Answer :
To determine which points Vera can use to graph the line with the given characteristics, we follow these steps:
1. Identify the line equation: The line passes through the point \((0, 2)\) and has a slope of \(\frac{2}{3}\). Therefore, the equation of the line in slope-intercept form \(y = mx + b\) is:
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
2. Check each point: We will substitute the \(x\) and \(y\) coordinates of each point into the line equation to see if they satisfy it.
- Point 1: \((-3, 0)\)
[tex]\[ y = \frac{2}{3}(-3) + 2 = -2 + 2 = 0 \][/tex]
Since \(0 = 0\), the point \((-3, 0)\) lies on the line.
- Point 2: \((-2, -3)\)
[tex]\[ y = \frac{2}{3}(-2) + 2 = -\frac{4}{3} + 2 = \frac{-4 + 6}{3} = \frac{2}{3} \][/tex]
Since \(\frac{2}{3} \neq -3\), the point \((-2, -3)\) does not lie on the line.
- Point 3: \((2, 5)\)
[tex]\[ y = \frac{2}{3}(2) + 2 = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \][/tex]
Since \(\frac{10}{3} \neq 5\), the point \((2, 5)\) does not lie on the line.
- Point 4: \((3, 4)\)
[tex]\[ y = \frac{2}{3}(3) + 2 = 2 + 2 = 4 \][/tex]
Since \(4 = 4\), the point \((3, 4)\) lies on the line.
- Point 5: \((6, 6)\)
[tex]\[ y = \frac{2}{3}(6) + 2 = 4 + 2 = 6 \][/tex]
Since \(6 = 6\), the point \((6, 6)\) lies on the line.
3. Conclusion: The points that lie on the line are:
[tex]\[ (-3, 0), \; (3, 4), \; (6, 6) \][/tex]
Vera could use these three points to graph the line successfully.
1. Identify the line equation: The line passes through the point \((0, 2)\) and has a slope of \(\frac{2}{3}\). Therefore, the equation of the line in slope-intercept form \(y = mx + b\) is:
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
2. Check each point: We will substitute the \(x\) and \(y\) coordinates of each point into the line equation to see if they satisfy it.
- Point 1: \((-3, 0)\)
[tex]\[ y = \frac{2}{3}(-3) + 2 = -2 + 2 = 0 \][/tex]
Since \(0 = 0\), the point \((-3, 0)\) lies on the line.
- Point 2: \((-2, -3)\)
[tex]\[ y = \frac{2}{3}(-2) + 2 = -\frac{4}{3} + 2 = \frac{-4 + 6}{3} = \frac{2}{3} \][/tex]
Since \(\frac{2}{3} \neq -3\), the point \((-2, -3)\) does not lie on the line.
- Point 3: \((2, 5)\)
[tex]\[ y = \frac{2}{3}(2) + 2 = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \][/tex]
Since \(\frac{10}{3} \neq 5\), the point \((2, 5)\) does not lie on the line.
- Point 4: \((3, 4)\)
[tex]\[ y = \frac{2}{3}(3) + 2 = 2 + 2 = 4 \][/tex]
Since \(4 = 4\), the point \((3, 4)\) lies on the line.
- Point 5: \((6, 6)\)
[tex]\[ y = \frac{2}{3}(6) + 2 = 4 + 2 = 6 \][/tex]
Since \(6 = 6\), the point \((6, 6)\) lies on the line.
3. Conclusion: The points that lie on the line are:
[tex]\[ (-3, 0), \; (3, 4), \; (6, 6) \][/tex]
Vera could use these three points to graph the line successfully.
