To determine which points Vera can use to graph the line, we need to verify if each point lies on the line that passes through [tex]\((0, 2)\)[/tex] with a slope of [tex]\(\frac{2}{3}\)[/tex].
The equation of the line can be written in the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept. Given:
- Slope [tex]\(m = \frac{2}{3}\)[/tex]
- The line passes through [tex]\((0, 2)\)[/tex], so the y-intercept [tex]\(b = 2\)[/tex]
Thus, the equation of the line is:
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
We will check if each given point satisfies this equation.
1. Point [tex]\((-3, 0)\)[/tex]
Substitute [tex]\(x = -3\)[/tex] into the equation:
[tex]\[
y = \frac{2}{3}(-3) + 2 = -2 + 2 = 0
\][/tex]
The point [tex]\((-3, 0)\)[/tex] is on the line because it satisfies the equation.
2. Point [tex]\((-2, -3)\)[/tex]
Substitute [tex]\(x = -2\)[/tex] into the equation:
[tex]\[
y = \frac{2}{3}(-2) + 2 = -\frac{4}{3} + 2 = -\frac{4}{3} + \frac{6}{3} = \frac{2}{3}
\][/tex]
This means [tex]\((-2, -3)\)[/tex] does not satisfy the equation as [tex]\(-3 \neq \frac{2}{3}\)[/tex].
3. Point [tex]\((2, 5)\)[/tex]
Substitute [tex]\(x = 2\)[/tex] into the equation:
[tex]\[
y = \frac{2}{3}(2) + 2 = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3}
\][/tex]
This means [tex]\((2, 5)\)[/tex] does not satisfy the equation as [tex]\(5 \neq \frac{10}{3}\)[/tex].
4. Point [tex]\((3, 4)\)[/tex]
Substitute [tex]\(x = 3\)[/tex] into the equation:
[tex]\[
y = \frac{2}{3}(3) + 2 = 2 + 2 = 4
\][/tex]
The point [tex]\((3, 4)\)[/tex] is on the line because it satisfies the equation.
5. Point [tex]\((6, 6)\)[/tex]
Substitute [tex]\(x = 6\)[/tex] into the equation:
[tex]\[
y = \frac{2}{3}(6) + 2 = 4 + 2 = 6
\][/tex]
The point [tex]\((6, 6)\)[/tex] is on the line because it satisfies the equation.
Thus, the points Vera could use to graph the line are:
[tex]\[
\boxed{(-3, 0), (3, 4), (6, 6)}
\][/tex]
