Question

Question 2
Find coordinates of the points of trisection of the line segment joining points (4, -1) and (-2, -3).

Answer 1

Let P (x₁, y₁) and Q (x₂, y₂) are the points of trisection of the line segment joining the given points.
$\Rightarrow$ AP = PQ = QB
Therefore, point P divides AB internally in the ratio 1:2 so using section formula.
(x = $\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2}$ y = $\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
In this case $(m_1=1,m_2=2and{x}_1=4,x_2=-2,y_1=-1,y_2{=}_3)$
$x_1=\frac{1\times (-2)+2\times 4}{1+2},y_1=\frac{1\times (-3)+2\times (-1)}{1+2}$
$x_1=\frac{-2+8}{3}=\frac{6}{3}=2,y_1=\frac{-3-2}{3}=\frac{-5}{3}$
Therefore,
$P(x_1,y_1)=(2,\frac{-5}{3})$
Point Q divides AB internally in the ratio 2:1
$x_2=\frac{2\times (-2)+1\times 4}{2+1},y_1=\frac{2\times (-3)+1\times (-1)}{2+1}$
$x_2=\frac{-4+4}{3}=0,y_2=\frac{-6-1}{3}=\frac{-7}{3}$
$P(x_2,y_2)=(0,\frac{-7}{3})$