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Find the coordinates of the points of trisection of the line segment joining \[\left( {4, - 1} \right)\] and \[\left( { - 2, - 3} \right)\].

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Last updated date: 04th Mar 2024
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IVSAT 2024
Answer
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Hint: Here, we will use the definition of trisection and get the ratio at which the line will trisect. Then by taking the cases according to the obtained trisection we will find the coordinate of the points using section formula.

Formula Used:
Section Formula: \[\left( {x,y} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\], where \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] are the two points of the line and \[m\] and \[n\] are the ratio in which the line is divided.

Complete step-by-step answer:
We have to find the coordinates of the points of trisection of the line segment \[\left( {4, - 1} \right)\] and \[\left( { - 2, - 3} \right)\].
We know that the trisection means dividing a figure or a quantity into three equal parts. Trisection points divide the line in the ratio 1:2 or 2:1 internally.
So, taking the first case as,
Case 1: If \[m:n = 1:2\]
The two point of the line segment is given as,
\[\left( {{x_1},{y_1}} \right) = \left( {4, - 1} \right)\]
\[\left( {{x_2},{y_2}} \right) = \left( { - 2, - 3} \right)\]
Using the above value in the formula \[\left( {x,y} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\], we get
\[\left( {x,y} \right) = \left( {\dfrac{{1 \times \left( { - 2} \right) + 2 \times 4}}{{1 + 2}},\dfrac{{1 \times \left( { - 3} \right) + 2 \times \left( { - 1} \right)}}{{1 + 2}}} \right)\]
Simplifying the expression, we get
\[ \Rightarrow \left( {x,y} \right) = \left( {\dfrac{{ - 2 + 8}}{3},\dfrac{{ - 3 - 2}}{3}} \right)\]
Adding and subtracting the terms, we get
\[ \Rightarrow \left( {x,y} \right) = \left( {\dfrac{6}{3},\dfrac{{ - 5}}{3}} \right)\]
Dividing the terms, we get
\[ \Rightarrow \left( {x,y} \right) = \left( {2, - \dfrac{5}{3}} \right)\]….\[\left( 1 \right)\]
Case 2: If \[m:n = 2:1\]
The two point of the line segment is given as,
\[\left( {{x_1},{y_1}} \right) = \left( {4, - 1} \right)\]
\[\left( {{x_2},{y_2}} \right) = \left( { - 2, - 3} \right)\]
Using the above value in the formula \[\left( {x,y} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\], we get
\[\left( {x,y} \right) = \left( {\dfrac{{2 \times \left( { - 2} \right) + 1 \times 4}}{{2 + 1}},\dfrac{{2 \times \left( { - 3} \right) + 1 \times \left( { - 1} \right)}}{{2 + 1}}} \right)\]
Simplifying the expression, we get
\[ \Rightarrow \left( {x,y} \right) = \left( {\dfrac{{ - 4 + 4}}{3},\dfrac{{ - 6 - 1}}{3}} \right)\]
Adding and subtracting the terms, we get
\[ \Rightarrow \left( {x,y} \right) = \left( {0, - \dfrac{7}{3}} \right)\]….\[\left( 2 \right)\]

So, we got the coordinate of the point of trisection of the line from equation \[\left( 1 \right)\] and \[\left( 2 \right)\] as \[\left( {x,y} \right) = \left( {2, - \dfrac{5}{3}} \right)\] and \[\left( {x,y} \right) = \left( {0, - \dfrac{7}{3}} \right)\].

Note:
Trisection of a line segment divides the line into three equal parts. Section formula is used to find the coordinate of points that divide a line segment in a ratio internally or externally. We can only use a section formula if two points are given to us and also the ratio at which the line is divided. Trisection is different from section formula as section formula is used to find the points for any ratio of the line divided but trisection has a common ratio that is 1:2 or 2:1 internally.
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