Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 23-28, find the slope of the line containing the given two points. \((2,3)\) and \((-5,-6)\)

Step-by-Step Solution

Answer

The slope of the line is \( \frac{9}{7} \).

Step 1: Identify the formula for slope

The slope of a line through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) can be calculated using the formula \[ m = \frac{y_2 - y_1}{x_2 - x_1}. \] This formula gives the ratio of the change in y to the change in x.

Step 2: Assign the points to variables

Assign the coordinates of the points to variables: Let \( (x_1, y_1) = (2, 3) \) and \( (x_2, y_2) = (-5, -6). \) This will help in substituting values into the slope formula.

Step 3: Substitute the values into the slope formula

Substitute the given points into the formula: \[ m = \frac{-6 - 3}{-5 - 2} = \frac{-9}{-7}. \] This results from subtracting the y-coordinates and x-coordinates respectively.

Step 4: Simplify the expression

Simplify the fraction \( \frac{-9}{-7} \) to get the slope. Since there are two negative signs in the fraction, they cancel each other out: \[ m = \frac{9}{7}. \]

Key Concepts

Slope calculation

Calculating the slope is essential in understanding how a line behaves. The slope is a measure of the line's steepness and direction. You can think of it as the "rise over run," where we compare how much the line rises vertically with how much it runs horizontally. To find the slope between two points, you use the slope formula:

• This formula is given by \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
• Here, \((x_1, y_1)\) and \((x_2, y_2)\) are your two points on the line.

Begin by identifying values from your points for substitution into the formula. Subtract the first point's \( y \) value, \( y_1 \), from the second point's \( y \) value, \( y_2 \), to find how much the line rises. Do the same for the \( x \) values to find how much the line runs horizontally. In our example, the points \((2, 3)\) and \((-5, -6)\) give a rise of \((-6) - 3 = -9\) and a run of \((-5) - 2 = -7\). Simplifying the slope fraction \( \frac{-9}{-7} \), we find the slope is \( \frac{9}{7} \). This indicates a positive slope, so the line ascends from left to right.

Coordinate geometry

Coordinate geometry, or analytic geometry, blends algebra and geometry to study geometric shapes using a coordinate plane. The coordinate plane consists of horizontal and vertical axes:

• The horizontal axis is called the x-axis.
• The vertical axis is called the y-axis.
• Each point on this plane is represented by a pair of coordinates \((x, y)\).

When plotting points, you move from the origin, \((0,0)\), sideways for the x value and vertically for the y value. Understanding how these points are connected through lines or curves forms the basis for many problems in calculus and pre-calculus. In our problem, we use points \((2, 3)\) and \((-5, -6)\) on a coordinate plane. These points describe a line, and by calculating their slope, we can ascertain how this line is oriented within the plane. Coordinate geometry not only helps visualize these points but also allows calculating distances and slopes, solving systems of equations, and analyzing shapes.

Linear equations

Linear equations form the core of algebra and calculus, representing lines in a formulaic way. A linear equation can often be written in slope-intercept form:

• This is expressed as \( y = mx + b \).
• Here, \( m \) is the slope and \( b \) represents the y-intercept, where the line crosses the y-axis.

When you have a slope and a point, you can write the equation of a line. Let's consider our points again, \((2, 3)\) and \((-5, -6)\). Using the slope \( m = \frac{9}{7} \) calculated in previous sections, you can insert one of these points into the slope-intercept equation to solve for \( b \). After calculations, you find that the line's equation is in the form \( y = \frac{9}{7}x + b \), where \( b \) is obtained by substitution. Linear equations allow us to describe relationships between variables graphically and algebraically, providing solutions to several kinds of problems in mathematics.