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Chapter 10: Problem 1

Find the equations of the lines that pass through the following points: (a) (1,-1),(2,2) (b) (0,1),(1,-1)

Short Answer

Expert verified
(a) y = 3x - 4; (b) y = -2x + 1

Step by step solution

01

Find the Slope (a)

To find the slope of the line passing through points (1, -1) and (2, 2), use the formula for slope: \( m = \frac{y_2 - y_1}{x_2 - x_1} \) Substituting the coordinates \((x_1, y_1) = (1, -1)\) and \((x_2, y_2) = (2, 2)\), we get: \( m = \frac{2 - (-1)}{2 - 1} = \frac{3}{1} = 3 \)
02

Write the Point-Slope Equation (a)

Using the point-slope form of a line \( y - y_1 = m(x - x_1) \), with slope \( m = 3 \) and point (1, -1), the equation becomes: \( y + 1 = 3(x - 1) \)
03

Convert to Slope-Intercept Form (a)

Simplify the point-slope equation to obtain the line equation in slope-intercept form \(y = mx + b\): \( y + 1 = 3x - 3 \) \( y = 3x - 4 \)
04

Find the Slope (b)

For the line through points (0, 1) and (1, -1), use the slope formula: \( m = \frac{-1 - 1}{1 - 0} = \frac{-2}{1} = -2 \)
05

Write the Point-Slope Equation (b)

With slope \( m = -2 \) and point (0, 1), the point-slope form is: \( y - 1 = -2(x - 0) \)This simplifies directly to: \( y - 1 = -2x \)
06

Convert to Slope-Intercept Form (b)

Rearrange the equation to the form \(y = mx + b\): \( y = -2x + 1 \)

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Key Concepts

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

Slope calculation
Calculating the slope of a line is a fundamental concept in algebra and geometry. It determines how steep a line is and in which direction it moves. To find the slope of a line passing through two points, \( (x_1, y_1) \) and \( (x_2, y_2) \), you use the slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). This equation essentially measures the change in the vertical direction (rise) over the change in the horizontal direction (run).
  • In our example, calculating the slope between the points \( (1, -1) \) and \( (2, 2) \), we substitute the coordinates into the formula, resulting in: \( m = \frac{2 - (-1)}{2 - 1} = \frac{3}{1} = 3 \). This means that for every 1 unit we move to the right, the line rises by 3 units.
  • To calculate the slope for points \( (0, 1) \) and \( (1, -1) \), substitute and solve: \( m = \frac{-1 - 1}{1 - 0} = -2 \). This negative slope indicates the line falls 2 units for every 1 unit it moves to the right.
Point-slope form equation
Once the slope is known, forming an equation in the point-slope form is the next step. The point-slope formula is given by \( y - y_1 = m(x - x_1) \). This form is useful because it clearly shows both the slope and a single point that the line passes through.
  • For a line with slope \( m = 3 \) passing through \( (1, -1) \), the point-slope equation becomes: \( y + 1 = 3(x - 1) \). This stage ties the line's position to the specific point given, making the equation tangible.
  • In another scenario, for a slope of \( m = -2 \) through \( (0, 1) \), the equation simplifies to \( y - 1 = -2(x - 0) \). It shows how easy it is to use point-slope form to establish the trajectory of a line.
Slope-intercept form conversion
The slope-intercept form of a line is popular because it directly shows the line's slope and y-intercept, making it easy to graph. This form is written as \( y = mx + b \).
  • To convert the point-slope form \( y + 1 = 3(x - 1) \) into slope-intercept form, distribute and rearrange to obtain: \( y = 3x - 4 \). Here, \( m = 3 \) indicates the steepness of the line, and \( b = -4 \) represents where the line crosses the y-axis.
  • For the equation \( y - 1 = -2x \), rearrange to find \( y = -2x + 1 \). Again, \( m = -2 \) describes the downward slope, while \( b = 1 \) is the y-intercept, where the line meets the y-axis.

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Most popular questions from this chapter

Suppose that a certain animal population is divided into two age classes and has a Leslie matrix $$ L=\left[\begin{array}{ll} 1 & \frac{3}{2} \\ \frac{1}{2} & 0 \end{array}\right] $$ (a) Calculate the positive eigenvalue \(\lambda_{1}\) of \(L\) and the corresponding eigenvector \(\mathbf{x}_{1}.\) (b) Beginning with the initial age distribution vector. $$\mathbf{x}^{(0)}=\left[\begin{array}{c} 100 \\ 0 \end{array}\right]$$ calculate \(x^{(1)}, x^{(2)}, x^{(3)}, x^{(4)},\) and \(x^{(S)},\) rounding off to the nearest integer when necessary. (c) Calculate \(x^{(6)}\) using the exact formula \(x^{(6)}=L x^{(S)}\) and using the approximation formula \(x^{(6)} \approx \lambda_{1} x^{(S)}.\) Answer: A. $$\lambda_{1}=\frac{3}{2}, \quad x_{1}=\left[\begin{array}{l} 1 \\ \frac{1}{3} \end{array}\right]$$ B. $$\mathbf{x}^{(1)}=\left[\begin{array}{c} 100 \\ 50 \end{array}\right], \mathbf{x}^{(2)}=\left[\begin{array}{c} 175 \\ 50 \end{array}\right], \mathbf{x}^{(3)}=\left[\begin{array}{c} 250 \\ 88 \end{array}\right], \mathbf{x}^{(4)}=\left[\begin{array}{l} 382 \\ 125 \end{array}\right], \mathbf{x}^{(5)}=\left[\begin{array}{l} 570 \\ 191 \end{array}\right]$$ C. $$\mathbf{x}^{(6)}=L \mathbf{x}^{(S)}=\left[\begin{array}{l} 857 \\ 285 \end{array}\right], \mathbf{x}^{(6)} \simeq \lambda_{1} \mathbf{x}^{(S)}=\left[\begin{array}{l} 855 \\ 287 \end{array}\right]$$

A problem on a Babylonian tablet requires finding the length and width of a rectangle given that the length and the width add up to 10 , while the length and one-fourth of the width add up to \(7\) . The solution provided on the tablet consists of the following four statements: Multiply \(7\) by \(4\) to obtain \(28\). Take away \(10\) from \(28\) to obtain \(18 .\). Take one-third of \(18\) to obtain \(6,\) the length. Take away \(6\) from \(10\) to obtain \(4,\) the width. Explain how these steps lead to the answer.

Let an affine transformation be given by a \(2 \times 2\) matrix \(M\) and a two- dimensional vector \(b\). Let \(\mathbf{v}=\mathbf{c}_{1} \mathbf{v}_{1}+\mathbf{c}_{2} \mathbf{v}_{2}+\mathbf{c}_{3} \mathbf{v}_{3},\) where \(c_{1}+c_{2}+c_{3}=1 ;\) let \(\mathbf{w}=M \mathbf{v}+\mathbf{b} ;\) and let \(\mathbf{w}_{i}=M \mathbf{v}_{i}+\mathbf{b}\) for \(i=1,2,3\) Show that \(\mathbf{w}=\mathbf{c}_{1} \mathbf{w}_{1}+\mathbf{c}_{2} \mathbf{w}_{2}+\mathbf{c}_{3} \mathbf{w}_{3}\). (This shows that an affine transformation maps a convex combination of vectors to the same convex combination of the images of the vectors.)

Show that the modular equation \(4 x=1\) (mod 26 ) has no solution in \(Z_{26}\) by successively substituting the values \(x=0,1,2, \ldots, 25\)

". All of the results of this section can be generalized to the case where the plaintext is a binary message; that is, it is a sequence of 0 's and 1 's. In this case we do all of our modular arithmetic using modulus 2 rather than modulus \(26 .\) Thus, for example, \(1+1=0\) (mod 2 ). Suppose we want to encrypt the message 110101111 . Let us first break it into triplets to form the three vectors \(\left[\begin{array}{l}1 \\ 1 \\\ 0\end{array}\right],\left[\begin{array}{l}1 \\ 0 \\\ 1\end{array}\right],\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]\), and let us take \(\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 1\end{array}\right]\) as our enciphering matrix. (a) Find the encoded message. (b) Find the inverse modulo 2 of the enciphering matrix, and verify that it decodes your encoded message.
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