To solve this problem, we need to identify which of the given points lie on the line that passes through [tex]\((0, 2)\)[/tex] with a slope of [tex]\(\frac{2}{3}\)[/tex].
The general equation of a line in slope-intercept form is [tex]\(y = mx + b\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\(b\)[/tex] is the y-intercept. For the line in question:
- The slope [tex]\(m = \frac{2}{3}\)[/tex]
- The y-intercept [tex]\(b = 2\)[/tex]
Thus, the equation of the line is:
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
Now, we will check each given point to determine if it satisfies this equation.
1. Point [tex]\((-3, 0)\)[/tex]
[tex]\[ y = \frac{2}{3}(-3) + 2 \][/tex]
[tex]\[ y = -2 + 2 \][/tex]
[tex]\[ y = 0 \][/tex]
Since the calculated [tex]\(y\)[/tex] value matches the given point's [tex]\(y\)[/tex] coordinate ([tex]\(0\)[/tex]), the point [tex]\((-3, 0)\)[/tex] lies on the line.
2. Point [tex]\((-2, -3)\)[/tex]
[tex]\[ y = \frac{2}{3}(-2) + 2 \][/tex]
[tex]\[ y = -\frac{4}{3} + 2 \][/tex]
[tex]\[ y = -\frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = \frac{2}{3} \][/tex]
Since [tex]\( \frac{2}{3} \neq -3 \)[/tex], the point [tex]\((-2, -3)\)[/tex] does not lie on the line.
3. Point [tex]\((2, 5)\)[/tex]
[tex]\[ y = \frac{2}{3}(2) + 2 \][/tex]
[tex]\[ y = \frac{4}{3} + 2 \][/tex]
[tex]\[ y = \frac{4}{3} + \frac{6}{3} \][/tex]
[tex]\[ y = \frac{10}{3} \][/tex]
Since [tex]\( \frac{10}{3} \neq 5 \)[/tex], the point [tex]\((2, 5)\)[/tex] does not lie on the line.
4. Point [tex]\((3, 4)\)[/tex]
[tex]\[ y = \frac{2}{3}(3) + 2 \][/tex]
[tex]\[ y = 2 + 2 \][/tex]
[tex]\[ y = 4 \][/tex]
Since the calculated [tex]\(y\)[/tex] value matches the given point's [tex]\(y\)[/tex] coordinate ([tex]\(4\)[/tex]), the point [tex]\((3, 4)\)[/tex] lies on the line.
5. Point [tex]\((6, 6)\)[/tex]
[tex]\[ y = \frac{2}{3}(6) + 2 \][/tex]
[tex]\[ y = 4 + 2 \][/tex]
[tex]\[ y = 6 \][/tex]
Since the calculated [tex]\(y\)[/tex] value matches the given point's [tex]\(y\)[/tex] coordinate ([tex]\(6\)[/tex]), the point [tex]\((6, 6)\)[/tex] lies on the line.
Therefore, the points that Vera could use to graph the line are:
[tex]\[
\boxed{(-3, 0), (3, 4), (6, 6)}
\][/tex]
