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Chapter 1: Problem 11

Determine the equation of the line that satisfies the stated requirements. Put the equation in standard form. The line passing through \((-2,-3)\) and \((1,4)\)

Short Answer

Expert verified
The equation of the line in standard form is \(7x - 3y = -5\).

Step by step solution

01

Find the Slope

First, calculate the slope of the line that passes through the points \((-2,-3)\) and \((1,4)\) using the slope formula:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]where \((x_1, y_1) = (-2, -3)\) and \((x_2, y_2) = (1, 4)\).Substitute these values into the formula:\[ m = \frac{4 - (-3)}{1 - (-2)} = \frac{4 + 3}{1 + 2} = \frac{7}{3} \]The slope (m) of the line is \(\frac{7}{3}\).
02

Use Point-Slope Form

Next, use the point-slope form to find the equation of the line:\[ y - y_1 = m(x - x_1) \]Using the slope \(m = \frac{7}{3}\) and one of the points, say \((-2,-3)\), substitute into the formula:\[ y - (-3) = \frac{7}{3}(x - (-2)) \]This simplifies to:\[ y + 3 = \frac{7}{3}(x + 2) \]
03

Simplify to Slope-Intercept Form

Distribute the slope into the equation:\[ y + 3 = \frac{7}{3}x + \frac{7}{3} \times 2 \]\[ y + 3 = \frac{7}{3}x + \frac{14}{3} \]Subtract 3 from both sides to isolate \(y\):\[ y = \frac{7}{3}x + \frac{14}{3} - 3 \]Convert \(-3\) into a fraction with a denominator of 3:\[ y = \frac{7}{3}x + \frac{14}{3} - \frac{9}{3} \]\[ y = \frac{7}{3}x + \frac{5}{3} \]
04

Convert to Standard Form

To convert to standard form \(Ax + By = C\), eliminate the fractions by multiplying the entire equation by 3:\[ 3y = 7x + 5 \]This becomes:\[ 7x - 3y = -5 \]Rearrange to standard form:\[ 7x - 3y = -5 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slope formula
The slope of a line measures its steepness. It's a vital component in understanding how lines behave, particularly when graphing them. To find the slope between two points, use the slope formula:
  • The formula is given by \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] where \( (x_1, y_1) \) and \( (x_2, y_2) \) are two distinct points on the line.
In our exercise, we calculated the slope between the points \((-2,-3)\) and \((1,4)\). Insert these points into the formula to find the slope which is \(\frac{7}{3}\). This positive value indicates the line rises as it moves from left to right. Understanding slope is crucial for graphing and determining the angles at which lines intersect.
Point-slope form
The point-slope form is a handy tool for writing the equation of a line when you have a point on the line and the slope. It is expressed as:
  • \[ y - y_1 = m(x - x_1) \]
This formula allows you to construct the equation of a line using a known point, \((x_1, y_1)\), and the slope \(m\). In our example, we selected the point \((-2, -3)\) and the previously calculated slope \(\frac{7}{3}\). Substituting these values into the formula gives us \[ y + 3 = \frac{7}{3}(x + 2) \], which guides us in forming the line's equation. This form is particularly useful for quickly transitioning a problem into an equation that can be worked with algebraically.
Standard form equation
The standard form of a linear equation is \(Ax + By = C\), where \(A\), \(B\), and \(C\) are integers, and \(A\) is non-negative. This form is beneficial for identifying intercepts and analyzing solutions in systems of equations. After acquiring the point-slope form, we simplify and rearrange the equation:
  • Initially, we had the form \(y = \frac{7}{3}x + \frac{5}{3}\).
  • To eliminate fractions and convert to standard form, multiply the entire equation by 3 to get \(3y = 7x + 5\).
  • This simplifies to \(7x - 3y = -5\), our final standard form representation.
While point-slope and slope-intercept forms are advantageous for specific contexts, the standard form is preferred for its universality and ease of use in various applications, including solving simultaneous equations.
Linear equations
Linear equations form the backbone of algebra. They represent straight lines on a graph and are typically characterized by constants and a degree of one. Here's what makes them indispensable:
  • They appear in the format \(y = mx + b\) or \(Ax + By = C\), depicting relationships between variables.
  • These equations help in modeling real-world scenarios such as motion, finance, and relationships between entities.
  • Being able to convert between forms, like slope-intercept to standard, provides flexibility and tools to handle different kinds of problems.
Grasping linear equations extends beyond just forming and solving them. It involves understanding their behavior on a graph, such as how the slope indicates direction, leading to deeper insights into variables' interactions. As you master these basics, tackling more complex algebraic and calculus problems becomes far more approachable.

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Most popular questions from this chapter

sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=-2 x^{2}-3 $$

Explain how the following functions can be obtained from \(y=e^{x}\) by basic transformations: (a) \(y=e^{x}+3\) (b) \(y=e^{-x}\) (c) \(y=2 e^{x-2}+3\)

Logistic Transformation Suppose that $$ f(x)=\frac{1}{1+e^{-(b+m x)}} $$ where \(b\) and \(m\) are constants. A function of the form (1.15) is called a logistic function. The logistic function was introduced by the Dutch mathematical biologist Verhulst around 1840 to describe the growth of populations with limited food resources. (a) Show that $$ \ln \frac{f(x)}{1-f(x)}=b+m x $$ This transformation is called the logistic transformation. (b) Given some data, a table of values of \(x\), and the function \(f(x)\) at each value \(x\), explain how to plot the data to produce a straight line, and to estimate the constant \(b\) and \(m\) from the model.

When \(\log y\) is graphed as a function of \(x\), a straight line results. Graph straight lines, each given by two points, on a log-linear plot, and determine the functional rela\mathrm{\\{} t i o n s h i p . ~ ( T h e ~ o r i g i n a l ~ \(x-y\) coordinates are given.) $$ \left(x_{1}, y_{1}\right)=(1,4),\left(x_{2}, y_{2}\right)=(4,1) $$

Find the amplitude and the period of \(f(x)\) $$ f(x)=-\frac{3}{2} \sin \left(\frac{\pi}{3} x\right), \quad x \in \mathbf{R} $$
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