☆ where

$\sf \: x_1 = 2,y_1 =7 ,x_2= - 4 ,y_2= - 8 ,m = 1,n =2$

☆ So,

$\bf \:( a, b) =( \dfrac{ - 4 \times 1 + 2 \times 2}{2 + 1} , \dfrac{ - 8 \times 1 + 2 \times 7}{1 + 2} )$

$\bf \:( a, b) =( \dfrac{0}{3} , \dfrac{6}{3} )$

$\bf \:( a, b) =( 0, 2)$

$\bf\implies \:Coordinates \: of \: P = ( a, b) =(0 , 2)$

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Case :- 2

$\bf \:When \: Q \: divides \: AB \: in \: the \: ratio \: 2 : 1.$

☆ Using Section Formula, we get

$\bf \:Coordinates \: of \: Q, \: are$

$\bf \:( c, d) = (\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} )$

☆ where,

$\sf \: x_1 = 2,y_1 =7 ,x_2= - 4 ,y_2= - 8 ,m = 2,n =1$

☆ So,

$\bf \:( c, d) =( \dfrac{ - 4 \times 2 + 1 \times 2}{1 + 2} , \dfrac{ - 8 \times 2 + 1 \times 7}{2 + 1} )$

$\bf \:( c, d) =( \dfrac{ - 6}{3} , \dfrac{ - 9}{3} )$

$\bf\implies \:( c, d) = ( - 2, - 3)$

$\bf\implies \:Coordinates \: of \: Q, ( c, d) = ( - 2, - 3)$

$\begin{gathered}\begin{gathered}\bf Hence = \begin{cases} &\sf{Coordinates \: of \: P \: be \: (0, 2)} \\ &\sf{Coordinates \: of \: Q \: be \: (-2, -3)} \end{cases}\end{gathered}\end{gathered}$

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