/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Find the slope of the line passi... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the slope of the line passing through the given points. Round to the nearest hundredth where necessary. \(\left(1, \frac{1}{2}\right)\) and \(\left(\frac{3}{4}, 2\right)\)

Short Answer

Expert verified
The slope of the line is \(-6\).

Step by step solution

01

Identify the Coordinates

Identify the given coordinates as \((x_1, y_1) = \left(1, \frac{1}{2}\right)\) and \((x_2, y_2) = \left(\frac{3}{4}, 2\right)\).
02

Recall the Slope Formula

The slope \(m\) of a line passing through points \((x_1, y_1)\) and \((x_2, y_2)\) is given by: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
03

Substitute the Coordinates

Substitute \((x_1, y_1)\) and \((x_2, y_2)\) into the slope formula: \[ m = \frac{2 - \frac{1}{2}}{\frac{3}{4} - 1} \]
04

Evaluate the Numerator

Calculate the difference in the y-values: \[ 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \]
05

Evaluate the Denominator

Calculate the difference in the x-values: \[ \frac{3}{4} - 1 = \frac{3}{4} - \frac{4}{4} = -\frac{1}{4} \]
06

Calculate the Slope

Using the values found for the numerator and the denominator, find the slope: \[ m = \frac{\frac{3}{2}}{-\frac{1}{4}} = \frac{3}{2} \times -4 = -6 \]
07

Round the Result

Since the slope is exactly an integer, rounding is not necessary. The slope is \(-6\).

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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 tells us how steep the line is. The formula to find the slope given two points, \((x_1, y_1)\) and \((x_2, y_2)\), is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] This formula is handy because it works for any two points on a straight line. Slope, denoted by \(m\), is the ratio of the change in y (vertical change) to the change in x (horizontal change). If you know the coordinates of any two points on a line, you can find the slope using this formula. Start by subtracting the y-values of the two points to get the numerator and the x-values to get the denominator. Simplify these differences if needed.

For example, given points \((1, \frac{1}{2})\) and \((\frac{3}{4}, 2)\), we use the slope formula as shown above. Constructing and simplifying our fractions correctly ensures we get the correct slope.
coordinate geometry
Coordinate geometry, or analytic geometry, is a method to study geometry using the coordinate system. It allows us to describe geometric properties and relationships algebraically. In the coordinate plane, any point can be represented by an ordered pair \((x, y)\), where \(x\) is the horizontal position and \(y\) is the vertical position.

When finding the slope, note how each point corresponds to \((x_1, y_1)\) and \((x_2, y_2)\). Recognizing these pairs is the first step. For our exercise, points \((1, \frac{1}{2})\) and \((\frac{3}{4}, 2)\) are translated directly into our slope formula. This approach harnesses both graphical and algebraic methods to solve problems involving lines and other shapes.
fraction arithmetic
Mastery of fraction arithmetic is crucial when dealing with slope calculations. The numerator and denominator in the slope formula often involves fractions. Knowing how to add, subtract, multiply, and divide fractions is essential. For the given points, when calculating \[ 2 - \frac{1}{2} = \frac{4}{2} - \frac{1}{2} = \frac{3}{2} \], you simplify by ensuring both terms have a common denominator.

Next, the expression \[ \frac{3}{4} - 1 \] involves converting 1 to a fraction with common denominators, \[ \ \frac{4}{4} \ \], to get \(-\frac{1}{4} \). Hence, fraction arithmetic is a vital skill.

Multiplying fractions is straightforward: \(\frac{\frac{3}{2}}{-\frac{1}{4}} = -\frac{3}{2} \times \frac{-4}{1} = -3 \times (-4) = -6 \). This step finalizes our slope calculation.
elementary algebra
Elementary algebra simplifies working with variables and known numbers, helping us solve equations or find specific values like our slope. This concept includes basic operations (addition, subtraction, multiplication, division) with algebraic expressions.

In slope calculations, we substitute known values into the formula: \[ m = \frac{2 - \frac{1}{2}}{\frac{3}{4} - 1} \]. By systematically substituting and simplifying, each step becomes less overwhelming. Here, algebra aids in clarifying seemingly complex manipulations.

Additionally, interpreting and simplifying negative and positive results correctly is another key algebraic skill. For instance, recognizing \(-6\) as our resulting slope, where the negative sign indicates the line descends from left to right, consolidates basic algebraic comprehension.

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Most popular questions from this chapter

A slanted roof on a house rises 2.8 meters over a horizontal distance of 6.2 meters. Find the grade (or pitch) of the roof.

Round off to the nearest hundredth when necessary. Bridges (and many concrete highways) are constructed with "expansion joints," which are small gaps in the roadway between one section of the bridge and the next. These expansion joints allow room for the roadway to expand during hot weather. Suppose that a bridge has a gap of \(1.5 \mathrm{cm}\) when the air temperature is \(24^{\circ} \mathrm{C},\) that the gap narrows to \(0.7 \mathrm{cm}\) when the air temperature is \(33^{\circ} \mathrm{C},\) and that the width of the gap is linearly related to the temperature. (a) Write an equation relating the width of the gap \(w\) and the temperature \(t\) (b) What would be the width of a gap in this roadway at \(28^{\circ} \mathrm{C} ?\) (c) At what temperature would the gap close completely? (d) If the temperature exceeds the value found in part (c) that causes the gap to close, it is possible that the roadway could buckle. Is this likely to occur? Explain.

Round off to the nearest hundredth when necessary. In physiology a jogger's heart rate \(N\), in beats per minute, is related linearly to the jogger's speed \(s .\) A certain jogger's heart rate is 80 beats per minute at a speed of \(15 \mathrm{ft} / \mathrm{sec}\) and 82 beats per minute at a speed of \(18 \mathrm{ft} / \mathrm{sec} .\) (a) Write an equation relating the jogger's speed and heart rate. (b) Predict this jogger's heart rate if she jogs at a speed of \(20 \mathrm{ft} / \mathrm{sec}\). (c) According to the equation obtained in part (a), what is the jogger's heart rate at rest? [Hint: At rest the jogger's speed is 0.]

Sketch the graph of the given equation. Label the intercepts. $$6(y-1)=3(x+2)$$

Round off to the nearest hundredth when necessary. Assume that the maximum speed your car can travel on a hill, whether going uphill or downhill, is linearly related to the steepness of the hill as measured by the angle the hill makes with a horizontal line. Suppose that your maximum speed on a \(6^{\circ}\) incline is 62 mph and your maximum speed on a \(2^{\circ}\) decline (that is, an incline of \(-2^{\circ}\) ) is 115 mph. (a) Write an equation relating the maximum speed \(s\) of the car and the number \(n\) of degrees of the incline. (b) What would be the maximum speed on a \(4.5^{\circ}\) incline? (c) What would be the maximum speed on a \(-2.8^{\circ}\) decline? (d) Sketch a graph of this equation using the horizontal axis for \(n\) and the vertical axis for \(s\) (e) What is the \(s\) -intercept of this graph? What is the significance of this \(s\) -intercept? (f) What is the \(n\) -intercept of this graph? What is the significance of this \(n\) -intercept?

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