Question

Let P and Q be the points of trisection of the line segment joining the points A( 2, -2 ) and B( - 7, 4 ) such that P is nearer to A. Find the coordinates of P and Q.

Answer 1

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Answer 2

Hey there !!


▶ Given points :-

→ A( 2 , -2 ).

→ B( - 7 , 4 ) .


▶ To find :-

→ Coordinates of P and Q .


▶ Solution :-

A/Q : - 

PA = PQ = QB { Due to trisection } 

And, 2PA = PB

=> AP / PB = 1 / 2 

➡ Using section formula 

$\begin{lgathered}Coordinates \:of \: P = \bigg( \dfrac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \dfrac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \bigg) = \bigg(\dfrac{(1 \times - 7) + (2 \times 2)}{1 + 2},\dfrac{(1 \times 4) + (2 \times - 2)}{3} \bigg) \\ \\ \implies \bigg( - \dfrac{3}{3} ,\frac{0}{3} \bigg)\end{lgathered}$

Therefore, coordinates of P = (- 1 , 0 ) 


▶ Now PQ = QB 

→ Q is the mid point of PB 

➡ Using mid point formula : 

$Coordinates \: of \:Q = \bigg( \dfrac{x_{1}+x_{2}}{2} , \dfrac{y_{1}+y_{2}}{2} \bigg) = \bigg ( \dfrac{-1 - 7}{2} , \dfrac{0 + 4}{2} \bigg)$

Therefore, co ordinates of Q = ( - 4 , 2 ).



✔✔ Hence, it is solved ✅✅.




THANKS



#BeBrainly.