Question

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

Solution

Let $$A(4,-1)$$ and $$B(-2,-3)$$ be the points of trisection of $$P$$ and $$Q$$so $$AP=PQ=QB$$For $$P$$:$$m_1:m_2=AP:PB=1:2$$$$(x_1,y_1)=(4,-1)$$ and $$(x_2,y_2)=(-2,-3)$$$$\therefore x=\dfrac{m_1 x_2+m_2x_1}{m_1+m_2}$$$$\Rightarrow x=\dfrac{1\times -2+2\times 4}{1+2}=\dfrac{-2+8}{3}=\dfrac{6}{3}=2$$$$\therefore y=\dfrac{m_1y_2+m_2y_1}{m_1+m_2}=\dfrac{1\times -3+2\times -1}{1+2}=\dfrac{-3-2}{3}=\dfrac{-5}{3}$$$$\therefore P=\left (2,\dfrac{-5}{3}\right )$$For $$Q$$:$$m_1:m_2=AQ:QB=2:1$$$$(x_1,y_1)=(4,-1)$$ and $$(x_2,y_2)=(-2,-3)$$$$\therefore x=\dfrac{m_1 x_2+m_2x_1}{m_1+m_2}$$$$\Rightarrow x=\dfrac{2\times -2+1\times 4}{1+2}=\dfrac{-4+4}{3}=\dfrac{0}{3}=0$$$$\therefore y=\dfrac{m_1y_2+m_2y_1}{m_1+m_2}=\dfrac{2\times -3+1\times -1}{1+2}=\dfrac{-6-1}{3}=\dfrac{-7}{3}$$$$\therefore Q=\left(0,\dfrac{-7}{3}\right)$$Mathematics

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