Problem 18 - Calculus

Question

Use the facts that parallel lines have equal slopes and that the slopes of perpendicular lines are negative reciprocals of one another. Find equations for the lines through the point (1,5) that are parallel to and perpendicular to the line with equation \(y+4 x=7\)

Step-by-Step Solution

Answer

Parallel line: \(y = -4x + 9\). Perpendicular line: \(y = \frac{1}{4}x + \frac{19}{4}\).

Step 1: Convert Given Equation to Slope-Intercept Form

The given equation is \(y + 4x = 7\). We first solve for \(y\) to convert it into the slope-intercept form \(y = mx + b\), where \(m\) is the slope. Subtract \(4x\) from both sides to get \(y = -4x + 7\). The slope \(m\) of the given line is \(-4\).

Step 2: Find the Slope of the Line Parallel to the Given Line

The slope of a line parallel to another line is the same as the slope of that line. Therefore, the slope of the line parallel to \(y = -4x + 7\) is also \(-4\).

Step 3: Use Point-Slope Form to Write the Parallel Line's Equation

The point-slope form of a line is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope. Substitute \(m = -4\) and \((x_1, y_1) = (1,5)\) to get \(y - 5 = -4(x - 1)\). Simplifying, we find \(y = -4x + 9\).

Step 4: Find Slope of the Perpendicular Line

The slope of a line perpendicular to another line is the negative reciprocal of the other line's slope. The original slope is \(-4\), so its negative reciprocal is \(\frac{1}{4}\).

Step 5: Use Point-Slope Form to Write the Perpendicular Line's Equation

Using the point-slope form again for the perpendicular line with slope \(\frac{1}{4}\) and point \((1,5)\), we have \(y - 5 = \frac{1}{4}(x - 1)\). Simplifying this, the equation becomes \(y = \frac{1}{4}x + \frac{19}{4}\).

Key Concepts

Slope-Intercept Form

The slope-intercept form of a linear equation is a fundamental aspect of analytic geometry. This form tells us how a line behaves on a graph and can be written as \(y = mx + b\). Here, \(m\) represents the slope of the line, which indicates how steep the line is. Meanwhile, \(b\) is the y-intercept, which is the point where the line crosses the y-axis.
For instance, when converting the given line equation \(y + 4x = 7\) into slope-intercept form, we rearrange it to \(y = -4x + 7\). Now, it's clear that the slope \(m\) is \(-4\), and the line crosses the y-axis at the point \((0, 7)\).

• Slope \(m\): measures the line's steepness (rise over run).
• Y-intercept \(b\): the point where the line hits the y-axis.

Understanding this form is key to analyzing lines and predicting their behavior graphically.

Parallel Lines

Parallel lines are an essential concept in geometry. They never intersect and have the same slope. This means that if two lines are parallel, they have identical directional growth.
For the problem at hand, the given line has a slope of \(-4\) (from its slope-intercept form \(y = -4x + 7\)). Therefore, any line parallel to it must also have a slope of \(-4\).
Creating an equation for a line parallel through a specific point, such as \((1,5)\), involves applying the same slope. Using the point-slope form (explained later), we find the equation of the parallel line: \(y = -4x + 9\). This line shares the same slope with the original line, confirming their parallel nature.

Perpendicular Lines

Perpendicular lines intersect at a right angle (90 degrees), which means their slopes are negative reciprocals. So, if one line has a slope \(a\), the other must have a slope \(-1/a\).
In our exercise, the slope of the original line is \(-4\). Thus, the slope of a perpendicular line must be the negative reciprocal, which is \(\frac{1}{4}\).

• Negative reciprocal: Invert the original slope and change the sign.
• Intersection at right angles: Creates a clear distinction from parallel lines.

Using the point \((1,5)\) and this new perpendicular slope, the point-slope form equation becomes \(y - 5 = \frac{1}{4}(x - 1)\). When simplified, this yields the equation \(y = \frac{1}{4}x + \frac{19}{4}\), illustrating the perpendicular relationship mathematically.

Point-Slope Form

The point-slope form is a useful tool in deriving the equation of a line. It is structured as \(y-y_1 = m(x-x_1)\), where \((x_1, y_1)\) is a known point on the line and \(m\) is the slope.
This form is particularly handy when you have a point and a slope and need to find the full equation of a line passing through the point. It allows you to structure an equation quickly without needing the y-intercept first.

• Flexibility: Easily adapt and find line equations with given mathematical information.
• Simplicity: Less rearrangement required compared to other forms.

In our example, we used point-slope form to find the equations for both parallel and perpendicular lines through the point \((1,5)\). For the parallel line, we substituted \(m = -4\) to get \(y - 5 = -4(x - 1)\). For the perpendicular line, \(m = \frac{1}{4}\) resulted in \(y - 5 = \frac{1}{4}(x - 1)\). This demonstrates the practicality and effectiveness of point-slope form.