Question

Given the four lines with the equation $x + 2y - 3 = 0,3x + 4y - 7 = 0,2x + 3y - 4 = 0,4x + 5y - 6 = 0$ , then
A.They are all concurrent
B.They are the sides of a quadrilateral
C.Only three lines are concurrent
D.None of the above

Answer 1

Here, we will find the point of intersection of two equations by solving two equations and then substituting the point of intersection in the other two equations. We will follow this method for each pair of equations and from there we will find the condition of the given lines.

Complete step-by-step answer:
We are given the four lines with the equation as
$x + 2y - 3 = 0$……………………………….$\left( 1 \right)$
$3x + 4y - 7 = 0$…………………………….$\left( 2 \right)$
$2x + 3y - 4 = 0$…………………………...$\left( 3 \right)$
$4x + 5y - 6 = 0$……………………………$\left( 4 \right)$
By multiplying equation$\left( 1 \right)$by 3 we get
$3x + 6y - 9 = 0$ …………………………….$\left( 5 \right)$
By subtracting the equation $\left( 2 \right)$ from $\left( 5 \right)$, we get
$2y = 2$
Dividing by 2 on both the sides, we get
$\Rightarrow y = 1$
By substituting $y = 1$ in equation $\left( 1 \right)$, we get
$x + 2\left( 1 \right) - 3 = 0$
By rewriting the equation, we get
$\Rightarrow x = 1$
Thus, the solution set is $\left( {1,1} \right)$ .

By substituting $\left( {1,1} \right)$ in equation $\left( 3 \right)$ and equation $\left( 4 \right)$ a we get
$\begin{array}{l}2x + 3y - 4 = 0\\\ \Rightarrow 2\left( 1 \right) + 3\left( 1 \right) - 4 = 1\end{array}$
$\begin{array}{l}4x + 5y - 6 = 0\\\ \Rightarrow 4\left( 1 \right) + 5\left( 1 \right) - 6 = 3\end{array}$
Thus the solution set does not satisfy the other equations.
By multiplying equation $\left( 3 \right)$ by 2, we get
$4x + 6y - 8 = 0$ ……………………………..$\left( 6 \right)$
By subtracting the equation $\left( 4 \right)$ from $\left( 6 \right)$, we get
$y = 2$
By substituting $y = 2$ in equation $\left( 3 \right)$, we get
$2x + 3\left( 2 \right) - 4 = 0$
By simplifying the equation, we get
$\Rightarrow 2x = - 2$
Dividing by 2 on both the sides, we get
$\Rightarrow x = - 1$
Thus, the solution set is $\left( { - 1,2} \right)$ .
By substituting $\left( { - 1,2} \right)$ in equation $\left( 1 \right)$ and equation $\left( 2 \right)$ a we get
$\begin{array}{l}x + 2y - 3 = 0\\\ \Rightarrow \left( { - 1} \right) + 2\left( 2 \right) - 3 = 0\end{array}$
$\begin{array}{l}3x + 4y - 7 = 0\\\ \Rightarrow 3\left( { - 1} \right) + 4\left( 2 \right) - 7 = - 2\end{array}$
Thus the solution set satisfies the equation $x + 2y - 3 = 0$.
By multiplying equation $\left( 2 \right)$ by 2, we get
$6x + 8y - 14 = 0$
By multiplying equation $\left( 3 \right)$ by 3, we get
$6x + 9y - 12 = 0$
By subtracting the above equations, we get
$- y = 2$
By rewriting the equation, we get
$\Rightarrow y = - 2$
By substituting $y = - 2$ in equation $\left( 2 \right)$, we get
$3x + 4\left( { - 2} \right) - 7 = 0$
$\Rightarrow 3x = 15$
Dividing both sides by 3, we get
$\Rightarrow x = \dfrac{{15}}{3}$
$\Rightarrow x = 5$
Thus, the solution set is $\left( {5, - 2} \right)$ .
By substituting $\left( {5, - 2} \right)$ in equation $\left( 1 \right)$and equation $\left( 4 \right)$ a we get
$\begin{array}{l}x + 2y - 3 = 0\\\ \Rightarrow \left( 5 \right) + 2\left( { - 2} \right) - 3 = - 2\end{array}$
$\begin{array}{l}4x + 5y - 6 = 0\\\ \Rightarrow 4\left( 5 \right) + 5\left( { - 2} \right) - 6 = 4\end{array}$
Thus the solution set does not satisfy the other equations.
Since the solution set $\left( { - 1,2} \right)$ satisfies the equation $x + 2y - 3 = 0$.
Therefore, among the four lines of the equation, only three lines are concurrent and thus Option (C) is the correct answer.

Note: We know that three straight lines are said to be concurrent if they pass through a point or all three straight lines meet at a point. If three lines are concurrent, then the point of intersection of two lines lies on the third line also. Three or more lines in a plane meet each other at the common point is known as concurrent lines whereas if two lines in a plane meet each other at the common point is known as intersecting lines.