Algebra / Graphs and Functions / Equations of Parallel and Perpendicular Lines
We have the line:
[tex]y=7x-1.[/tex]We must find the equation:
0. of the perpendicular line,
,1. and the parallel line,
to the given line that passes through the point (-2, 3).
1) Perpendicular line
The equation of the perpendicular line has the form:
[tex]y=m_p\cdot(x-x_0)+y_0.[/tex]Where mₚ is the slope, and (x₀, y₀) = (-2, 3).
From the equation of the given line, we see that its slope is m = 7. The slope of the perpendicular line mₚ is given by the equation:
[tex]\begin{gathered} m\cdot m_p=-1, \\ 7\cdot m_p=-1, \\ m_p=-\frac{1}{7}. \end{gathered}[/tex]Replacing mₚ = -1/7 and (x₀, y₀) = (-2, 3) in the equation of the perpendicular line, we get:
[tex]y=-\frac{1}{7}\cdot(x-(-2))+3=-\frac{1}{7}\cdot(x+2)+3=-\frac{1}{7}\cdot x-\frac{2}{7}+3=-\frac{1}{7}\cdot x+\frac{19}{7}.[/tex]2) Parallel line
The equation of the perpendicular line has the form:
[tex]y=m_p\cdot(x-x_0)+y_0.[/tex]Where mₚ is the slope, and (x₀, y₀) = (-2, 3).
From the equation of the given line, we see that its slope is m = 7. The parallel line has the same slope as the given line, so we have:
[tex]\begin{gathered} m_p=m, \\ m_p=7. \end{gathered}[/tex]Replacing mₚ = 7 and (x₀, y₀) = (-2, 3) in the equation of the parallel line, we get:
[tex]y=7\cdot(x-(-2))+3=7\cdot(x+2)+3=7x+14+3=7x+17.[/tex]3) Graph
Plotting the equations obtained, we get the following graph:
Answer1) Equation of the perpendicular line:
[tex]y=-\frac{x}{7}+\frac{19}{7}[/tex]2) Equation of the parallel line:
[tex]y=7x+17[/tex]
