Question

Find the coordinates of the points of trisection of the line segment joining the points A(7, -2) and B(1, -5).

Answer 1

Trisection means dividing a line segment into three equal parts or dividing a line segment in the ratio 1:2 and 2:1 internally.

We know that if a point P(x,y) lies on a line segment AB between points A and B and satisfies AP: PB = m: n then we say that P divides AB internally in the ratio m: n The point of division has the coordinates, P(x,y) = $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Here ratio m:n = 1:2 and points are A(7, -2) and B(1, -5).

$\therefore P(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})=(\frac{1\times 1+2\times 7}{1+2},\frac{1\times -5+2\times -2}{1+2})=(\frac{1+14}{3},\frac{-5-4}{3})=(\frac{15}{3},\frac{-9}{3})=(5,-3)$

Here ratio m:n = 2:1 and points are A(7, -2) and B(1, -5).

$\therefore Q(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})=(\frac{2\times 1+1\times 7}{1+2},\frac{2\times -5+1\times -2}{1+2})=(\frac{2+7}{3},\frac{-10-2}{3})=(\frac{9}{3},\frac{-12}{3})=(3,-4)$

The trisection points are $(5,-3)and(3,-4)$