Question

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

A(-2, 8) and B(4,12)

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

Answer 1

The correct option is B

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

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

$(x_1,y_1)=(-2,8)(x_2,y_2)=(4,12)$

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

$\therefore y=\frac{m_1{y}_2+m_2{y}_1}{m+n}=\frac{1\left(12\right)+2\left(8\right)}{3}=\frac{28}{3}$

$P=(0,\frac{28}{3})$

$m_1:m_2=2:1$

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

$\therefore y=\frac{m_1{y}_2+m_2{y}_1}{m+n}=\frac{2\left(12\right)+1\left(8\right)}{3}=\frac{32}{3}$

$Q=(2,\frac{32}{3})$