Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find an equation of the line perpendicular to \(y=-\frac{2}{3} x+7\) that contains the point (-2,-5).

Short answer

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

Step-by-step solution

Identify the slope of the given line

The equation of the given line is in the slope-intercept form: \[ y = mx + b \] Where \(m\) is the slope. In the equation \(y = -\frac{2}{3}x + 7\), the slope \(m\) is \(-\frac{2}{3}\).

Find the slope of the perpendicular line

The slope of a line perpendicular to another is the negative reciprocal of the original slope. The negative reciprocal of \(-\frac{2}{3}\) is \(\frac{3}{2}\). Therefore, the slope of the perpendicular line is \(\frac{3}{2}\).

Use point-slope form to find the equation

The point-slope form of a line's equation is: \[ y - y_1 = m(x - x_1) \]Where \(m\) is the slope, and \((x_1, y_1)\) is a point on the line. Here, \(m = \frac{3}{2}\) and \((x_1, y_1) = (-2, -5)\). Substituting the values into the point-slope form: \[ y - (-5) = \frac{3}{2}(x - (-2)) \]Which simplifies to: \[ y + 5 = \frac{3}{2}(x + 2) \]

Simplify to find the equation in slope-intercept form

Distribute \(\frac{3}{2}\) on the right side: \[ y + 5 = \frac{3}{2}x + \frac{3}{2} \times 2 \] \[ y + 5 = \frac{3}{2}x + 3 \]Then, subtract 5 from both sides to solve for \(y\): \[ y = \frac{3}{2}x + 3 - 5 \] \[ y = \frac{3}{2}x - 2 \]

Key concepts

slope-intercept form

Understanding the slope-intercept form is vital. It is written as \( y = mx + b \). Here, \( m \) denotes the slope of the line, while \( b \) stands for the y-intercept. The slope-intercept form is widely used because it makes graphing easier. For example, in the equation \( y = -\frac{2}{3}x + 7 \), the slope (\( m \)) is \( -\frac{2}{3} \) and the y-intercept (\( b \)) is 7. This form allows us to immediately understand how steep the line is and where it crosses the y-axis.

point-slope form

The point-slope form, given by \( y - y_1 = m(x - x_1) \), is incredibly useful when you know one point on the line and the slope. It focuses on the specific point \( (x_1, y_1) \) and adjusts the line accordingly. For instance, if a line passes through \((-2, -5)\) and has a slope of \( \frac{3}{2} \), you would use \( y + 5 = \frac{3}{2}(x + 2) \) to describe that line.

negative reciprocal

To find a perpendicular line, you need to use the concept of negative reciprocals. The negative reciprocal of a given slope \( m \) is \( -\frac{1}{m} \). This flips the fraction and changes its sign. For instance, the negative reciprocal of \( -\frac{2}{3} \) is \( \frac{3}{2} \). This means if you have a line with slope \( -\frac{2}{3} \), a line perpendicular to it will have a slope of \( \frac{3}{2} \).

finding equations of lines

Finding equations of lines generally involves determining the slope and y-intercept, or using a known point and slope. For instance, if asked to find a line perpendicular to \( y = -\frac{2}{3}x + 7 \) containing the point \( (-2, -5) \), you first identify the original slope as \( -\frac{2}{3} \). The perpendicular slope is \( \frac{3}{2} \). Using point-slope form makes it simpler: \( y + 5 = \frac{3}{2}(x + 2) \). Then, convert it to slope-intercept form for clarity: \( y = \frac{3}{2}x - 2 \).

linear equations

Linear equations, like \( y = mx + b \), describe straight lines on a graph. These equations are paramount in algebra because they model relationships with constant rates of change. In a linear equation, each pair of \( x \) and \( y \) values forms a straight line when plotted. For example, \( y = -\frac{2}{3}x + 7 \) is a linear equation where the slope of \( -\frac{2}{3} \) indicates a decline and 7 is where the line crosses the y-axis. Understanding linear equations helps in solving problems related to rates, motions, and many real-world scenarios.