Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

Write the equation for the line through \((-2,-1)\) that is perpendicular to the line \(y+3=-\frac{2}{3}(x-5)\).

Step-by-Step Solution

Answer

The equation of the line is \(y = \frac{3}{2}x + 2\).

Step 1: Identify the Slope of the Given Line

The given line equation is in the form \(y + 3 = -\frac{2}{3}(x - 5)\). Rearrange it to slope-intercept form \(y = mx + b\) by solving for \(y\). The equation becomes \(y = -\frac{2}{3}x + \frac{10}{3} - 3\), which simplifies to \(y = -\frac{2}{3}x + \frac{1}{3}\). The slope \(m\) of the given line is \(-\frac{2}{3}\).

Step 2: Determine the Slope of the Perpendicular Line

Two lines are perpendicular if the product of their slopes is \(-1\). If the slope of the first line is \(-\frac{2}{3}\), set up the equation \(-\frac{2}{3} \times m_\text{perpendicular} = -1\). Solve for \(m_\text{perpendicular}\): \(m_\text{perpendicular} = \frac{3}{2}\).

Step 3: Use Point-Slope Form to Write the Equation

Now that we have a slope \(m = \frac{3}{2}\) and a point \((-2, -1)\), use the point-slope form of the equation: \(y - y_1 = m(x - x_1)\). Substitute the point and slope: \(y - (-1) = \frac{3}{2}(x + 2)\). This becomes \(y + 1 = \frac{3}{2}x + 3\).

Step 4: Convert to Slope-Intercept Form

To make it easier to interpret, convert the equation to slope-intercept form \(y = mx + b\). Subtract \(1\) from both sides: \(y = \frac{3}{2}x + 3 - 1\). Simplify to get \(y = \frac{3}{2}x + 2\).

Key Concepts

Slope-Intercept Form

The slope-intercept form of a line is a handy way to express the equation of a straight line. It is written as \( y = mx + b \). Here, \( m \) represents the slope of the line, while \( b \) is the y-intercept of the line. The y-intercept is the point where the line crosses the y-axis.

Using slope-intercept form is great because it gives you a quick snapshot of the line's behavior. The slope \( m \) tells us how steep the line is. A positive slope means the line rises as it moves from left to right, while a negative slope indicates it falls. The y-intercept \( b \) offers another crucial piece by showing where the line hits the y-axis.

For example, in the step-by-step solution of our exercise, we turned the provided equation into slope-intercept form to find the slope. This is important for determining the slope of the perpendicular line, a crucial step in writing a new line's equation.

Equation of a Line

An equation of a line is a mathematical description of a straight line on a graph. It tells us everything about the line, like its slope and position. There are different ways to write the equation of a line, such as the slope-intercept form and the point-slope form.

The importance of a line equation lies in its ability to pinpoint how a line behaves on a coordinate plane. It allows us to understand the line's direction and where it lies relative to the x and y axes. For solving geometric problems, this becomes invaluable.

In our exercise, finding the new line's equation was crucial because it indicated how a perpendicular line behaves compared to the original line. Understanding the equation enables you to draw or analyze the line without needing to directly visualize it.

Point-Slope Form

The point-slope form is another way to write the equation of a line, especially useful when you know a point on the line and its slope. The formula is \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the known point, and \( m \) is the slope of the line.

This form is particularly helpful in problems like the exercise where you need to find the equation of a line passing through a certain point with a given slope. Once you have a slope and a specific point, the rest is just plugging in values!

In the exercise, we started with a known point and a newly calculated slope for the perpendicular line. Using the point-slope form made it easy to establish the equation of the new line. Ultimately, this form provided a simple method to navigate from known slopes and points to the line's full equation.