Question

The coordinates of the point of trisection of the line segment joining the points A(4, 8) and B(-2, 4) are

• A. $\begin{array}{c}P=(2,\frac{20}{3})\\ Q=(0,\frac{16}{3})\end{array}$
• B. $\begin{array}{c}P=(-4,\frac{8}{3})\\ Q=(10,-9)\end{array}$
• C. $\begin{array}{c}P=(8,-9)\\ Q=(4,\frac{-2}{3})\end{array}$
• D. $\begin{array}{c}P=(4,9)\\ Q=(-5,\frac{4}{5})\end{array}$

Answer 1

The correct option is A

$\begin{array}{c}P=(2,\frac{20}{3})\\ Q=(0,\frac{16}{3})\end{array}$

Let, P and Q be point of Trisection.
AP = PQ = QB and AP:PB = 1:2

The part of the line from (4, 8) to P is one third of AB.

Coordinates by section formula $(x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

$\therefore$ x-coordinate of P $=\frac{4\times 2-2\times 1}{3}=\frac{8-2}{3}=2$

y-coordinate of P $=\frac{8\times 2+4\times 1}{3}=\frac{16+4}{3}=\frac{20}{3}$

Therefore coordinates of point P are $(2,\frac{20}{3})$

Also since AP = PQ = QB, AQ:QB = 2:1

The part of the line from (4,8) to Q is two third the length of AB.

Again, by section formula we have,

x-coordinate of Q $=\frac{4\times 1-2\times 2}{3}=0$

y-coordinate of Q $=\frac{8\times 1+4\times 2}{3}=\frac{16}{3}$
Therefore coordinates of point Q are $(0,\frac{16}{3})$