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Sagot :
To find which points Vera can use to graph the line passing through [tex]\((0, 2)\)[/tex] with a slope of [tex]\(\frac{2}{3}\)[/tex], we need to determine if each point satisfies the line's equation.
Given:
- The slope [tex]\(m\)[/tex] is [tex]\(\frac{2}{3}\)[/tex].
- The line passes through the point [tex]\((0, 2)\)[/tex].
We use the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substituting the given point [tex]\((0, 2)\)[/tex] and the slope [tex]\(\frac{2}{3}\)[/tex] into the equation:
[tex]\[ y - 2 = \frac{2}{3}(x - 0) \][/tex]
[tex]\[ y - 2 = \frac{2}{3}x \][/tex]
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
We will check each point to see if it lies on the line by substituting the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] coordinates into this equation:
1. Checking the point [tex]\((-3, 0)\)[/tex]:
[tex]\[ y = \frac{2}{3}(-3) + 2 \][/tex]
[tex]\[ y = -2 + 2 \][/tex]
[tex]\[ y = 0 \][/tex]
This matches the [tex]\(y\)[/tex]-coordinate of the point [tex]\((-3, 0)\)[/tex], so this point is on the line.
2. Checking the 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]
This does not match the [tex]\(y\)[/tex]-coordinate of [tex]\(-3\)[/tex], so this point is not on the line.
3. Checking the 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]
This does not match the [tex]\(y\)[/tex]-coordinate of [tex]\(5\)[/tex], so this point is not on the line.
4. Checking the point [tex]\((3, 4)\)[/tex]:
[tex]\[ y = \frac{2}{3}(3) + 2 \][/tex]
[tex]\[ y = 2 + 2 \][/tex]
[tex]\[ y = 4 \][/tex]
This matches the [tex]\(y\)[/tex]-coordinate of the point [tex]\((3, 4)\)[/tex], so this point is on the line.
5. Checking the point [tex]\((6, 6)\)[/tex]:
[tex]\[ y = \frac{2}{3}(6) + 2 \][/tex]
[tex]\[ y = 4 + 2 \][/tex]
[tex]\[ y = 6 \][/tex]
This matches the [tex]\(y\)[/tex]-coordinate of the point [tex]\((6, 6)\)[/tex], so this point is on the line.
Based on the calculations, the points that lie on the line are:
[tex]\[ \boxed{(-3, 0), (3, 4), (6, 6)} \][/tex]
Given:
- The slope [tex]\(m\)[/tex] is [tex]\(\frac{2}{3}\)[/tex].
- The line passes through the point [tex]\((0, 2)\)[/tex].
We use the point-slope form of the equation of a line:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Substituting the given point [tex]\((0, 2)\)[/tex] and the slope [tex]\(\frac{2}{3}\)[/tex] into the equation:
[tex]\[ y - 2 = \frac{2}{3}(x - 0) \][/tex]
[tex]\[ y - 2 = \frac{2}{3}x \][/tex]
[tex]\[ y = \frac{2}{3}x + 2 \][/tex]
We will check each point to see if it lies on the line by substituting the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] coordinates into this equation:
1. Checking the point [tex]\((-3, 0)\)[/tex]:
[tex]\[ y = \frac{2}{3}(-3) + 2 \][/tex]
[tex]\[ y = -2 + 2 \][/tex]
[tex]\[ y = 0 \][/tex]
This matches the [tex]\(y\)[/tex]-coordinate of the point [tex]\((-3, 0)\)[/tex], so this point is on the line.
2. Checking the 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]
This does not match the [tex]\(y\)[/tex]-coordinate of [tex]\(-3\)[/tex], so this point is not on the line.
3. Checking the 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]
This does not match the [tex]\(y\)[/tex]-coordinate of [tex]\(5\)[/tex], so this point is not on the line.
4. Checking the point [tex]\((3, 4)\)[/tex]:
[tex]\[ y = \frac{2}{3}(3) + 2 \][/tex]
[tex]\[ y = 2 + 2 \][/tex]
[tex]\[ y = 4 \][/tex]
This matches the [tex]\(y\)[/tex]-coordinate of the point [tex]\((3, 4)\)[/tex], so this point is on the line.
5. Checking the point [tex]\((6, 6)\)[/tex]:
[tex]\[ y = \frac{2}{3}(6) + 2 \][/tex]
[tex]\[ y = 4 + 2 \][/tex]
[tex]\[ y = 6 \][/tex]
This matches the [tex]\(y\)[/tex]-coordinate of the point [tex]\((6, 6)\)[/tex], so this point is on the line.
Based on the calculations, the points that lie on the line are:
[tex]\[ \boxed{(-3, 0), (3, 4), (6, 6)} \][/tex]
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