- Identify the slope of the original line \(-4x + 3y = 2\) as \(m = \frac{4}{3}\).
- For the parallel line, use the same slope \(m = \frac{4}{3}\) and the point \((-2, 4)\) to derive the equation \(y = \frac{4}{3}x + \frac{20}{3}\) or \(-4x + 3y = 20\).
- For the perpendicular line, use the negative reciprocal slope \(m = -\frac{3}{4}\) and the point \((-2, 4)\) to derive the equation \(y = -\frac{3}{4}x + \frac{5}{2}\) or \(3x + 4y = 10\).
- Final answers: \(\boxed{y = \frac{4}{3}x + \frac{20}{3}}\) and \(\boxed{y = -\frac{3}{4}x + \frac{5}{2}}\)
Explanation
- Analyzing the Given Line
The given equation of the line is \(-4x + 3y = 2\). To determine the slope of this line, we rewrite it in the slope-intercept form, \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
Starting with \(-4x + 3y = 2\):
\[3y = 4x + 2\]
Dividing by 3, we get:
\[y = \frac{4}{3}x + \frac{2}{3}\]
The slope of the given line is \(m = \frac{4}{3}\).
- Finding the Parallel Line
A line parallel to the given line must have the same slope, so \(m_{parallel} = \frac{4}{3}\). Using the point-slope form \(y - y_1 = m(x - x_1)\) with the point \((-2, 4)\):
\[y - 4 = \frac{4}{3}(x - (-2))\]
\[y - 4 = \frac{4}{3}(x + 2)\]
\[y - 4 = \frac{4}{3}x + \frac{8}{3}\]
Adding 4 (which is \(\frac{12}{3}\)) to both sides:
\[y = \frac{4}{3}x + \frac{8}{3} + \frac{12}{3}\]
\[y = \frac{4}{3}x + \frac{20}{3}\]
To express this in standard form (Ax + By = C), we multiply by 3:
\[3y = 4x + 20\]
\[-4x + 3y = 20\]
- Finding the Perpendicular Line
A line perpendicular to the given line must have a slope that is the negative reciprocal of the original slope. Since the original slope is \(m = \frac{4}{3}\), the perpendicular slope is:
\[m_{perp} = -\frac{1}{4/3} = -\frac{3}{4}\]
Using the point-slope form \(y - y_1 = m(x - x_1)\) with the point \((-2, 4)\):
\[y - 4 = -\frac{3}{4}(x - (-2))\]
\[y - 4 = -\frac{3}{4}(x + 2)\]
\[y - 4 = -\frac{3}{4}x - \frac{6}{4}\]
Adding 4 (which is \(\frac{16}{4}\)) to both sides:
\[y = -\frac{3}{4}x - \frac{6}{4} + \frac{16}{4}\]
\[y = -\frac{3}{4}x + \frac{10}{4} = -\frac{3}{4}x + \frac{5}{2}\]
To express this in standard form, multiply by 4:
\[4y = -3x + 10\]
\[3x + 4y = 10\]
Examples
Parallel and perpendicular lines are fundamental in fields like urban planning and architecture. For example, when laying out city streets, planners often use parallel lines to create consistent traffic flow, while perpendicular streets help in creating efficient intersection designs. Similarly, in construction, structural beams are often designed to be parallel to distribute weight evenly or perpendicular to provide maximum bracing and support for a structure. Understanding these relationships allows professionals to build safe, functional, and organized spaces.