To graph a line passing through the point \((0, 2)\) with a slope of \( \frac{2}{3} \), Vera should identify other points that satisfy the equation of the line \( y = \frac{2}{3}x + 2 \).
Let's determine which of the given points lie on this line by substituting the \((x, y)\) coordinates into the equation \( y = \frac{2}{3}x + 2 \):
1. For \((-3, 0)\):
\[ y = \frac{2}{3}(-3) + 2 = -2 + 2 = 0 \]
This point is on the line.
2. For \((-2, -3)\):
\[ y = \frac{2}{3}(-2) + 2 = -\frac{4}{3} + 2 = -\frac{4}{3} + \frac{6}{3} = \frac{2}{3} \neq -3 \]
This point is not on the line.
3. For \((2, 5)\):
\[ y = \frac{2}{3}(2) + 2 = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \neq 5 \]
This point is not on the line.
4. For \((3, 4)\):
\[ y = \frac{2}{3}(3) + 2 = 2 + 2 = 4 \]
This point is on the line.
5. For \((6, 6)\):
\[ y = \frac{2}{3}(6) + 2 = 4 + 2 = 6 \]
This point is on the line.
Hence, the points Vera could use to graph the line are:
- \((-3, 0)\)
- \((3, 4)\)
- \((6, 6)\)
A coordinate plane with a point plotted at and labeled (0, 2).
Vera wants to graph a line that passes through (0, 2) and has a slope of StartFraction 2 Over 3 EndFraction. Which points could Vera use to graph the line? Select three options.
(–3, 0)
(–2, –3)
(2, 5)
(3, 4)
(6, 6)
1 answer