Question

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

Answer 1

Here, we need to find the coordinates of the trisection of the line joining the given points. We will use the midpoint formula to form linear equations in two variables. We will solve these equations to find the abscissa and ordinates of the points of trisection, and hence, obtain the coordinates of the points of trisection.

Formula Used:
According to the midpoint formula, the coordinates of the mid-point of the line segment joining two points $P\left( {{x_1},{y_1}} \right)$ and $Q\left( {{x_2},{y_2}} \right)$ are given by $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$.

Complete step by step solution:
Let the given points be $A\left( {4, - 1} \right)$ and $B\left( { - 2, - 3} \right)$.
Let the co-ordinates of the points of trisection be $P\left( {a,b} \right)$ and $Q\left( {p,q} \right)$.
First, we will draw the line segment.

Here, P and Q are the points of intersection. Therefore, P is the mid-point of AQ, and Q is the mid-point of PB.
Now, we will use the mid-point formula in the line segment AQ.
The co-ordinates of the mid-point of the line segment joining the points $A\left( {4, - 1} \right)$ and $Q\left( {p,q} \right)$ are $P\left( {a,b} \right)$.
Therefore, substituting ${x_1} = 4$, ${y_1} = - 1$, ${x_2} = p$, and ${y_2} = q$ in the mid-point formula, $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$, we get
$\Rightarrow \left( {a,b} \right) = \left( {\dfrac{{4 + p}}{2},\dfrac{{ - 1 + q}}{2}} \right)$
Comparing the ordinate and the abscissa, we get
$\Rightarrow a = \dfrac{{4 + p}}{2}$ and $b = \dfrac{{ - 1 + q}}{2}$
Now, we will use the mid-point formula in the line segment PB.
The co-ordinates of the mid-point of the line segment joining the points $P\left( {a,b} \right)$ and $B\left( { - 2, - 3} \right)$ are $Q\left( {p,q} \right)$.
Therefore, substituting ${x_1} = a$, ${y_1} = b$, ${x_2} = - 2$, and ${y_2} = - 3$ in the mid-point formula, we get
$\Rightarrow \left( {p,q} \right) = \left( {\dfrac{{a - 2}}{2},\dfrac{{b - 3}}{2}} \right)$
Comparing the ordinate and the abscissa, we get
$\Rightarrow p = \dfrac{{a - 2}}{2}$ and $q = \dfrac{{b - 3}}{2}$
We have the equations $a = \dfrac{{4 + p}}{2}$, $b = \dfrac{{ - 1 + q}}{2}$, $p = \dfrac{{a - 2}}{2}$, and $q = \dfrac{{b - 3}}{2}$.
We can observe that these four equations are linear equations in two variables.
First, we will solve the equations $a = \dfrac{{4 + p}}{2}$ and $p = \dfrac{{a - 2}}{2}$ to find the value of the abscissa of the points of trisection.
Substituting $p = \dfrac{{a - 2}}{2}$ in the equation $a = \dfrac{{4 + p}}{2}$, we get
$\Rightarrow a = \dfrac{{4 + \dfrac{{a - 2}}{2}}}{2}$
Adding the terms in the numerator, we get
$\Rightarrow a = \dfrac{{\dfrac{{8 + a - 2}}{2}}}{2}$
Simplifying the expression, we get
$\Rightarrow a = \dfrac{{6 + a}}{4}$
Multiplying both sides of the equation by 4, we get
$\Rightarrow 4a = 6 + a$
Subtracting $a$ from both sides of the equation, we get
$\Rightarrow 3a = 6$
Dividing both sides by 3, we get
$\Rightarrow a = 2$
Substituting $a = 2$ in the equation $p = \dfrac{{a - 2}}{2}$, we get
$\Rightarrow p = \dfrac{{2 - 2}}{2}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow p = \dfrac{0}{2}\\\ \Rightarrow p = 0\end{array}$
Now, we will solve the equations $b = \dfrac{{ - 1 + q}}{2}$ and $q = \dfrac{{b - 3}}{2}$ to find the value of the ordinates of the points of trisection.
Substituting $q = \dfrac{{b - 3}}{2}$ in the equation $b = \dfrac{{ - 1 + q}}{2}$, we get
$\Rightarrow b = \dfrac{{ - 1 + \dfrac{{b - 3}}{2}}}{2}$
Adding the terms in the numerator, we get
$\Rightarrow b = \dfrac{{\dfrac{{ - 2 + b - 3}}{2}}}{2}$
Simplifying the expression, we get
$\Rightarrow b = \dfrac{{b - 5}}{4}$
Multiplying both sides of the equation by 4, we get
$\Rightarrow 4b = b - 5$
Subtracting $b$ from both sides of the equation, we get
$\Rightarrow 3b = - 5$
Dividing both sides by 3, we get
$\Rightarrow b = - \dfrac{5}{3}$
Substituting $b = - \dfrac{5}{3}$ in the equation $q = \dfrac{{b - 3}}{2}$, we get
$\Rightarrow q = \dfrac{{ - \dfrac{5}{3} - 3}}{2}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow q = \dfrac{{\dfrac{{ - 5 - 9}}{3}}}{2}\\\ \Rightarrow q = \dfrac{{\dfrac{{ - 14}}{3}}}{2}\end{array}$
Dividing both denominator and numerator by 2, we get
$\begin{array}{l} \Rightarrow q = - \dfrac{{14}}{6}\\\ \Rightarrow q = - \dfrac{7}{3}\end{array}$
Therefore, we get the points of trisection as
$P\left( {a,b} \right) = \left( {2, - \dfrac{5}{3}} \right)$ and $Q\left( {p,q} \right) = \left( {0, - \dfrac{7}{3}} \right)$

$\therefore$ The coordinates of the points of trisection are $\left( {2, - \dfrac{5}{3}} \right)$ and $\left( {0, - \dfrac{7}{3}} \right)$.

Note:
We used the terms ‘abscissa’ and ‘ordinate’ in the solution. The abscissa of a point $\left( {x,y} \right)$ is $x$, and the ordinate of a point $\left( {x,y} \right)$ is $y$.
We can also solve the problem using the section formula.
According to the section formula, the coordinates of a point dividing the line segment joining two points $P\left( {{x_1},{y_1}} \right)$ and $Q\left( {{x_2},{y_2}} \right)$ in the ratio $m:n$, are given by $\left( {\dfrac{{m{x_1} + n{x_2}}}{{m + n}},\dfrac{{m{y_1} + n{y_2}}}{{m + n}}} \right)$.
The point $P\left( {a,b} \right)$ divides the line segment joining $A\left( {4, - 1} \right)$ and $B\left( { - 2, - 3} \right)$ in the ratio 1 : 2.
Similarly, the point $Q\left( {p,q} \right)$ divides the line segment joining $A\left( {4, - 1} \right)$ and $B\left( { - 2, - 3} \right)$ in the ratio $2:1$.
Using this, we can apply the section formula and obtain the coordinates of the points of trisection.