The equation of a line in point-slope form is:
![y=mx+b](https://img.qammunity.org/2023/formulas/mathematics/high-school/smsb8cbft03lwblmi49nf2l6jby2ofxzws.png)
Any point that belongs to this line is a solution to this equation. Looking at the graph we can see two points that belong to the line:
![(0,1)\text{ and }(5,3)](https://img.qammunity.org/2023/formulas/mathematics/college/opzio9b0laaljlz538vv5osp8emvnbfo4u.png)
By replacing x and y in the equation of the line with the values given by the points we can build two equations:
![\begin{gathered} y=mx+b \\ \text{From point (x,y)=(0,1) we get:} \\ 1=m\cdot0+b \\ 1=b \\ \text{From point (x,y)=(5,3) we get:} \\ 3=m\cdot5+b \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/rkkdvjjucaddnelj15sfywktljln3pmk3e.png)
So we have these two equations:
![\begin{gathered} 1=b \\ 3=5m+b \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/xscwx796gz35fdvmi6hq1aepijldi4ar0i.png)
So according to the first equation b=1. If we replace this value in the second equation we can also find m:
![\begin{gathered} 3=5m+b \\ 3=5m+1 \\ 3-1=5m \\ 2=5m \\ m=(2)/(5) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/iwult2hsclcejm6voixdag0zgki6oz0zqb.png)
Then the equation of the line that we are looking for and the answer to this question is:
![y=(2)/(5)x+1](https://img.qammunity.org/2023/formulas/mathematics/college/ce8tvypsk1zwrwgxmpwtds7uy6fruv12l4.png)