Question

Find the coordinates of points of trisection of the line segment joining the points A(-2, 8) and B(4,12)

• A. $\begin{array}{c}P=(8,\frac{5}{3})\\ Q=(4,\frac{5}{9})\end{array}$
• B. $\begin{array}{c}P=(4,\frac{5}{3})\\ Q=(9,\frac{2}{3})\end{array}$
• C. $\begin{array}{c}P=(0,\frac{28}{3})\\ Q=(2,\frac{32}{3})\end{array}$
• D. $\begin{array}{c}P=(9,\frac{4}{3})\\ Q=(4,\frac{9}{5})\end{array}$

Answer 1

The correct option is C

$\begin{array}{c}P=(0,\frac{28}{3})\\ Q=(2,\frac{32}{3})\end{array}$

Since AP = PQ = QB, AP:PB = 1:2

The part of the line from (-2,8) to P is one-third of AB.

So, by section formula we have,

x-coordinate of P $=-2+\frac{1}{3}(4-(-2\left)\right)=0$

y-coordinate of P $=8+\frac{1}{3}(12-8)=\frac{28}{3}$

Therefore coordinates of point P are $(0,\frac{28}{3})$

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

The part of the line from (-2,8) to Q is two third the length of AB.

Again, by section formula we have,

x-coordinate of Q $=-2+\frac{2}{3}(4-(-2\left)\right)=2$

y-coordinate of Q $=8+\frac{2}{3}(12-8)=\frac{32}{3}$

Therefore coordinates of point Q are $(2,\frac{32}{3})$