The diagonal passing through the origin of a quadrilateral formed by \[x=0\], \[y=0\], \[x+y=1\], and \[6x+y=3\] is
A. \[3x-2y=0\]
B. \[2x-3y=0\]
C. \[3x+2y=0\]
D. None of these

Hint: In this question, we are to find the equation of a diagonal passing through the origin of a quadrilateral formed by the given lines. To find this, the vertices of the quadrilateral are calculated by the given lines, since we know that the point of intersection between two lines results in a vertex. Then, the required diagonal is obtained by the standard form of the equation of a line with two points.

 Formula Used: The equation of the line, that is passing through $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$
Where $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$ is said to be the slope of the line.

Complete step by step solution: It is given that, a quadrilateral is formed by the lines,
\[\begin{align}
  & x=0 \\
 & y=0 \\
 & x+y=1 \\
 & 6x+y=3 \\
\end{align}\]
Finding the vertices by calculating the intersection of these lines as below:
First vertex from \[x=0\] and \[y=0\] is $A(0,0)$
Second vertex from \[y=0\] and \[x+y=1\] is
\[\begin{align}
  & y=0 \\
 & x+y=1 \\
 & \Rightarrow x+0=1 \\
 & \Rightarrow x=1 \\
\end{align}\]
Thus, the second vertex is $B(0,1)$
Third vertex from \[x+y=1\] and \[6x+y=3\] is
\[\begin{align}
  & x+y=1 \\
 & \Rightarrow x=1-y \\
\end{align}\]
\[\begin{align}
  & 6x+y=3 \\
 & \Rightarrow 6(1-y)+y=3 \\
 & \Rightarrow 6-6y+y=3 \\
 & \Rightarrow -5y=-3 \\
 & \Rightarrow y=\dfrac{3}{5} \\
\end{align}\]
Then,
$\begin{align}
  & x=1-y \\
 & \Rightarrow x=1-\dfrac{3}{5} \\
 & \Rightarrow x=\dfrac{2}{5} \\
\end{align}$
Thus, the third vertex is $C(\dfrac{2}{5},\dfrac{3}{5})$
Fourth vertex from \[6x+y=3\] and \[x=0\]is
\[\begin{align}
  & x=0 \\
 & 6x+y=3 \\
 & \Rightarrow 6(0)+y=3 \\
 & \Rightarrow y=3 \\
\end{align}\]
Thus, the fourth vertex is $D(0,3)$
The diagonal passing through the origin in the given quadrilateral is $\overleftrightarrow{AC}$.
Then, the equation of the diagonal $\overleftrightarrow{AC}$ is
$A(0,0)=({{x}_{1}},{{y}_{1}})$ and $C(\dfrac{2}{5},\dfrac{3}{5})=({{x}_{2}},{{y}_{2}})$
$\begin{align}
  & y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}}) \\
 & y-0=\dfrac{\dfrac{3}{5}-0}{\dfrac{2}{5}-0}(x-0) \\
 & \Rightarrow y=\dfrac{3}{2}x \\
 & \Rightarrow 2y=3x \\
 & \, therefore, 3x-2y=0 \\
\end{align}$

Option ‘A’ is correct

Note: Here we may confuse with the intersection of lines. In order to get the vertex, we can also use a graph by plotting the given lines and forming a quadrilateral. So, easily we can recognize the vertices. After that, the diagonal is obtained and its equation is framed.