To determine which points Vera can use to graph a line that passes through [tex]\((0, 2)\)[/tex] and has a slope of [tex]\(\frac{2}{3}\)[/tex], we need to use the slope-intercept form of the equation of a line, which is [tex]\(y = mx + b\)[/tex].
1. Identify the Parameters:
- Slope [tex]\(m\)[/tex]: [tex]\(\frac{2}{3}\)[/tex]
- Y-intercept [tex]\(b\)[/tex]: 2 (because the line passes through [tex]\((0, 2)\)[/tex])
Thus, the equation of the line is:
[tex]\[
y = \frac{2}{3}x + 2
\][/tex]
2. Substitute [tex]\(x\)[/tex] Values of Given Points: For each of the provided points, substitute the [tex]\(x\)[/tex] value into the equation and see if the resulting [tex]\(y\)[/tex] value matches the given [tex]\(y\)[/tex] value.
- Point [tex]\((-3, 0)\)[/tex]:
[tex]\[
y = \frac{2}{3}(-3) + 2 = -2 + 2 = 0
\][/tex]
This point lies on the line.
- Point [tex]\((-2, -3)\)[/tex]:
[tex]\[
y = \frac{2}{3}(-2) + 2 = -\frac{4}{3} + 2 = -\frac{4}{3} + \frac{6}{3} = \frac{2}{3}
\][/tex]
Since [tex]\(\frac{2}{3} \neq -3\)[/tex], this point does not lie on the line.
- Point [tex]\((2, 5)\)[/tex]:
[tex]\[
y = \frac{2}{3}(2) + 2 = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}
\][/tex]
Since [tex]\(\frac{10}{3} \neq 5\)[/tex], this point does not lie on the line.
- Point [tex]\((3, 4)\)[/tex]:
[tex]\[
y = \frac{2}{3}(3) + 2 = 2 + 2 = 4
\][/tex]
This point lies on the line.
- Point [tex]\((6, 6)\)[/tex]:
[tex]\[
y = \frac{2}{3}(6) + 2 = 4 + 2 = 6
\][/tex]
This point lies on the line.
3. Conclusion: Based on the calculations, the points that Vera can use to graph the line are:
[tex]\[
(-3, 0), (3, 4), (6, 6)
\][/tex]
