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Chapter 1: Problem 13

Find an equation of the given line. \((0,0)\) and \((1,0)\) on line

Short Answer

Expert verified
The equation of the line is \(y = 0\).

Step by step solution

01

Identify the points

The given points on the line are \((0,0)\) and \((1,0)\).
02

Find the slope

The slope of a line passing through the points \(x_1, y_1\) and \(x_2, y_2\) is given by: \(\text{slope} = \frac{y_2 - y_1}{x_2 - x_1}\). Substituting the given points \((0,0)\) and \((1,0)\), we get: \(\text{slope} = \frac{0 - 0}{1 - 0} = 0\).
03

Use the point-slope form

The equation of a line can be written in point-slope form: \(y - y_1 = \text{slope} \times (x - x_1)\). Using point \((0,0)\) and \(\text{slope} = 0\), we substitute into the formula: \(y - 0 = 0 \times (x - 0)\).
04

Simplify the equation

Simplify the equation from Step 3 to get the standard linear form: \(y = 0\).

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Key Concepts

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

Slope
The slope of a line indicates its steepness and direction. It's calculated as the 'rise' over the 'run' between two points on the line.
In mathematical terms, the slope (m) between two points \(x_1, y_1\) and \(x_2, y_2\) is given by \[m = \frac{y_2 - y_1}{x_2 - x_1}\].
For example, in the exercise, the points provided are \(0,0\) and \(1,0\).
Substituting these into the formula, we have:
\[m = \frac{0 - 0}{1 - 0} = 0\]
Therefore, the line is horizontal, as the slope is zero.
A zero slope indicates no vertical change as we move along the line.
Point-Slope Form
Point-slope form is a way to write the equation of a line given a point and the slope.
The general formula is: \[y - y_1 = m \cdot (x - x_1)\]
Here, \(x_1, y_1\) is a point on the line, and \(m\) is the slope.
From the exercise, we have the point \(0,0\) and the slope \(0\).
Substituting these values, we get:
\[y - 0 = 0 \cdot (x - 0)\]
This simplifies to:
\[y = 0\]
This equation tells us that for every x-value, y is always 0, reinforcing that the line is horizontal.
Standard Linear Form
Another common way to write the equation of a line is the standard linear form.
It is usually written as \[Ax + By = C\], where A, B, and C are constants.
In the exercise, we found the slope to be 0 and derived the equation \[y = 0\] using point-slope form.
Converting \[y = 0\] into standard linear form, we get:
\[0x + 1y = 0\]
Here, A = 0, B = 1, and C = 0.
This form provides an alternative way to represent the same horizontal line, where it's clear that there is no dependence on x (since A = 0).
Coordinate Geometry
Coordinate geometry allows us to study geometry using a coordinate system.
Key concepts include points, lines, and curves represented within an x-y plane.
For lines, we use equations like the ones derived in the exercise.
With the points \(0,0\) and \(1,0\), drawing these on the coordinate plane shows a horizontal line along the x-axis.
Using coordinate geometry, any point on this line would satisfy the equation \[y = 0\].
Understanding how to use points and slopes to derive equations is fundamental for solving geometric problems in the coordinate system.

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Most popular questions from this chapter

Use a derivative routine to obtain the value of the derivative. Give the value to 5 decimal places. $$f^{\prime}(1), \text { where } f(x)=\sqrt{1+x^{2}}$$

If possible, define \(f(x)\) at the exceptional point in a way that makes \(f(x)\) continuous for all \(x.\) $$f(x)=\frac{\sqrt{9+x}-\sqrt{9}}{x}, x \neq 0$$

Differentiate the function \(f(x)=\left(3 x^{2}+x-2\right)^{2}\) in two ways. (a) Use the general power rule. (b) Multiply \(3 x^{2}+x-2\) by itself and then differentiate the resulting polynomial.

Use limits to compute \(f^{\prime}(x) .\) $$f(x)=\frac{1}{x^{2}+1}$$

Determine which of the following limits exist. Compute the limits that exist. Compute the limits that exist, given that $$\lim _{x \rightarrow 0} f(x)=-\frac{1}{2} \quad\( and \)\quad \lim _{x \rightarrow 0} g(x)=\frac{1}{2}.$$ (a) \(\lim _{x \rightarrow 0}(f(x)+g(x))\) (b) \(\lim _{x \rightarrow 0}(f(x)-2 g(x))\) (c) \(\lim _{x \rightarrow 0} f(x) \cdot g(x)\) (d) \(\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}\)
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