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Chapter 4: Problem 12

Find the slope of the line through the first two points given. Assume the third point is also on the line and find the missing coordinate. a. \((-1,5)\) and \((3,1) ;(5, \square)\) b. \((2,-5)\) and \((2,-2) ;(\square, 3)\) c. \((-10,22)\) and \((-2,2) ;(\square,-3)\)

Short Answer

Expert verified
a: -1; b: 2; c: -4/5

Step by step solution

01

Find the Slope (a)

The formula for the slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is \( m = \frac{y_2 - y_1}{x_2 - x_1} \). For points \((-1, 5)\) and \((3, 1)\), the slope is \( m = \frac{1 - 5}{3 - (-1)} = \frac{-4}{4} = -1 \).
02

Use the Slope to Find the Missing Coordinate (a)

Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), we substitute one of the known points, say \((3, 1)\), and the slope \(-1\) to get: \( y - 1 = -1(x - 3) \). Simplifying, \( y = -x + 4 \). For point \((5, \square)\), set \(x = 5\) to get \( y = -5 + 4 = -1 \). So, the missing coordinate is \(-1\).
03

Find the Slope (b)

The slope between points \((2, -5)\) and \((2, -2)\) is undefined because both points have the same x-coordinate. This indicates a vertical line, so all points have the same x-coordinate. The missing coordinate for point \((\square, 3)\) is also \(2\). The coordinate is \(2\).
04

Find the Slope (c)

Calculate the slope for points \((-10, 22)\) and \((-2, 2)\) using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Here, \( m = \frac{2 - 22}{-2 - (-10)} = \frac{-20}{8} = -\frac{5}{2} \).
05

Use the Slope to Find the Missing Coordinate (c)

Using point-slope form with point \((-2, 2)\) and slope \(-\frac{5}{2}\), we have \( y - 2 = -\frac{5}{2}(x + 2) \). Distributing and simplifying, \( y = -\frac{5}{2}x - 5 \). For point \((\square, -3)\), set \( y = -3 \): \(-3 = -\frac{5}{2}x - 5 \). Solving for \(x\), add 5 to both sides: \(2 = -\frac{5}{2}x\). Multiply both sides by \(-\frac{2}{5}\) to get \(x = -\frac{4}{5}\).

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Key Concepts

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

Linear Equations
Linear equations describe lines on a graph using a simple algebraic formula. Each linear equation can be written in different forms, such as the standard form, slope-intercept form, and point-slope form.
A linear equation in the slope-intercept form, for example, looks like this:
  • \( y = mx + b \), where \( m \) is the slope of the line and \( b \) is the y-intercept (the point where the line crosses the y-axis).
Every linear equation will graph as a straight line if plotted on a coordinate plane because the relationship between \(x\) and \(y\) is constant. This property makes understanding linear equations crucial for interpreting graphs and data.

By identifying the slope and y-intercept, you can quickly sketch the graph of the line, making linear equations a powerful tool for predicting and understanding real-world scenarios.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, lets us observe and analyze geometric properties using coordinates on a graph. Each point on a graph is defined by its coordinates, often written as \((x, y)\), which specify its position relative to the x-axis and y-axis.
This system allows us to find the distance between two points, the slope of a line, and other important geometric measures directly from numerical coordinates.
Key Concepts:
  • Distance Formula: The formula \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \) calculates how far one point is from another in the coordinate plane.
  • Slope Formula: The formula \( m = \frac{y_2-y_1}{x_2-x_1} \) finds the direction and steepness between two points.
These formulas help translate geometric concepts into algebraic expressions, making complex shapes and their properties manageable to study through algebra.
Point-Slope Form
The point-slope form is a linear equation format that highlights the slope of a line and a specific point on the line. It is especially useful when you know the slope and a single point but not the y-intercept. The formula for point-slope form is:
  • \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \((x_1, y_1)\) is an exact point on the line.
This form allows you to quickly write the equation of a line given minimal information, which is extremely handy in problem-solving.
Steps to Use Point-Slope Form:
  • Identify a point on the line \((x_1, y_1)\).
  • Use the slope \( m \) of the line.
  • Plug the values into \( y - y_1 = m(x - x_1) \) to form the equation.
The point-slope form is essential for finding the equation based on a slope and a point, rather than using graph-drawn lines or plotted points alone.

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Most popular questions from this chapter

At 2:00 P.M., elevator B passes the 94th floor of the same building going down. The table shows the floors and the times in seconds after 2:00. \begin{tabular}{|l|c|c|c|c|c|c|c|} \hline Floor \(x\) & 94 & 92 & 90 & 88 & 86 & 84 & 80 \\ \hline Time after 2:00 (s) \(y\) & 0 & \(1.3\) & \(2.5\) & \(3.8\) & 5 & \(6.3\) & \(8.6\) \\\ \hline \end{tabular} a. What is the line of fit based on Q-points for the data? b. Give a real-world meaning of the slope. c. About what time will this elevator pass the 10 th floor if it makes no stops? d. Where will this elevator be at \(2: 00: 34\) if it makes no stops?

Plot the points \((4,2),(1,3.5)\), and \((10,-1)\) on graph paper. These points are on the same line, or collinear, so you can draw a line through them. a. Draw a slope triangle between \((4,2)\) and \((1,3.5)\), and calculate the slope from the change in \(y\) and the change in \(x\). b. Draw another slope triangle between \((10,-1)\) and \((4,2)\), and calculate the slope from the change in \(y\) and the change in \(x\). c. Compare the slope triangles and the slopes you calculated. What do you notice? d. What would happen if you made a slope triangle between \((10,-1)\) and \((1,3.5)\) ?

Rewrite each equation in intercept form. Show your steps. Check your answer by using a calculator graph or table. a. \(y=14+3(x-5)\) b. \(y=-5-2(x+5)\) (a) c. \(6 x+2 y=24\)

Write each equation in the form requested. Check your answers by graphing on your calculator. a. Write \(y=13.6(x-1902)+158.2\) in intercept form. b. Write \(y=-5.2 x+15\) in point-slope form using \(x=10\) as the first coordinate of the point.

The base of a triangle was recorded as \(18.3 \pm 0.1 \mathrm{~cm}\) and the height was recorded as \(7.4 \pm 0.1 \mathrm{~cm}\). These measurements indicate the measured value and an accuracy component. a. Use the formula \(A=0.5 b h\) and the measured values for base and height to calculate the area of the triangle. b. Use the smallest possible lengths for base and height to calculate an area. (a) c. Use the largest possible lengths for base and height to calculate an area. d. Use your answers to \(12 \mathrm{a}-\mathrm{c}\) to express the range of possible area values as a number \(\pm\) an accuracy component. (h)
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