Question

Find coordinates of the points of trisection of the line segment joining the points A(-4,2) and B (-4, 9).

• A. $\begin{array}{c}P=(4,9)\\ Q=(9,\frac{7}{4})\end{array}$
• B. $\begin{array}{c}P=(-4,6)\\ Q=(-4,8)\end{array}$
• C. $\begin{array}{c}P=(-4,\frac{20}{3})\\ Q=(-4,\frac{13}{3})\end{array}$
• D. $\begin{array}{c}P=(-4,\frac{13}{3})\\ Q=(-4,\frac{20}{3})\end{array}$

Answer 1

The correct option is D

$\begin{array}{c}P=(-4,\frac{13}{3})\\ Q=(-4,\frac{20}{3})\end{array}$

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

The part of the line from (-4,2) to P is one-third of AB.

So, by section formula we have,

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

y-coordinate of P $=2+\frac{1}{3}(9-2)=\frac{13}{3}$

Therefore coordinates of point P are $(-4,\frac{13}{3})$

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

The part of the line from (-4,2) to Q is two third the length of AB.

Again, by section formula we have,

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

y-coordinate of Q $=2+\frac{2}{3}(9-2)=\frac{20}{3}$

Therefore coordinates of point Q are $(-4,\frac{20}{3})$