So we have to find the equation of a line that passes through (2,4) and has a slope of 1/2 in standard form. First, we should write it in point-slope form because it will be easier to find it. Then we can change it into the standard form. The point-slope form of a line with a slope m that passes through point (a,b) is:
![y-b=m(x-a)](https://img.qammunity.org/2023/formulas/mathematics/college/6ptzkyv3jnl37vfv4lfbpm0m05p41b0vtk.png)
In this case we are told that the slope is 1/2 and the point is (2,4) so we get:
![y-4=(1)/(2)(x-2)](https://img.qammunity.org/2023/formulas/mathematics/college/hlvh0booy1p20lchvqytjh5d58lymbjaxk.png)
Now that we have the equation in point-slope form we must convert it into the standard form. This form looks like the following:
![Ax+By=C](https://img.qammunity.org/2023/formulas/mathematics/high-school/75j0pzqy8f6gtpgzjampxw030qc85p0hp7.png)
Where A, B and C are numbers. So let's take the equation we found before and distribute the product in the right side:
![\begin{gathered} y-4=(1)/(2)(x-2) \\ y-4=(x)/(2)-(2)/(2) \\ y-4=(x)/(2)-1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8y2dg8simu7eljp2mpnj3tlztjvf0ztnac.png)
Now let's add 4 to both sides:
![\begin{gathered} y-4=(x)/(2)-1 \\ y-4+4=(x)/(2)-1+4 \\ y=(x)/(2)+3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/bki6094fzfinf77idx3iz1cact3ocsujw3.png)
We substract x/2 from both sides:
![\begin{gathered} y=(x)/(2)+3 \\ y-(x)/(2)=(x)/(2)+3-(x)/(2) \\ -(x)/(2)+y=3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/th35xvnht51p3y3ai0af8dppnh15jz4sy8.png)
And finally we multiply both sides by 2:
![\begin{gathered} 2\cdot(-(x)/(2)+y)=2\cdot3 \\ -x+2y=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yodvk3v415kvoiy80l3nqd7j6pscx6x70n.png)
Then the answer is option b.