Question

Find the points of trisection of the line segment joining the points A(-4, 3) and B (2, -1).

• A. $(2,\frac{5}{3})\text{and}(0,\frac{1}{3})$
• B. $(-2,\frac{5}{3})\text{and}(2,\frac{1}{3})$
• C. $(-2,\frac{5}{3})\text{and}(0,\frac{1}{3})$
• D. $(-2,\frac{5}{3})\text{and}(0,\frac{4}{3})$

Answer 1

The correct option is C $(-2,\frac{5}{3})\text{and}(0,\frac{1}{3})$

Let the points of trisection be $P(x,y)\text{and}Q(h,k)$

We know, $AP=PQ=QB\therefore AP:PB=1:2$
Co-ordinate by Section formula, $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

$P(x,y)=(\frac{1\left(2\right)+2(-4)}{1+2},\frac{1(-1)+2\left(3\right)}{1+2})=(\frac{2-8}{3},\frac{-1+6}{3})=(\frac{-6}{3},\frac{5}{3})\therefore P=(-2,\frac{5}{3})$

Point Q divides segment AB in the ratio 2 : 1.
Co-ordinate of point $Q(x,y)=(\frac{2\times 2+1\times (-4)}{2+1},\frac{2\times (-1)+1\times 3}{2+1})Q(x,y)=(\frac{0}{3},\frac{1}{3})=(0,\frac{1}{3})$

$\therefore$ Coordinates of the points of trisection are $(-2,\frac{5}{3})\&(0,\frac{1}{3})$.