Question

find the co-ordinates of the points of trisection of the line segment joining the points A (-4,3) and B (2.1)​

Answer 1

$\textbf{Given:}$

$\textsf{Points are A(-4,3) and B(2,1)}$

$\textbf{To find:}$

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

$\textbf{Solution:}$

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

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

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

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

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

$\mathsf{P\,\left(\dfrac{1(2)+2(-4)}{1+2},\dfrac{1(1)+2(3)}{1+2}\right)}$

$\mathsf{P\,\left(\dfrac{2-8}{3},\dfrac{1+6}{3}\right)}$

$\mathsf{P\,\left(\dfrac{-6}{3},\dfrac{7}{3}\right)}$

$\implies\boxed{\mathsf{P\,\left(-2,\dfrac{7}{3}\right)}}$

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

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

$\mathsf{Q\,\left(\dfrac{2(2)+1(-4)}{2+1},\dfrac{2(1)+1(3)}{2+1}\right)}$

$\mathsf{Q\,\left(\dfrac{4-4}{3},\dfrac{2+3}{3}\right)}$

$\mathsf{Q\,\left(\dfrac{0}{3},\dfrac{5}{3}\right)}$

$\implies\boxed{\mathsf{Q\,\left(0,\dfrac{5}{3}\right)}}$

Find more:

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

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