Question

Find the coordinates of points which trisect the line segment joining $\left( {1, - 2} \right)$ and $\left( { - 3,4} \right)$.
A) $\left( { - \dfrac{1}{3},0} \right)$ and $\left( { - \dfrac{5}{3},2} \right)$
B) $\left( {\dfrac{1}{3},0} \right)$ and $\left( { - \dfrac{5}{3},2} \right)$
C) $\left( { - \dfrac{1}{3},0} \right)$ and $\left( {\dfrac{5}{3},2} \right)$
D) None of the above

Answer 1

Here we will first draw the line segment which is trisected. Then we will find the ratio in which the points trisect the line joining the given points. Further, we will apply the section formula with internal division to get the value of the coordinates.

Formula used:
We will use the section formula with internal division is given by
$\left( {\dfrac{{\left( {m \times c} \right) + \left( {n \times a} \right)}}{{m + n}},\dfrac{{\left( {m \times d} \right) + \left( {n \times b} \right)}}{{m + n}}} \right)$, where $\left( {a,b} \right)$ and $\left( {c,d} \right)$ are the points and $m$ and $n$ are the ratios.

Complete step by step solution:
Given points are $\left( {1, - 2} \right)$ and $\left( { - 3,4} \right)$.
Let A be the point with coordinates $\left( {1, - 2} \right)$ and B be the point with coordinates $\left( { - 3,4} \right)$.
Let P and Q be the point which trisects the line segment AB.
Firstly we will draw a figure which is trisected. Therefore, we get

Now according to the figure we can say that
$AP = PQ = QB = \alpha$
Now we will find the ratio in which the point P will divide the line segment AB. Therefore, we get
$\dfrac{{AP}}{{PB}} = \dfrac{{AP}}{{PQ + QB}}$
Substituting $AP = PQ = QB = \alpha$ in the above equation, we get
$\Rightarrow \dfrac{{AP}}{{PB}} = \dfrac{\alpha }{{2\alpha }} = \dfrac{1}{2}$
Hence, point P divides the line segment AB in the ratio $\dfrac{1}{2}$.
Now we will find the ratio in which the point Q will divide the line segment AB. Therefore, we get
$\dfrac{{AQ}}{{QB}} = \dfrac{{AP + PQ}}{{QB}}$
Substituting $AP = PQ = QB = \alpha$ in the above equation, we get
$\Rightarrow \dfrac{{AQ}}{{QB}} = \dfrac{{2\alpha }}{\alpha } = \dfrac{2}{1}$
Hence, point Q divides the line segment AB in the ratio $\dfrac{2}{1}$.
Now we will use the section formula with the internal division.
So, we will find the coordinates of the point P.
Hence, substituting $m = 1,n = 2,a = 1,b = - 2,c = - 3$ and $d = 4$ in the formula $\left( {\dfrac{{\left( {m \times c} \right) + \left( {n \times a} \right)}}{{m + n}},\dfrac{{\left( {m \times d} \right) + \left( {n \times b} \right)}}{{m + n}}} \right)$, we get
$\Rightarrow P\left( {\dfrac{{\left( {1 \times - 3} \right) + \left( {2 \times 1} \right)}}{{1 + 2}},\dfrac{{\left( {1 \times 4} \right) + \left( {2 \times - 2} \right)}}{{1 + 2}}} \right) = P\left( {\dfrac{{ - 3 + 2}}{3},\dfrac{{4 - 4}}{3}} \right) = P\left( { - \dfrac{1}{3},0} \right)$
Now, we will find the coordinates of the point Q.
Hence, substituting$m = 2,n = 1,a = 1,b = - 2,c = - 3$ and $d = 4$in the formula $\left( {\dfrac{{\left( {m \times c} \right) + \left( {n \times a} \right)}}{{m + n}},\dfrac{{\left( {m \times d} \right) + \left( {n \times b} \right)}}{{m + n}}} \right)$, we get
$\Rightarrow Q\left( {\dfrac{{\left( {2 \times - 3} \right) + \left( {1 \times 1} \right)}}{{2 + 1}},\dfrac{{\left( {2 \times 4} \right) + \left( {1 \times - 2} \right)}}{{2 + 1}}} \right) = Q\left( {\dfrac{{ - 6 + 1}}{3},\dfrac{{8 - 2}}{3}} \right)$
Simplifying the terms, we get
$\Rightarrow Q\left( {\dfrac{{ - 5}}{3},\dfrac{6}{3}} \right) = Q\left( { - \dfrac{5}{3},2} \right)$

Hence, the coordinates of the points which trisect the line segment joining $\left( {1, - 2} \right)$ and $\left( { - 3,4} \right)$ is $\left( { - \dfrac{1}{3},0} \right)$ and $\left( { - \dfrac{5}{3},2} \right)$.

Note:
We should note that the trisect means that the line segment is divided into three parts and all the three parts are equal i.e. equally divided. Coordinates system is represented in the Cartesian plane and the coordinates are written in such a way that the X intercept is written firstly and then the Y intercept is written after X coordinate in the form of $\left( {x,y} \right)$.
We should also remember the formula for the coordinates of the point which divides the line segment externally. If a point externally divides a line formed by joining the points $\left( {a,b} \right),\left( {c,d} \right)$ in the ratio of $m:n$, then the coordinates of that point is given by $\left( {\dfrac{{\left( {m \times c} \right) - \left( {n \times a} \right)}}{{m - n}},\dfrac{{\left( {m \times d} \right) - \left( {n \times b} \right)}}{{m - n}}} \right)$.