Given the four lines with equations $x + 2y = 3$, $3x + 4y = 7$, $2x + 3y = 4$, and $4x + 5y = 6$. Then find the relation between the lines.
A. Concurrent
B. Perpendicular
C. The sides of a rectangle
D. None of these

Hint: First find the intersection points of the given equations and substitute the point in the other equations. From this, we will check whether the lines are concurrent or perpendicular or the sides of a rectangle.

Complete step by step solution:
Given equations of lines are
$x + 2y = 3$ ………(i)
$3x + 4y = 7$ ………(ii)
$2x + 3y = 4$ ………(iii)
$4x + 5y = 6$ ………(iv)
Multiply 2 with equation (i) and subtract it from equation (ii)
 $3x + 4y = 7$
$2x + 4y = 6$
$( - )$$( - )$$( - )$
$\overline {x\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1} $
Putting $x = 1$ in equation (i)
$1 + 2y = 3$
$ \Rightarrow 2y = 2$
$ \Rightarrow y = 1$
Putting $x = 1$ and $y = 1$ in equation $2x + 3y = 4$
$2 \cdot 1 + 3 \cdot 1 = 4$
$ \Rightarrow 5 \ne 4$
$\left( {1,1} \right)$ does not satisfy $2x + 3y = 4$.
Putting $x = 1$ and $y = 1$ in equation $4x + 5y = 6$
$4 \cdot 1 + 5 \cdot 1 = 6$
$ \Rightarrow 9 \ne 4$
$\left( {1,1} \right)$ does not satisfy $4x + 5y = 6$.
Multiply 2 with equation (iii) and subtract it from equation (iv)
$4x + 5y = 6$
$4x + 6y = 8$
$( - )$$( - )$$( - )$
$\overline {\,\,\,\,\,\, - y\,\, = - 2} $
$ \Rightarrow y = 2$
Putting $y = 2$ in the equation (iv)
$4x + 5 \cdot 2 = 6$
$ \Rightarrow 4x + 10 = 6$
$ \Rightarrow 4x = - 4$
$ \Rightarrow x = - 1$
The intersection point of $2x + 3y = 4$ and $4x + 5y = 6$ is $\left( { - 1,2} \right)$.
Putting $x = - 1$ and $y = 2$ in the equation $x + 2y = 3$
$ - 1 + 2 \cdot 2 = 3$
$3 = 3$
So, the equation $x + 2y = 3$ satisfies the point $\left( { - 1,2} \right)$.
Putting $x = - 1$ and $y = 2$ in the equation $3x + 4y = 7$
$3 \cdot \left( { - 1} \right) + 4 \cdot 2 = 7$
$ \Rightarrow 5 = 7$
So, the equation $3x + 4y = 7$ does not satisfy the point $\left( { - 1,2} \right)$.
Therefore $x + 2y = 3$, $2x + 3y = 4$ and $4x + 5y = 6$ pass through the point $\left( { - 1,2} \right)$.
Thus $x + 2y = 3$, $2x + 3y = 4$ and $4x + 5y = 6$ are concurrent.
Three lines are concurrent. So, no option is correct.

Option ‘D’ is correct

Note: Often tend to calculate the intersection point of $x + 2y = 3$ and $3x + 4y = 7$. But the intersection point does not satisfy the other two lines. They conclude that the equations are not concurrent. But it is an incomplete solution. The correct answer is there are three lines which are concurrent.