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Chapter 10: Problem 1

To find an equation of a line, what two pieces of information are needed?

Short Answer

Expert verified
You need either the slope and y-intercept or the slope and a point on the line.

Step by step solution

01

Understanding a Line's Equation

The equation of a line in a two-dimensional plane can be expressed in the slope-intercept form, which is \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept.
02

Identifying Necessary Components

To write the equation of a line, you need two critical pieces of information. These are the slope of the line (\( m \)) and the y-intercept (\( b \)).
03

Alternative Formula

Alternatively, if you do not know the y-intercept, you can still find the equation of a line using the point-slope form, which is \( y - y_1 = m(x - x_1) \), requiring a point \((x_1, y_1)\) on the line and the slope \( m \).
04

Confirming Key Information

Summarizing, the two pieces of information required are either a) the slope and the y-intercept or b) the slope and a point on the line.

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Key Concepts

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

Understanding Slope-Intercept Form
The slope-intercept form is a commonly used way to express the equation of a line in the two-dimensional plane. This form is written as \( y = mx + b \), where:
  • \( m \) represents the slope of the line.
  • \( b \) is the y-intercept, the point where the line crosses the y-axis.
This formula is helpful for quickly identifying the slope and y-intercept from the equation. The slope \( m \) tells you how steep the line is. If the slope is positive, the line rises from left to right. If the slope is negative, the line falls. The y-intercept \( b \) gives the starting point of the line along the y-axis. Understanding these components makes it much easier to plot the line on a graph and understand its orientation and position.
Exploring Point-Slope Form
When the y-intercept is not available, the point-slope form offers a powerful alternative for writing the equation of a line. This form is expressed as \( y - y_1 = m(x - x_1) \). Here:
  • \( m \) is the slope of the line.
  • \((x_1, y_1)\) is a known point on the line.
This structure of the equation serves well when you have the slope and any point located on the line. It allows you to derive the equation by merely plugging in those values. It's especially handy in real-world scenarios where measuring or determining the exact y-intercept might be challenging. With this method, knowing any specific point and the slope leads you directly to the line's equation.
Navigating the Two-Dimensional Plane
The two-dimensional plane is a flat surface where you graphically plot equations of lines and other geometric figures. It consists of two axes:
  • The x-axis, running horizontally.
  • The y-axis, running vertically.
Together, these axes intersect at the origin, marked as \((0, 0)\). On this plane, each point is defined by a pair of numerical coordinates \((x, y)\). The structure makes it possible to visually present lines using mathematical equations like those in slope-intercept or point-slope forms. Graphing these equations effectively helps visualize their relationships and changes. Understanding how to navigate the two-dimensional plane is essential for not just plotting lines, but for a wide range of geometrical evaluations and problem-solving.

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Most popular questions from this chapter

Find the distance from the point to the line. \(Q=(0,3), \quad \vec{\ell}(t)=\langle 2,0\rangle+t\langle 1,1\rangle\)

Write the vector, parametric and symmetric equations of the lines described. Passes through the point of intersection of \(\ell_{1}(t)\) and \(\ell_{2}(t)\) and orthogonal to both lines, where \(\ell_{1}=\left\\{\begin{array}{l}x=t \\\ y=-2+2 t \\ z=1+t\end{array}\right.\) and \(\ell_{2}=\left\\{\begin{array}{l}x=2+t \\ y=2-t \\ z=3+2 t\end{array}\right.\)

Determine if the described lines are the same line, parallel lines, intersecting or skew lines. If intersecting, give the point of intersection. $$ \begin{array}{l} \overrightarrow{\ell_{1}}(t)=\langle 1,1,1\rangle+t\langle 3,1,3\rangle \\ \vec{\ell}_{2}(t)=\langle 7,3,7\rangle+t\langle 6,2,6\rangle . \end{array} $$

Determine if the described lines are the same line, parallel lines, intersecting or skew lines. If intersecting, give the point of intersection. $$ \ell_{1}=\left\\{\begin{array}{l} x=1+2 t \\ y=3-2 t \\ z=t \end{array}\right. \text { and } \ell_{2}=\left\\{\begin{array}{l} x=3-t \\ y=3+5 t \\ z=2+7 t \end{array}\right. $$

Find a unit vector orthogonal to both \(\vec{u}\) and \(\vec{v} .\) \(\vec{u}=\langle 1,-2,1\rangle, \quad \vec{v}=\langle-2,4,-2\rangle\)
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