Question

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

Answer 1

Hint : In this question, we will use the concept of section formulae of coordinate geometry. This states that the coordinate of the point which divides the line segment joining the points $({x_1},{y_1})$and $({x_2},{y_2})$ internally in the ratio m:n is given by $\left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$. …….(i)

Complete step-by-step solution -
Here , we have given points of the line segment , (3,-2) and (-3,-4)
Comparing this with $({x_1},{y_1})$and $({x_2},{y_2})$, we get
${x_1} = 3,{x_2} = - 3,{y_1} = - 2$ and ${y_2} = - 4$.
We know that trisection means dividing a line segment in three equal parts or dividing a line segment in the ratio 1:2 and 2:1 .
Case 1: when the line segment is divided into m:n as 1:2.
So, m=1 and n=2
Now putting these values in equation (i), we get
$
   \Rightarrow \left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) \\
   \Rightarrow \left( {x = \dfrac{{(1)( - 3) + (2)(3)}}{{1 + 2}},y = \dfrac{{(1)( - 4) + (2)( - 2)}}{{1 + 2}}} \right) \\
   \Rightarrow \left( {x = \dfrac{{( - 3) + (6)}}{3},y = \dfrac{{( - 4) + ( - 4)}}{3}} \right) \\
   \Rightarrow \left( {x = \dfrac{3}{3},y = \dfrac{{ - 8}}{3}} \right) \\
   \Rightarrow \left( {x = 1,y = \dfrac{{ - 8}}{3}} \right) \\
$
Hence, the point $\left( {1,\dfrac{{ - 8}}{3}} \right)$ divides the line segment in 1:2.
Case 2: when the line segment is divided into 2:1.
So we have , m=2 and n=1.
Now put these values in equation (i), we get
$
   \Rightarrow \left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) \\
   \Rightarrow \left( {x = \dfrac{{(2)( - 3) + (1)(3)}}{{2 + 1}},y = \dfrac{{(2)( - 4) + (1)( - 2)}}{{2 + 1}}} \right) \\
   \Rightarrow \left( {x = \dfrac{{( - 6) + (3)}}{3},y = \dfrac{{( - 8) + ( - 2)}}{3}} \right) \\
   \Rightarrow \left( {x = \dfrac{{ - 3}}{3},y = \dfrac{{ - 10}}{3}} \right) \\
   \Rightarrow \left( {x = - 1,y = \dfrac{{ - 10}}{3}} \right) \\
$
Hence , point $\left( { - 1,\dfrac{{ - 10}}{3}} \right)$ divides the line segment in 2:1 .
Therefore the points of trisection joining the given line segments are $\left( {1,\dfrac{{ - 8}}{3}} \right)$ and $\left( { - 1,\dfrac{{ - 10}}{3}} \right)$ .

Note : In this type of question we have to remember the concept of the section formulae .first we have to find out the required values and then we will make two case , in one case the ratio m:n is 1:2 and in other case the ratio m:n is 2:1 after that by putting those values in section formulae i.e. $\left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$ we will get the required points .