IDNLearn.com makes it easy to get reliable answers from experts and enthusiasts alike. Explore thousands of verified answers from experts and find the solutions you need, no matter the topic.
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]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
