Question

find the co ordinate of the point of trisection of the line segment joining the points P(3,-1) and Q(-6,5)​

Answer 1

$\textbf{Given:}$

$\textsf{Points are P(3,-1) and Q(-6,5)}$

$\textbf{To find:}$

$\textsf{Point of trisection of line segment PQ}$

$\textbf{Solution:}$

$\textsf{Let P and Q be the points of trisection of line segment PQ}$

$\textsf{Then, P and Q divide the line segment AB internally}$

$\textsf{in the ratio 1:2 and 2:1 respectively}$

$\textsf{Since P divides AB internally in the ratio 1:2, we have}$

$P\left(\,\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

$P\left(\,\dfrac{1(-6)+2(3)}{1+2},\dfrac{1(5)+2(-1)}{1+2}\right)$

$P\left(\,\dfrac{6-6}{3},\dfrac{5-2}{3}\right)$

$P\left(0,\dfrac{3}{3}\right)$

$\implies\boxed{P\,(0,1)}$

$\text{Since Q divides AB internally in the ratio 2:1, we have}$

$Q\left(\,\dfrac{2(-6)+1(3)}{2+1},\dfrac{2(5)+1(-1)}{2+1}\right)$

$Q\left(\dfrac{-12+3}{3},\dfrac{10-1}{3}\right)$

$Q\left(\dfrac{-9}{3},\dfrac{9}{3}\right)$

$Q(-3,3)$

$\implies\boxed{Q(-3,3)}$

Find more:

Find the point of trisection of the line segment AB, where A (-6, 11) and B (10, -3).

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