Answer :
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]
