hello
the equation of a straight line is given as
![\begin{gathered} y=mx+b \\ b=\text{intercept} \\ m=\text{slope} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d9drfj60y2tlp28d0n8b78j0sm1u86wzz7.png)
![\text{slope}=(y_2-y_1)/(x_2-x_1)](https://img.qammunity.org/2023/formulas/mathematics/college/x8v52dcr18u40wxqmdrkiv958nucq58v09.png)
from the graph given the values of the co-ordinates are given as
![\begin{gathered} y_2=280 \\ x_2=20 \\ y_1=120 \\ x_1=0 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6djk1mrllu3t46v1z0kqpfl5dne0xyzw2b.png)
now let's substitute the values into the equation and solve
![\begin{gathered} \text{slope}=(y_2-y_1)/(x_2-x_1) \\ \text{slope}=(280-120)/(20-0) \\ \text{slope(m)}=(160)/(20) \\ \text{slope(m)}=8 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/g54y2cgrxr9334f390myd0xohex381qz2m.png)
from the calculations above, the slope of the line is equal to 8 and also the y-intercept is equal to 120
the equation of the line =
![\begin{gathered} y=mx+b \\ m=8 \\ b=120 \\ y=8x+120 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ba7otcn9bwtcrqgjjbl4qsz5y93xdpm5kd.png)
the equation of the line is y = 8x + 120 and this correspo