The equation of a line passing through a point X having its coordinates to be (x₁, y₂) is given as
![\begin{gathered} y-y_1=m(x-x_1) \\ \text{where} \\ m\text{ is the slope of the line} \\ (x_(1,)y_1)\text{ is the coordinate of the point X} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/sd0i2zutyfust7u4ab36611y38ypovfwns.png)
Taking a nonzero number for the slope m and a point in the plane,
![\begin{gathered} \text{let} \\ m=5 \\ (x_(1,)y_1)\text{ = (3,2)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nxsvwvmut61zemvyeuh3rawj2sdeamyb9u.png)
Thus, the equation of the line becomes
![\begin{gathered} y-2=5(x-3) \\ \text{open the brackets} \\ y-2=5x-15 \\ make\text{ y the subject of the formula} \\ y=5x-15+2 \\ \Rightarrow y=5x-13 \end{gathered}]()
Hence, the equation of the line with the chosen slope (5) passing through the chosen point (3,2) is evaluated to be
![y=5x-13](https://img.qammunity.org/2023/formulas/mathematics/college/t2p7cwy9qozz2nbs98qxxphxnxlp1l2a1n.png)