The equation of a line passing through a point X having its coordinates to be (x₁, y₂) is given as
[tex]\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}[/tex]Taking a nonzero number for the slope m and a point in the plane,
[tex]\begin{gathered} \text{let} \\ m=5 \\ (x_{1,}y_1)\text{ = (3,2)} \end{gathered}[/tex]Thus, the equation of the line becomes
[tex]\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}[/tex]Hence, the equation of the line with the chosen slope (5) passing through the chosen point (3,2) is evaluated to be
[tex]y=5x-13[/tex]
