Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
Let's find the equation of the line that passes through the points given in the table.
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -10 & 2 \\ \hline -4 & 1 \\ \hline 8 & -1 \\ \hline 14 & -2 \\ \hline \end{array} \][/tex]
### Step 1: Calculate the Slope (m)
To find the slope of the line, we can use the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Let's use the first point [tex]\((-10, 2)\)[/tex] and the last point [tex]\((14, -2)\)[/tex]:
[tex]\[ m = \frac{-2 - 2}{14 - (-10)} = \frac{-4}{24} = -\frac{1}{6} \][/tex]
So, the slope [tex]\(m\)[/tex] is [tex]\(-\frac{1}{6}\)[/tex].
### Step 2: Find the y-intercept (b)
We use the point-slope form of the line equation [tex]\(y = mx + b\)[/tex] to find [tex]\(b\)[/tex]:
Using the point [tex]\((-10, 2)\)[/tex]:
[tex]\[ 2 = -\frac{1}{6}(-10) + b \][/tex]
[tex]\[ 2 = \frac{10}{6} + b \][/tex]
[tex]\[ 2 = \frac{5}{3} + b \][/tex]
[tex]\[ 2 - \frac{5}{3} = b \][/tex]
[tex]\[ \frac{6}{3} - \frac{5}{3} = b \][/tex]
[tex]\[ \frac{1}{3} = b \][/tex]
So, the y-intercept [tex]\(b\)[/tex] is [tex]\(\frac{1}{3}\)[/tex].
### Step 3: Write the Equation of the Line
Now that we have the slope [tex]\(m = -\frac{1}{6}\)[/tex] and the y-intercept [tex]\(b = \frac{1}{3}\)[/tex], we can write the equation of the line:
[tex]\[ y = -\frac{1}{6}x + \frac{1}{3} \][/tex]
### Step 4: Check the Given Equations
Let’s verify which of the given equations match our line’s equation [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex].
1. [tex]\(y - 2 = -6(x + 10)\)[/tex]
2. [tex]\(y - 2 = -\frac{1}{6}(x + 10)\)[/tex]
3. [tex]\(y - 1 = -\frac{1}{6}(x + 4)\)[/tex]
4. [tex]\(y = -6x - 58\)[/tex]
5. [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex]
6. [tex]\(y = -\frac{1}{6}x + 5\)[/tex]
### Verifying Each Equation:
1. [tex]\(y - 2 = -6(x + 10)\)[/tex]
- This implies a slope of [tex]\(-6\)[/tex], which is not [tex]\(-\frac{1}{6}\)[/tex]
2. [tex]\(y - 2 = -\frac{1}{6}(x + 10)\)[/tex]
- Rearranging: [tex]\(y - 2 = -\frac{1}{6}x - \frac{10}{6}\)[/tex]
- [tex]\(y = -\frac{1}{6}x - \frac{5}{3} + 2\)[/tex]
- [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex], this matches our equation.
3. [tex]\(y - 1 = -\frac{1}{6}(x + 4)\)[/tex]
- Rearranging: [tex]\(y - 1 = -\frac{1}{6}x - \frac{4}{6}\)[/tex]
- [tex]\(y = -\frac{1}{6}x - \frac{2}{3} + 1\)[/tex]
- [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex], this too matches our equation.
4. [tex]\(y = -6x - 58\)[/tex]
- This equation has a slope of [tex]\(-6\)[/tex], which is not [tex]\(-\frac{1}{6}\)[/tex]
5. [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex]
- This directly matches our derived equation.
6. [tex]\(y = -\frac{1}{6}x + 5\)[/tex]
- This has the correct slope but the intercept of [tex]\(5\)[/tex] does not match [tex]\(\frac{1}{3}\)[/tex]
### Step 5: Conclusion
The equations that match a line passing through the points in the table are:
- [tex]\(y - 2 = -\frac{1}{6}(x + 10)\)[/tex]
- [tex]\(y - 1 = -\frac{1}{6}(x + 4)\)[/tex]
- [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -10 & 2 \\ \hline -4 & 1 \\ \hline 8 & -1 \\ \hline 14 & -2 \\ \hline \end{array} \][/tex]
### Step 1: Calculate the Slope (m)
To find the slope of the line, we can use the formula for the slope between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Let's use the first point [tex]\((-10, 2)\)[/tex] and the last point [tex]\((14, -2)\)[/tex]:
[tex]\[ m = \frac{-2 - 2}{14 - (-10)} = \frac{-4}{24} = -\frac{1}{6} \][/tex]
So, the slope [tex]\(m\)[/tex] is [tex]\(-\frac{1}{6}\)[/tex].
### Step 2: Find the y-intercept (b)
We use the point-slope form of the line equation [tex]\(y = mx + b\)[/tex] to find [tex]\(b\)[/tex]:
Using the point [tex]\((-10, 2)\)[/tex]:
[tex]\[ 2 = -\frac{1}{6}(-10) + b \][/tex]
[tex]\[ 2 = \frac{10}{6} + b \][/tex]
[tex]\[ 2 = \frac{5}{3} + b \][/tex]
[tex]\[ 2 - \frac{5}{3} = b \][/tex]
[tex]\[ \frac{6}{3} - \frac{5}{3} = b \][/tex]
[tex]\[ \frac{1}{3} = b \][/tex]
So, the y-intercept [tex]\(b\)[/tex] is [tex]\(\frac{1}{3}\)[/tex].
### Step 3: Write the Equation of the Line
Now that we have the slope [tex]\(m = -\frac{1}{6}\)[/tex] and the y-intercept [tex]\(b = \frac{1}{3}\)[/tex], we can write the equation of the line:
[tex]\[ y = -\frac{1}{6}x + \frac{1}{3} \][/tex]
### Step 4: Check the Given Equations
Let’s verify which of the given equations match our line’s equation [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex].
1. [tex]\(y - 2 = -6(x + 10)\)[/tex]
2. [tex]\(y - 2 = -\frac{1}{6}(x + 10)\)[/tex]
3. [tex]\(y - 1 = -\frac{1}{6}(x + 4)\)[/tex]
4. [tex]\(y = -6x - 58\)[/tex]
5. [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex]
6. [tex]\(y = -\frac{1}{6}x + 5\)[/tex]
### Verifying Each Equation:
1. [tex]\(y - 2 = -6(x + 10)\)[/tex]
- This implies a slope of [tex]\(-6\)[/tex], which is not [tex]\(-\frac{1}{6}\)[/tex]
2. [tex]\(y - 2 = -\frac{1}{6}(x + 10)\)[/tex]
- Rearranging: [tex]\(y - 2 = -\frac{1}{6}x - \frac{10}{6}\)[/tex]
- [tex]\(y = -\frac{1}{6}x - \frac{5}{3} + 2\)[/tex]
- [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex], this matches our equation.
3. [tex]\(y - 1 = -\frac{1}{6}(x + 4)\)[/tex]
- Rearranging: [tex]\(y - 1 = -\frac{1}{6}x - \frac{4}{6}\)[/tex]
- [tex]\(y = -\frac{1}{6}x - \frac{2}{3} + 1\)[/tex]
- [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex], this too matches our equation.
4. [tex]\(y = -6x - 58\)[/tex]
- This equation has a slope of [tex]\(-6\)[/tex], which is not [tex]\(-\frac{1}{6}\)[/tex]
5. [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex]
- This directly matches our derived equation.
6. [tex]\(y = -\frac{1}{6}x + 5\)[/tex]
- This has the correct slope but the intercept of [tex]\(5\)[/tex] does not match [tex]\(\frac{1}{3}\)[/tex]
### Step 5: Conclusion
The equations that match a line passing through the points in the table are:
- [tex]\(y - 2 = -\frac{1}{6}(x + 10)\)[/tex]
- [tex]\(y - 1 = -\frac{1}{6}(x + 4)\)[/tex]
- [tex]\(y = -\frac{1}{6}x + \frac{1}{3}\)[/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
