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Chapter 0: Problem 34

Find an equation for each line. Then write your answer in the form \(A x+B y+C=0 .\) Through (4,1) and (8,2)

Short Answer

Expert verified
The equation in standard form is \(x - 4y = 0\).

Step by step solution

01

Determine the Slope

To find the slope of the line passing through these points, use the formula for the slope between two points: \[m = \frac{y_2-y_1}{x_2-x_1}\] Substitute the points (4,1) and (8,2) into the formula:\[m = \frac{2-1}{8-4} = \frac{1}{4}\] So, the slope is \(\frac{1}{4}\).
02

Use Point-Slope Form

Next, use the point-slope form of a line equation which is: \[y - y_1 = m(x - x_1)\].Plug in the slope (\(\frac{1}{4}\)) and one of the points, say (4,1):\[y - 1 = \frac{1}{4}(x - 4)\]
03

Simplify to Slope-Intercept Form

Simplify the equation from Step 2:\[y - 1 = \frac{1}{4}x - 1\]Add 1 to both sides:\[y = \frac{1}{4}x\]
04

Convert to Standard Form

We need to rearrange the equation to the form \(Ax + By + C = 0\):Multiply through by 4 to eliminate the fraction:\[4y = x\]Subtract x from both sides:\[-x + 4y = 0\]This is equivalent to \(x - 4y = 0\) in standard form.

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Key Concepts

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

Slope Calculation
Understanding how to calculate the slope of a line is essential in algebra and geometry. The slope of a line measures its steepness and direction. To find it, especially between two points, use the slope formula:
  • The formula is: \(m = \frac{y_2-y_1}{x_2-x_1}\)
  • Here, \((x_1, y_1)\) and \((x_2, y_2)\) are coordinates of the two given points.
  • "m" indicates the slope.
To use this with points (4,1) and (8,2):
  • Calculate \(y_2 - y_1\) which is \(2-1 = 1\).
  • Next, calculate \(x_2 - x_1\) which is \(8-4 = 4\).
  • The slope \(m = \frac{1}{4}\).
The final slope result, \(\frac{1}{4}\), tells us that for every 4 units the line moves horizontally, it rises by 1 unit.
Point-Slope Form
The point-slope form of a line is incredibly useful when you know a point on the line and its slope. This form is given by:
  • \(y - y_1 = m(x - x_1)\)
Here, \(m\) represents the slope, and \((x_1, y_1)\) is a known point.
To express a line using point-slope form:
  • Take the slope \(\frac{1}{4}\).
  • Use a point from the line, for example, (4,1).
Plugging these values into the formula, we have:
  • \(y - 1 = \frac{1}{4}(x - 4)\)
This equation represents the same line through a specific point with the given slope.
Slope-Intercept Form
To delve further, you can convert a point-slope form equation into the slope-intercept form. This form is \(y = mx + b\), where:
  • "m" is the slope.
  • "b" is the y-intercept, the point where the line crosses the y-axis.
Starting with our point-slope form:
  • \(y - 1 = \frac{1}{4}x - 1\)
Simply add 1 to both sides to get the slope-intercept form:
  • \(y = \frac{1}{4}x\)
In this line equation \(y = \frac{1}{4}x\), there is no constant "b" after \(\frac{1}{4}x\), signifying the line passes through the origin (0,0).
Standard Form of a Line
The standard form of a line equation is \(Ax + By + C = 0\). Converting from slope-intercept form requires a few manipulations.
Starting with \(y = \frac{1}{4}x\), multiply through by 4 to clear the fraction:
  • \(4y = x\)
Reorganize the terms by subtracting x from both sides to arrange it in standard form:
  • \(-x + 4y = 0\)
However, to ensure the leading coefficient is positive:
  • Reorder terms to \(x - 4y = 0\)
This final arrangement of terms is a neat and orderly way to bring the line equation into a unified standard form.

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Most popular questions from this chapter

Plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\) -intercepts. \(3 x^{2}+4 y^{2}=12\)

Sketch the graph of each of the following, making use of translations. (a) \(y=\frac{1}{4} x^{2}\) (b) \(y=\frac{1}{4}(x+2)^{2}\) (c) \(y=-1+\frac{1}{4}(x+2)^{2}\)

Circular motion can be modeled by using the parametric representations of the form \(x(t)=\sin t\) and \(y(t)=\cos t\) (A parametric representation means that a variable, \(t\) in this case, determines both \(x(t)\) and \(y(t) .\) This will give the full circle for \(0 \leq t \leq 2 \pi .\) If we consider a 4 -foot-diameter wheel making one complete rotation clockwise once every 10 seconds, show that the motion of a point on the rim of the wheel can be represented by \(x(t)=2 \sin (\pi t / 5)\) and \(y(t)=2 \cos (\pi t / 5)\) (a) Find the positions of the point on the rim of the wheel when \(t=2\) seconds, 6 seconds, and 10 seconds. Where was this point when the wheel started to rotate at \(t=0 ?\) (b) How will the formulas giving the motion of the point change if the wheel is rotating counterclockwise. (c) At what value of \(t\) is the point at (2,0) for the first time?

Plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs (see Example 4). $$ \begin{array}{l} y=-2 x+3 \\ y=-2(x-4)^{2} \end{array} $$

Plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\) -intercepts. \(4 x^{2}+3 y^{2}=12\)
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