Question

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

Answer 1

$\large\pmb{\sf{Question : }}$

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

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$\large\pmb{\sf{Solution :}}$

Let $\sf{P (x_1, y_1) \: and \: Q (x_2, y_2) }$are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

$\red \bigstar$Therefore, point P divides AB internally in the ratio 1:2

$\sf{x_1 = \frac{1 \times ( - 2) + 2 \times 4}{1 + 2} \: , \: y_1 = \frac{1 \times ( - 3) + 2 \times ( - 1)}{1 + 2} }$

$\sf{x_1 = \frac{ - 2 + 8}{3} = \frac{6}{3} = 2 \: , \: y_1 = \frac{ - 3 - 2}{3} = \frac{ - 5}{3} }$

$\underline{ \boxed{\sf {\pink{\therefore \: P(x_1,y_1) = \left(2, \frac{ - 5}{3} \right) }}}}$

$\red \bigstar$ Point Q divides AB internally in the ratio 1:2

$\sf{x_2 = \frac{2 \times ( - 2) + 1 \times 4}{2 + 1} \: , \: y_2 = \frac{2 \times ( - 3) + 1 \times ( - 1)}{2 + 1} }$

$\sf{x_2 = \frac{ - 4 + 4}{3} = 0 \: , \: y_2 = \frac{ -6 - 1}{3} = \frac{ - 7}{3} }$

$\underline{ \boxed{\sf {\pink{\therefore \: Q(x_2,y_2) = \left(0, \frac{ - 7}{3} \right) }}}}$

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Answer 2

Solution :

• 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).