Question

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

Answer 1

Solution :

• Let Let us suppose the given points are A (4, -1) and B(-2, -3).
• Let E and F be the point of trisection. Therefore, we have : AE = EF = FB
• Points of trisection means two points between the segment which divide the segment in three equal parts.
• First part divide the segment in 1 : 2 and second part divide the segment in 2 : 1.
• Hence, we can say that E divides AB in the ratio of 1:2 and F divides in 2:1 .
• Thus coordinates of E is given by :

$: \implies\sf E= \bigg(\dfrac{mx_2+nx_1}{m + n} ,\dfrac{my_2+ny_1}{m + n}\bigg)$

$:\implies\sf E = \bigg(\dfrac{1 \times ( - 2)+2 \times 4}{1+ 2} ,\dfrac{1 \times ( - 3)+2 \times ( - 1)}{1 + 2}\bigg)$

$: \implies\sf E = \bigg(\dfrac{ - 2+8}{3} ,\dfrac{ - 3 + (- 2)}{3}\bigg)$

$: \implies\sf E =\bigg(\dfrac{6}{3} ,\dfrac{ - 3 - 2}{3}\bigg)$

$: \implies\sf E =\bigg(\dfrac{6}{3} ,\dfrac{ - 5}{3}\bigg)$

$: \implies\sf E =\bigg(2 ,\dfrac{ - 5}{3}\bigg)$

• Similarly the coordinate of F is given by :

$\\: \implies\sf F = \bigg(\dfrac{mx_2+nx_1}{m + n} ,\dfrac{my_2+ny_1}{m + n}\bigg)$

$:\implies\sf F= \bigg(\dfrac{2 \times ( - 2)+1 \times 4}{2+ 1} ,\dfrac{2 \times ( - 3)+1 \times ( - 1)}{2 + 1}\bigg)$

$:\implies\sf F= \bigg( \dfrac{ - 4+4}{3} ,\dfrac{ - 6- 1}{3}\bigg)$

$:\implies\sf F= \bigg( \dfrac{ 0}{3} ,\dfrac{ - 7}{3}\bigg)$

$:\implies\sf F= \bigg( 0 ,\dfrac{ - 7}{3}\bigg)$

Therefore, the coordinates of the point of trisection are (2,-5/3) and (0,-7/3).