Answer:
The equation parallel to the given equation and passing through the point (8, 3) is:
[tex]y\text{ = }\frac{5}{2}x\text{ - 17}[/tex]
The equation perpendicular to the given equation and passing through the point (8, 3) is:
[tex]y\text{ = }\frac{-2}{5}x\text{ + }\frac{31}{5}[/tex]Explanations:
The equation of the line parallel to the line y = mx + c and passing through the point (x₁, y₁) is given as:
[tex]y-y_1=m(x-x_1)[/tex]
The equation of the line perpendicular to the line y = mx + c and passing through the point (x₁, y₁) is given as:
[tex]y-y_1\text{ = }\frac{-1}{m}(x-x_1)[/tex]
Now, for the equation:
[tex]\begin{gathered} y\text{ = }\frac{5}{2}x\text{ - 7} \\ m\text{ = }\frac{5}{2} \end{gathered}[/tex]
The line parallel to the equation and passing through the point (8, 3) will be:
[tex]\begin{gathered} y\text{ - 3 = }\frac{5}{2}(x\text{ - 8)} \\ y\text{ - 3 = }\frac{5}{2}x\text{ - 20} \\ y\text{ = }\frac{5}{2}x\text{ - 20 + 3} \\ y\text{ = }\frac{5}{2}x\text{ - 17} \end{gathered}[/tex]
The line perpendicular to the given equation and passing through the point (8, 3) will be:
[tex]\begin{gathered} y\text{ - 3 = }\frac{-2}{5}(x\text{ - 8)} \\ y\text{ - 3 = }\frac{-2}{5}x\text{ + }\frac{16}{5} \\ y\text{ = }\frac{-2}{5}x\text{ + }\frac{16}{5}+3 \\ y\text{ = }\frac{-2}{5}x\text{ + }\frac{31}{5} \end{gathered}[/tex]
