To determine whether the four straight lines given by the equations [tex]\(x + 2y - 3 = 0\)[/tex], [tex]\(3x + 4y - 7 = 0\)[/tex], [tex]\(2x + 3y - 4 = 0\)[/tex], and [tex]\(4x + 5y - 6 = 0\)[/tex] are concurrent, we need to check if there exists a common point where all four lines intersect simultaneously.
1. Finding Intersection Point of the First Three Lines:
We start by solving the system of three equations for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
[tex]\[
\begin{cases}
x + 2y - 3 = 0 \\
3x + 4y - 7 = 0 \\
2x + 3y - 4 = 0
\end{cases}
\][/tex]
Shortcut Explanation: Solve these equations algebraically or using substitution/elimination methods to find the solution for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]. However, in this problem, there is no common solution for these three equations, i.e., there is no intersection point from the first three lines.
2. Intersection Check:
Since there is no common intersection point of the first three lines, it implies that they do not meet at a single point. Without a common intersection point, it is impossible for all four lines to intersect at one point simultaneously.
3. Conclusion:
Given that the first three lines do not intersect at a point, the question of checking the intersection with the fourth line becomes irrelevant. Therefore, the four lines are not concurrent.
The final step is to summarize our conclusion based on the absence of a solution from the intersection of the first three lines. The result confirms that the four lines do not converge at a single point.
Result: The lines given by the equations [tex]\(x + 2y - 3 = 0\)[/tex], [tex]\(3x + 4y - 7 = 0\)[/tex], [tex]\(2x + 3y - 4 = 0\)[/tex], and [tex]\(4x + 5y - 6 = 0\)[/tex] are not concurrent.
