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Chapter 2: Problem 31

Exer. \(21-32:\) Find a general form of an equation of the IIne through the point \(A\) that satisfies the given condition. \(A(7,-3) ;\) perpendicular to the line \(2 x-5 y=8\)

Short Answer

Expert verified
The general form of the equation is \(5x + 2y = 29\).

Step by step solution

01

Identify the slope of the given line

Start by rewriting the given equation of the line into the slope-intercept form, which is \(y = mx + b\). The given line is \(2x - 5y = 8\). Rearrange it to find \(y\):\[-5y = -2x + 8\]\[y = \frac{2}{5}x - \frac{8}{5}\]. The slope \(m\) of the line is \(\frac{2}{5}\).
02

Find the perpendicular slope

To find a line perpendicular to the given line, we take the negative reciprocal of the slope. If the slope of the given line is \(\frac{2}{5}\), the perpendicular slope \(m'\) will be \(-\frac{5}{2}\).
03

Use point-slope form to write the equation

We have the point \((7, -3)\) and the slope \(-\frac{5}{2}\). Use the point-slope formula \(y - y_1 = m'(x - x_1)\) where \((x_1, y_1) = (7, -3)\) and \(m' = -\frac{5}{2}\):\[y + 3 = -\frac{5}{2}(x - 7)\].
04

Convert to a general form equation

Expand and rearrange the equation \(y + 3 = -\frac{5}{2}(x - 7)\):\[y + 3 = -\frac{5}{2}x + \frac{35}{2}\]. Simplify it to get:\[2y + 6 = -5x + 35\], which simplifies to \[5x + 2y = 29\]. This is the general form of the equation of the line.

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Key Concepts

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

Slope-Intercept Form
The slope-intercept form of a line is a popular way to express the equation of a line. It's represented as \(y = mx + b\), where \(m\) is the slope of the line, and \(b\) is the y-intercept, which is the point where the line crosses the y-axis.

To determine the slope from an equation given in a different form, say standard form \(Ax + By = C\) like in the exercise, rearrange it to solve for \(y\). This way, you can easily identify the slope \(m\). For example, in the equation \(2x - 5y = 8\), we rearranged to find \(y = \frac{2}{5}x - \frac{8}{5}\).

Knowing the slope-intercept form helps simplify finding relationships and intersections between lines.
Point-Slope Form
Point-slope form is a valuable tool when you know a point on the line and the slope but not the y-intercept. The formula is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope, and \((x_1, y_1)\) is a known point on the line.

This form is especially useful when writing the equation of a line through a particular point as it directly uses the point in its structure. In our example, we had a point \((7, -3)\) and a slope \(-\frac{5}{2}\), leading to the equation \(y + 3 = -\frac{5}{2}(x - 7)\).

This format can then be expanded or converted into other forms, such as slope-intercept or standard form for specific applications.
Negative Reciprocal
The concept of the negative reciprocal is crucial when dealing with perpendicular lines. If one line has a slope \(m\), a line perpendicular to it will have a slope that is the negative reciprocal of \(m\).

To find the negative reciprocal, you simply flip the fraction and change the sign. If the slope is \(\frac{2}{5}\), like in our problem, the negative reciprocal is \(-\frac{5}{2}\).

This understanding aids in quickly identifying the slope needed for perpendicularity, which is a key skill in geometry and trigonometry.
General Form of a Line
The general form of a line is written as \(Ax + By = C\). This format is useful because it can easily handle horizontal, vertical, and other standard lines within coordinate space.

In the provided exercise, we started with point-slope form and rearranged to arrive at the general form \(5x + 2y = 29\). This was done by expanding the equation \(y + 3 = -\frac{5}{2}(x - 7)\) and solving for a form where both \(x\) and \(y\) appear on one side with a constant on the other.

The general form highlights the relationship between \(x\) and \(y\) while being a universally friendly format for further manipulations.

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Most popular questions from this chapter

The symbol \([x]\) denotes values of the greatest Integer function. Sketch the graph of \(f\) (a) \(f(x)=[| x+2] \quad\) (b) \(f(x)=|x|+2\) (c) \(f(x)=\frac{1}{2}[x]\) (d) \(f(x)=\left[\frac{1}{2} x\right]\) (e) \(f(x)=-[-x]\)

Sketch the graph of the equation. $$y=|| x|-1|$$

Explain why the graph of the equation is not the graph of a function. $$x=y^{2}$$

An object is projected vertically upward with an initial velocity of \(v_{0}\) ft/sec, and its distance \(s(t)\) in feet above the ground after \(t\) seconds is given by the formula \(s(t)=-16 t^{2}+v_{0} t\) (a) If the object hits the ground after 12 seconds, find its initial velocity \(v_{0}\) (b) Find its maximum distance above the ground.

An object is projected vertically upward from the top of a building with an initial velocity of \(144 \mathrm{ft} / \mathrm{sec} .\) Its distance \(s(t)\) in feet above the ground after \(t\) seconds is given by the equation $$ s(t)=-16 t^{2}+144 t+100 $$ (a) Find its maximum distance above the ground. (b) Find the height of the building.
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