Question

Find coordinates of the point of trisection of the line segment joining the points

A(4,8) and B(-2,4).

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

Answer 1

The correct option is A

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

$m_1:m_2=1:2=AP:PB$

$(x_1,y_1)=(4,8)$

$(x_2,y_2)=(-2,4)$

$\therefore x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}=\frac{\left(1\right)(-2)+2\left(4\right)}{1+2}=\frac{6}{3}=2$

$\therefore y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}=\frac{\left(1\right)4+2\left(8\right)}{3}=\frac{20}{3}$

$\therefore PointP=(2,\frac{20}{3})$

For Q

$m_1:m_2=2:1=AQ:QB$

$(x_1,y_1)=(4,8):\left(x_2{y}_2\right)=(-2,4)$

$\therefore x=\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}=\frac{2(-2)+1\left(4\right)}{3}=0$

$\therefore y=\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}=\frac{2\left(4\right)+1\left(8\right)}{3}=\frac{16}{3}$

$\therefore Q=(x,y)=(0,\frac{16}{3})$